#+TITLE: \zallman Advanced Modern Algebra
#+AUTHOR: Joseph J. Rotman

#+OPTIONS: tex:imagemagick
#+OPTIONS: toc:2
#+LATEX_HEADER: \input{preamble.tex}
#+LATEX_HEADER: \setcounter{secnumdepth}{2}
#+LATEX_HEADER: \setcounter{tocdepth}{2}
#+EXPORT_FILE_NAME: ../latex/AdvancedModernAlgebra/AdvancedModernAlgebra.tex



* Things Past
** Some Number Theory
   *Least Integer Axiom* (*Well-ordering Principle*). There is a smallest integer in
   every nonempty subset $C$ of $\N$
** Roots of Unity
   #+ATTR_LATEX: :options [Polar Decomposition]
   #+BEGIN_proposition
   Every complex number $z$ has a factorization
   \begin{equation*}
   z=r(\cos\theta+i\sin\theta)
   \end{equation*}
   where $r=\abs{z}\ge0$ and $0\le\theta\le 2\pi$
   #+END_proposition

   #+ATTR_LATEX: :options [Addition Theorem]
   #+BEGIN_proposition
   If $z=\cos\theta+i\sin\theta$ and $w=\cos\psi+i\sin\psi$, then
   \begin{equation*}
   zw=\cos(\theta+\psi)+i\sin(\theta+\psi)
   \end{equation*}
   #+END_proposition
   
   #+ATTR_LATEX: :options [De Moivre]
   #+BEGIN_theorem
   $\forall x\in\R,n\in\N$
   \begin{equation*}
   \cos(nx)+i\sin(nx)=(\cos x+i\sin x)^n
   \end{equation*}
   #+END_theorem

   #+ATTR_LATEX: :options [Euler]
   #+BEGIN_theorem
   $e^{ix}=\cos x+i\sin x$
   #+END_theorem

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $n\in\N\ge 1$ , an *nth root of unity* is a complex number $\xi$ with
   $\xi^n=1$

   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Every nth root of unity is equal to
   \begin{equation*}
   e^{2\pi ik/n}=\cos(\frac{2\pi k}{n})+i\sin(\frac{2\pi k}{n})
   \end{equation*}
   for $k=0,1,\dots,n-1$
   #+END_corollary

   \begin{equation*}
   x^n-1=\displaystyle\prod_{\xi^n=1}(x-\xi)
   \end{equation*}

   If $\xi$ is an nth root of unity and if $n$ is the smallest, then $\xi$ is a
   *primitive* *\(n\)th root of unity*

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $d\in\N^+$ , then the \(d\)th *cyclotomic polynomial* is 
   \begin{equation*}
   \Phi_d(x)=\displaystyle\prod(x-\xi)
   \end{equation*}
   where $\xi$ ranges over all the /primitive dth/ roots of unity
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   For every integer $n\ge 1$
   \begin{equation*}
   x^n-1=\displaystyle\prod_{d|n}\Phi_d(x)
   \end{equation*}
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Define *Euler \phi-function* as the degree of the nth cyclotomic
   polynomial
   \begin{equation*}
   \phi(n)=\deg(\Phi_n(x))
   \end{equation*}
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $n\ge1$ is an integer, then $\phi(n)$ is the number of integers $k$ with
   $1\le k\le n$ and $(k,n)=1$
   #+END_proposition

   #+BEGIN_proof
   Suffice to prove $e^{2\pi ik/n}$ is a primitive nth root of unity if and only
   if $k$ and $n$ are relatively prime
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   For every integer $n\ge 1$, we have
   \begin{equation*}
   n=\displaystyle\sum_{d|n}\phi(d)
   \end{equation*}
   #+END_corollary
** Some Set Theory
   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop1.47
   1. If \(f:X\to Y\) and \(g:Y\to X\) are functions s.t. \(g\circ f=1_X\), then
      $f$ is injective and $g$ is surjective
   2. A function $f:X\to Y$ has an inverse \(g:Y\to X\) if and only if $f$ is a bijection
   #+END_proposition
* Group \rom{1}
** Permutations
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A *permutation* of a set $X$ is a bijection from $X$ to itself.
   #+END_definition


   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   The family of all the permutations of a set $X$, denoted by $S_X$ is called
   the *symmetric group* on $X$. When $X=\lb 1,2,\dots,n\rb$, $S_X$ is
   usually denoted by $X_n$ and is called the *symmetric group on* $n$
   *letters*
   #+END_definition
   
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let $i_1,i_2,\dots,i_r$ be distinct integers in $\lb 1,2,\dots,n\rb$. If
   $\alpha\in S_n$ fixes the other integers and if
   \begin{equation*}
   \alpha(i_1)=i_2,\alpha(i_2)=i_3,\dots,\alpha(i_{r-1})=i_r,\alpha(i_r)=i_1
   \end{equation*}
   then \alpha is called an *r-cycle*. \alpha is a cycle of
   *length* $r$ and denoted by
   \begin{equation*}
   \alpha=(i_1\; i_2\;\dots\; i_r)
   \end{equation*}
   #+END_definition

   2-cycles are also called the *transpositions*.

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Two permutations $\alpha,\beta\in S_n$ are *disjoint* if every $i$
   moved by one is fixed by the other.
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   Disjoint permutations $\alpha,\beta\in S_n$ commute
   #+END_lemma
   
   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Every permutation $\alpha\in S_n$ is either a cycle or a product of disjoint cycles.
   #+END_proposition

   #+BEGIN_proof
   Induction on the number $k$ of points moved by $\alpha$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A *complete factorization* of a permutation $\alpha$ is a
   factorization of $\alpha$ into disjoint cycles that contains exactly one
   1-cycle $(i)$ for every $i$ fixed by $\alpha$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   Let $\alpha\in S_n$ and let $\alpha=\beta_1\dots\beta_t$ be a complete
   factorization into disjoint cycles. This factorization is unique except for
   the order in which the cycles occur
   #+END_theorem

   #+BEGIN_proof
   for all $i$, if $\beta_t(i)\neq i$, then $\beta_t^k(i)\neq\beta_t^{k-1}(i)$
   for any $k\ge 1$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   If $\gamma,\alpha\in S_n$, then $\alpha\gamma\alpha^{-1}$ has the same cycle
   structure as $\gamma$. In more detail, if the complete factorization of
   $\gamma$ is
   \begin{equation*}
   \gamma=\beta_1\beta_2\dots(i_1\; i_2\;\dots)\dots\beta_t
   \end{equation*}
   then $\alpha\gamma\alpha^{-1}$ is permutation that is obtained from $\gamma$
   by applying $\alpha$ to the symbols in the cycles of $\gamma$
   #+END_lemma

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   label:example2.8
   Suppose
   \begin{gather*}
   \beta=(1\;2\;3)(4)(5)\\
   \gamma=(5\;2\;4)(1)(3)
   \end{gather*}
   then we can easily find the $\alpha$
   \begin{equation*}
   \alpha=
   \begin{pmatrix}
   1&2&3&4&5\\
   5&2&4&1&3
   \end{pmatrix}
   \end{equation*}
   and so \(\alpha=(1\;5\;3\;4)\). Now $\alpha\in S_5$ and $\gamma=(\alpha 1\;\alpha 2\;\alpha 3)$
   #+END_examplle
   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   Permutations $\gamma$ and $\sigma$ in $S_n$ has the same cycle structure if
   and only if there exists $\alpha\in S_n$ with $\sigma=\alpha\gamma\alpha^{-1}$
   #+END_theorem


   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $n\ge 2$ then every $\alpha\in S_n$ is a product of tranpositions
   #+END_proposition
   #+BEGIN_proof
   $(1\;2\;\dots\; r)=(1\; r)(1\; r-1)\dots(1\; 2)$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   The *15-puzzle* has a *starting position* that is a $4\times 4$ array of the
   numbers between 1 and 15 and a symbol #, which we interpret as "blank". For
   example, consider the following starting position

   #+ATTR_LATEX: :align |c|c|c|c|
   |----+----+----+----|
   |  3 | 15 |  4 |  8 |
   |----+----+----+----|
   | 10 | 11 |  1 |  9 |
   |----+----+----+----|
   |  2 |  5 | 13 | 12 |
   |----+----+----+----|
   |  6 |  7 | 14 |  # |
   |----+----+----+----|

   A *simple move* interchanges the blank with a symbol adjacent to it. We win the
   game if after a sequence of simple moves, the starting position is
   transformed into the standard array $1,2,\dots,15,\#$. 

   To analyze this game, note that the given array is really a permutation
   $\alpha\in S_{16}$. For example, the given starting position is
   #+ATTR_LATEX: :mode math :environment pmatrix
   | 1 |  2 | 3 | 4 |  5 |  6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
   | 3 | 15 | 4 | 8 | 10 | 11 | 1 | 9 | 2 |  5 | 13 | 12 |  6 |  7 | 14 | 16 |

   To win the game, we need special transpositions $\tau_1,\dots,\tau_m$ sot
   that
   \begin{equation*}
   \tau_m\dots\tau_1\alpha=(1)
   \end{equation*}
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A permutation $\alpha\in S_n$ is *even* if it can be factored into a
   product of an even number of transpositions. Otherwise *odd*
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $\alpha\in S_n$ and $\alpha=\beta_1\dots\beta_t$ is a complete
   factorization, then *signum* $\alpha$ is defined by
   \begin{equation*}
   \sgn(\alpha)=(-1)^{n-t}
   \end{equation*}
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   For all $\alpha,\beta\in S_n$
   \begin{equation*}
   \sgn(\alpha\beta)=\sgn(\alpha)\sgn(\beta)
   \end{equation*}
   #+END_theorem

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   1. Let $\alpha\in S_n$; if $\sgn(\alpha)=1$ then $\alpha$ is even. otherwise
      odd
   2. A permutation $\alpha$ is odd if and only if it's a product of an odd
      number of transpositions
   #+END_theorem

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Let $\alpha,\beta\in S_n$. If $\alpha$ and $\beta$ have the same parity, then
   $\alpha\beta$ is even while if $\alpha$ and $\beta$ have distinct parity,
   $\alpha\beta$ is odd
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   An analysis of the 15-puzzle shows that if $\alpha\in S_{16}$ is the starting
   position, then the game can be won if and only if \alpha is an even permutation
   that fixes 16.

   The blank 16 starts in position 16. Each simple move takes 16 up, down, left
   or right. Thus the total number $m$ of moves is $u+d+l+r$. If 16 is to return
   home, each one of these must be undone. Thus the total number of moves is
   even: $m=2u+2r$. Hence $\alpha=\tau_1\dots\tau_m$ and so $\alpha$ is an even
   permutation. In example
   \begin{equation*}
   \alpha=(1\;3\;4\;8\;9\;2\;15\;14\;7)(5\;10)(6\;11\;13)(12)(16)
   \end{equation*}
   Now $\sgn(\alpha)=(-1)^{16-5}=-1$.
   #+END_examplle
** Groups
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A *binary operation* on a set $G$ is a function
   \begin{equation*}
   *:G\times G\to G
   \end{equation*}
   #+END_definition
   
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A *group* is a set $G$ equipped with a binary operation * s.t.
   1. the *associative law* holds
   2. *identity*
   3. every $x\in G$ has an *inverse*, there is a $x'\in G$  with 
      $x*x'=e=x'*x$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A group $G$ is called *abelian* if it satisfies the
   *commutative law*
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   Let $G$ be a group
   1. The *cancellation laws* holds: if either $x*a=x*b$ or $a*x=b*x$, then
      $a=b$
   2. $e$ is unique
   3. Each $x\in G$ has a unique inverse
   4. $(x^{-1})^{-1}=x$
   #+END_lemma

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   An expression $a_1a_2\dots a_n$ *needs no parentheses* if all the ultimate
   products it yields are equal
   #+END_definition

   #+ATTR_LATEX: :options [Generalized Associativity]
   #+BEGIN_theorem
   If $G$ is a group and $a_1,a_2,\dots,a_n\in G$ then the expression
   $a_1a_2\dots a_n$ needs no parentheses
   #+END_theorem

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let $G$ be a group and let $a\in G$. If $a^k=1$ for some $k>1$ then the
   smallest such exponent $k\ge 1$ is called the *order* or $a$; if no such
   power exists, then one says that $a$ has *infinite order*
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $G$ is a finite group, then every $x\in G$ has finite order
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A *motion* is a distance preserving bijection $\varphi:\R^2\to\R^2$. If
   \pi is a polygon in the plane, then its *symmetry group* $\Sigma(\pi)$
   consists of all the motions $\varphi$ for which $\varphi(\pi)=\pi$. The
   elements of $\Sigma(\pi)$ are called the *symmetries* of \pi
   #+END_definition

   Let $\pi_4$ be a square. Then the group $\Sigma(\pi_4)$ is called the
   *dihedral group* with 8 elements, denoted by $D_8$

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $\pi_n$ is a regular polygon with $n$ vertices $v_1,\dots,v_n$ and center
   $O$, then the symmetry group $\Sigma(\pi_n)$ is called the *dihedral*
   *group* with $2n$ elements, and it's denoted by $D_{2n}$
   #+END_definition

   #+BEGIN_exercise
   label:ex2.26
   If $G$ is a group in which \(x^2=1\) for every \(x\in G\), prove that $G$
   must be abelian
   #+END_exercise

   #+BEGIN_exercise
   label:ex2.27
   If $G$ is a group with an even number of elements, prove that the number of
   elements in $G$ of order 2 is odd. In particular, $G$ must contain an element of
   order 2.
   #+END_exercise

   #+BEGIN_proof
   1 is an element of order 1.
   #+END_proof
** Lagrange's Theorem


   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A subset $H$ of a group $G$ is a *subgroup* if
   1. $1\in H$
   2. if $x,y\in H$, then $xy\in H$
   3. if $x\in H$, then $x^{-1}\in H$
   #+END_definition

   If $H$ is a subgroup of $G$, we write $H\le G$. If $H$ is a proper subgroup,
   then we write $H<G$

   The four permutations
   \begin{equation*}
   \bV=\{(1),(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)\}
   \end{equation*}
   form a group because $\bV\le S_4$

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   A subset $H$ of a group $G$ is a subgroup if and only if $H$ is nonempty and
   whenever $x,y\in H$, $xy^{-1}\in H$
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   A nonempty subset $H$ of a finite group $G$ is a subgroup if and only if $H$
   is closed; that is, if $a,b\in H$, then $ab\in H$
   #+END_proposition

   [[index:alternating group]]
   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   The subset $A_n$ of $S_n$, consisting of all the even permutations, is a
   subgroup called the *alternating group* on $n$ letters
   #+END_examplle

   [[index:cyclic group]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $G$ is a group and $a\in G$
   \begin{equation*}
   \langle a\rangle=\{a^n:n\in\Z\}=\{\text{all powers of } a\}
   \end{equation*}
   $\la a\ra$ is called the *cyclic subgroup* of $G$ *generated* by $a$. A
   group $G$ is called *cyclic* if there exists $a\in G$ s.t. $G=\la a\ra$,
   in which case $a$ is called the *generator*
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   The *integers mod $m$*, denoted by $\I_m$ is the family of all congruence
   classes mod $m$
   #+END_definition


   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let $m\ge 2$ be a fixed integer
   1. If $a\in \Z$, then $[a]=[r]$ for some $r$ with $0\le r<m$
   2. If $0\le r'<r<m$, then $[r']\neq[r]$
   3. $\I_m$ has exactly $m$ elements
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:nthm1.39
   1. If $G=\la a\ra$ is a cyclic group of order $n$, then $a^k$ is a generator
      of $G$ if and only if $(k,n)=1$
   2. If $G$ is a cyclic group of order $n$ and $\gen(G)=\{\text{all generators
      of } G\}$, then
      \begin{equation*}
      \abs{\gen(G)}=\phi(n)
      \end{equation*}
      where $\phi$ is the Euler \phi-function
   #+END_theorem
   #+BEGIN_proof
   1. there is $t\in\N$ s.t. $a^{kt}=a$ hence $a^{kt-1}=1$ and $n\mid kt-1$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let $G$ be a finite group and let $a\in G$. Then the order of $a$ is
   $\abs{\la a\ra}$.
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $G$ is a finite group, then the number of elements in $G$, denoted by
   $\abs{G}$ is called the *order* of $G$
   #+END_definition


   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   The intersection $\bigcap_{i\in I}H_i$ of any family of subgroups of a group
   $G$ is again a subgroup of $G$
   #+END_proposition


   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $X$ is a subset of a group $G$, then there is a subgroup $\la X\ra$ of $G$
   containing $X$ tHhat is *smallest* in the sense that $\la X\ra\le H$ for
   every subgroup $H$ 
   of $G$ that contains $X$
   
   #+END_corollary


   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $X$ is a subset of a group $G$, then $\la X\ra$ is called the *subgroup*
   *generated by* $X$
   #+END_definition

   A *word* on $X$ is an element $g\in G$ of the form $g=x_1^{e_1}\dots
   x_n^{e_n}$ where $x_i\in X$ and $e_i=\pm 1$ for all $i$

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $X$ is a nonempty subset of a group $G$, then $\la X\ra$ is the set of all
   words on $X$
   #+END_proposition

   [[index:coset]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $H\le G$ and $a\in G$, then the *coset* $aH$ is the subset $aH$ of $G$,
   where
   \begin{equation*}
   aH=\{ah:h\in H\}
   \end{equation*}
   #+END_definition
   $aH$ *left coset*, $Ha$ *right coset*

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   $H\le G,a,b\in G$
   1. $aH=bH$ if and only if $b^{-1}a\in H$
   2. if $aH\cap bH\neq\emptyset$, then $aH=bH$
   3. $\abs{aH}=\abs{H}$ for all $a\in G$
   #+END_lemma
   #+BEGIN_proof
   define a relation $a\equiv b$ if $b^{-1}a\in H$
   #+END_proof


   #+ATTR_LATEX: :options [Lagrange's Theorem]
   #+BEGIN_theorem
   If $H$ is a subgroup of a finite group $G$, then $\abs{H}$ is a divisor of $\abs{G}$
   #+END_theorem

   #+BEGIN_proof
   Let $\{a_1H,a_2H,\dots,a_tH\}$ be the family of all the distinct cosets of
   $H$ in $G$. Then
   \begin{equation*}
   G=a_1H\cup a_2H\cup\dots\cup a_tH
   \end{equation*}
   hence
   \begin{equation*}
   \abs{G}=\abs{a_1H}+\dots+\abs{a_tH}
   \end{equation*}
   But $\abs{a_iH}=\abs{H}$ for all $i$. Hence $\abs{G}=t\abs{H}$
   #+END_proof

   [[index:index]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   The *index* of a subgroup $H$ in $G$ denoted by $[G:H]$, is the number of
   left cosets of $H$ in $G$
   #+END_definition

   Note that $\abs{G}=[G:H]\abs{H}$

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $G$ is a finite group and $a\in G$, then the order of $a$ is a divisor of
   $\abs{G}$ 
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $G$ is a finite group, then $a^{\abs{G}}=1$ for all $a\in G$
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $p$ is a prime, then every group $G$ of order $p$ is cyclic
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   The set $U(\I_m)$, defined by
   \begin{equation*}
    U(\I_m)=\{[r]\in\I_m:(r,m)=1\}
   \end{equation*}
   is a multiplicative group of order $\phi(m)$. If $p$ is a prime, then
   $U(\I_p)=\I_p^{\times}$, the nonzero elements of $\I_p$.
   #+END_proposition
   
   #+BEGIN_proof
   $(r,m)=1=(r',m)$ implies $(rr',m)=1$. Hence $U(\I_m)$ is closed under
   multiplication. If $(x,m)=1$, then $rs+sm=1$. Therefore $(r,m)=1$. Each of
   them have inverse.
   #+END_proof

   [[index:Fermat's Theorem]]
   #+ATTR_LATEX: :options [Fermat]
   #+BEGIN_corollary
   label:Fermat
   If $p$ is a prime and $a\in\Z$, then
   \begin{equation*}
   a^p\equiv a\mod p
   \end{equation*}
   #+END_corollary

   #+BEGIN_proof
   suffices to show $[a^p]=[a]$ in $\I_p$. If $[a]=[0]$, then $[a^p]=[a]^p=[0]$.
   Else, since $\abs{\I_p^\times}=p-1$, $[a]^{p-1}=[1]$
   #+END_proof


   #+ATTR_LATEX: :options [Euler]
   #+BEGIN_theorem
   If $(r,m)=1$, then
   \begin{equation*}
   r^{\phi(m)}\equiv 1\mod m
   \end{equation*}
   #+END_theorem
   #+BEGIN_proof
   Since $\abs{U(\I_m)}=\phi(m)$. Lagrange's theorem gives
   $[r]^{\phi(m)}=[1]$ for all $[r]\in U(\I_m)$.

   In fact we construct a group to prove this.
   #+END_proof

   #+ATTR_LATEX: :options [Wilson's Theorem]
   #+BEGIN_theorem
   An integer $p$ is a prime if and only if
   \begin{equation*}
   (p-1)!\equiv -1\mod p
   \end{equation*}
   #+END_theorem

   #+BEGIN_proof
   Assume that $p$ is a prime. If $a_1,\dots,a_n$ is a list of all the elements
   of finite abelian group, then product $a_1a_2\dots a_n$ is the same as the
   product of all elements $a$ with $a^2=1$. Since $p$ is prime, $\I_p^\times$
   has only one element of order 2, namely $[-1]$. It follows that the product
   of all the elements in $\I_p^\times$ namely $[(p-1)!]$ is equal to $[-1]$.

   Conversly assume that $m$ is composite: there are integers $a$ and $b$ with
   $m=ab$ and $1<a\le b<m$. If $a<b$ then $m=ab$ is a divisor of $(m-1)!$. If
   $a=b$, then $m=a^2$. if $a=2$, then $(a^2-1)!\equiv 2\mod 4$. If $2<a$, then
   $2a<a^2$ and so $a$ and $2a$ are factors of $(a^2-1)!$
   #+END_proof

   #+BEGIN_exercise
   label:ex2.36
   Let $G$ be a group of order 4. Prove that either $G$ is cyclic or \(x^2=1\)
   for every \(x\in G\). Conclude, using Exercise ref:ex2.26 that $G$ must be abelian.
   #+END_exercise

   #+BEGIN_proof
   
   #+END_proof
** Homomorphisms
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $(G,*)$ and $(H,\circ)$ are groups, then a function $f:G\to H$ is a
   *homomorphism* if
   \begin{equation*}
   f(x*y)=f(x)\circ f(y)
   \end{equation*}
   for all $x,y\in G$. If $f$ is also a bijection, then $f$ is called an
   *isomorphism*. $G$ and $H$ are called *isomorphic*, denoted by $G\cong H$
   #+END_definition
   
   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   Let $f:G\to H$ be a homomorphism
   1. $f(1)=1$
   2. $f(x^{-1})=f(x)^{-1}$
   3. $f(x^n)=f(x)^n$ for all $n\in\Z$
   #+END_lemma

   

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $f:G\to H$ is a homomorphism, define
   \begin{equation*}
   \ker f=\{x\in G:f(x)=1
   \end{equation*}
   and
   \begin{equation*}
   \im f=\{h\in H:h=f(x)\text{ for some } x\in G\
   \end{equation*}
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let $f:G\to H$ be a homomorphism
   1. $\ker f$ is a subgroup of $G$ and $\im f$ is a subgroup of $H$
   2. if $x\in\ker f$ and if $a\in G$, then $axa^{-1}\in\ker f$
   3. $f$ is an injection if and only if $\ker f=\{1\}$
   #+END_proposition

   #+BEGIN_proof
   3. [@3] $f(a)=f(b)\Leftrightarrow f(ab^{-1})=1$
   #+END_proof
   
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A subgroup $K$ of a group $G$ is called a *normal subgroup* if $k\in K$
   and $g\in G$ imply $gkg^{-1}\in K$, denoted by $K\triangleleft G$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $G$ is a group and $a\in G$, then a *conjugate* of $a$ is any element
   in $G$ of the form
   \begin{equation*}
   gag^{-1}
   \end{equation*}
   where $g\in G$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $G$ is a group and $g\in G$, define *conjugation* $\gamma_g:G\to G$ by
   \begin{equation*}
   \gamma_g(a)=gag^{-1}
   \end{equation*}
   for all $a\in G$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   1. If $G$ is a group and $g\in G$, then conjugation $\gamma_g:G\to G$ is an
      isomorphism
   2. Conjugate elements have the same order
   #+END_proposition

   #+BEGIN_proof
   1. bijection: $\gamma_g\circ\gamma_{g^{-1}}=1=\gamma_{g^{-1}}\circ\gamma_g$.
      
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   Define the *center* of a group $G$, denoted by $Z(G)$, to be
   \begin{equation*}
   Z(G)=\{z\in G:zg=gz\text{ for all }g\in G\}
   \end{equation*}
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   If $G$ is a group, then an *automorphism* of $G$ is an isomorphism $f:G\to G$.
   For example, every conjugation $\gamma_g$ is an automorphism of $G$ (it is
   called an *inner automorphism*), for its inverse is conjugation by $g^{-1}$.
   The set $\Aut(G)$ of all the automorphism of $G$ is itself a group.
   \begin{equation*}
   \inn(G)=\{\gamma_g:g\in G\}
   \end{equation*}
   is a subgroup of $\Aut(G)$
   #+END_examplle
   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   1. If $H$ is a subgroup of index 2 in a group $G$, then $g^2\in H$ for every
      $g\in G$
   2. If $H$ is a subgroup of index 2 in a group $G$, then $H$ is a normal
      subgroup of $G$
   #+END_proposition


   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   The group of *quaternions* is the group $\bQ$ of order 8 consisting of the
   following matrices in $GL(2, \C)$
   \begin{equation*}
   \bQ=\{I,A,A^2,A^3,B,BA,BA^2,BA^3\}
   \end{equation*}
   where $I$ is the identity matrix
   \begin{equation*}
   A=
   \begin{pmatrix}
   0&1\\
   -1&0
   \end{pmatrix}, \text{ and }
   B=\begin{pmatrix}
   0&i\\
   i&0
     \end{pmatrix}
   \end{equation*}
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   $\bQ$ is normal. By Lagrange's theorem the only possible orders of subgroups
   are 1,2,4 or 8. The only subgroup of order 2 is $\la -I\ra$ since $-I$ is the
   only element of order 2
   #+END_examplle
   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   The alternating group $A_4$ is a group of order 12 having no subgroup of
   order 6
   #+END_proposition

   #+BEGIN_exercise
   Show that if there is a bijection $f:X\to Y$, then there is an isomorphism
   $\varphi:S_X\to S_Y$
   #+END_exercise
   #+BEGIN_proof
   If $\alpha\in S_X$, define $\varphi(\alpha)=f\circ\alpha\circ f^{-1}$. Since
   $f,\alpha,f^{-1}$ are bijections, $\varphi(\alpha)$ is an bijection. \varphi is a
   homomorphism. $\forall \beta\in S_Y$, we have $\alpha=f^{-1}\circ\beta\circ f$
   #+END_proof
** Quotient group
   $\cals(G)$ is the set of all nonempty subsets of a group $G$. If
   $X,Y\in\cals(G)$, define
   \begin{equation*}
   XY=\{xy:x\in X\text{ and } y\in Y\}
   \end{equation*}

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   $K\le G$ is normal if and only if
   \begin{equation*}
   gK=Kg
   \end{equation*}
   #+END_lemma

   A natural question is that whether $HK$ is a subgroup when $H$ and $K$ are
   subgroups. The answer is no. Let $G=S_3,H=\la(1\;2)\ra,K=\la(1\;3)\ra$


   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   1. If $H$ and $K$ are subgroups of a group $G$, and if one of them is normal,
      then $HK\le G$ and $HK=KH$
   2. If $H,K\tril G$, then $HK\tril G$
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   Let $G/K$ denote the family of all the left cosets of a subgroup $K$ of $G$.
   If $K\tril G$, then
   \begin{equation*}
   aKbK=abK
   \end{equation*}
   for all $a,b\in G$ and $G/K$ is a group under this operation
   #+END_theorem

   #+BEGIN_proof
   $aKbK=abKK=abK$
   #+END_proof

   $G/K$ is called the *quotient group* $G$ mod $K$

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Every $K\tril G$ is the kernel of some homomorphism
   #+END_corollary
   
   #+BEGIN_proof
   Define the *natural map* $\pi:G\to G/K$, $a\mapsto aK$
   #+END_proof

   #+ATTR_LATEX: :options [First Isomorphism Theorem]
   #+BEGIN_theorem
   If $f:G\to H$ is a homomorphism, then
   \begin{equation*}
   \ker f\tril G\quad\text{ and }\quad G/\ker f\cong\im f
   \end{equation*}
   If $\ker f=K$ and $\varphi:G/K\to\im f\le H,aK\mapsto f(a)$, then $\varphi$
   is an isomorphism
   #+END_theorem
   
   #+BEGIN_remark
   \begin{center}
   \begin{tikzcd}
   G \arrow[rr,"f"] \arrow[dr,"\pi"]& &
   H\\ 
   &G/K \arrow[ur,"\varphi"]&
   \end{tikzcd}
   \end{center}
   #+END_remark

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   What's the quotient group $\R/\Z$? Define $f:\R\to S^1$ where $S^1$ is the
   circle group by
   \begin{equation*}
   f:x\mapsto e^{2\pi ix}
   \end{equation*}
   $\R/\Z\cong S^1$
   #+END_examplle

   #+ATTR_LATEX: :options [Product Formula]
   #+BEGIN_proposition
   If $H$ and $K$ are subgroups of a finite group $G$, then
   \begin{equation*}
   \abs{HK}\abs{H\cap K}=\abs{H}\abs{K}
   \end{equation*}
   #+END_proposition

   #+BEGIN_proof
   Define a function $f:H\times K\to HK,(h,k)\mapsto hk$. Show that
   $\abs{f^{-1}(x)}=\abs{H\cap K}$. 

   Claim that if $x=hk$, then
   \begin{equation*}
   f^{-1}(x)=\{(hd,d^{-1}k):d\in H\cap K\}
   \end{equation*}
   #+END_proof

   #+ATTR_LATEX: :options [Second Isomorphism Theorem]
   #+BEGIN_theorem
   If $H\tril G, K\le G$, then $HK\le G,H\cap K\tril G$ and
   \begin{equation*}
   K/(H\cap K)\cong HK/H
   \end{equation*}
   #+END_theorem

   #+BEGIN_proof
   $hkH=kk^{-1}hkH=kh'H=kH$
   #+END_proof

   #+ATTR_LATEX: :options [Third Isomorphism Theorem]
   #+BEGIN_theorem
   If $H,K\tril G$ with $K\le H$, then $H/K\tril G/K$ and
   \begin{equation*}
   (G/K)/(H/K)\cong G/H
   \end{equation*}
   #+END_theorem

   #+ATTR_LATEX: :options [Correspondence Theorem]
   #+BEGIN_theorem
   label:thm2.76
   If $K\tril G, \pi:G\to G/K$ is the natural map, then
   \begin{equation*}
   S\mapsto \pi(S)=S/K
   \end{equation*}
   is a bijection between $Sub(G;K)$, the family of all those subgroups $S$ of
   $G$ that contain $K$, and $Sub(G/K)$, the family of all the subgroups of
   $G/K$. If we denote $S/K$ by $S^*$, then
   1. $T\le S\le G$ if and only if $T^*\le S^*$, in which case $[S:T]=[S^*:T^*]$
   2. $T\tril S$ if and only if $T^*\tril S^*$, in which case $S/T\cong S^*/T^*$
   #+END_theorem

   \begin{center}
   \begin{tikzcd}
   G \arrow[d,dash] \arrow[rd]&\\
   S \arrow[d,dash] \arrow[rd] & G/K \arrow[d,dash]\\
   T \arrow[d,dash] \arrow[rd] & S/K=S^* \arrow[d,dash]\\
   K  \arrow[rd] & T/K=T^* \arrow[d,dash]\\
   & \{1\}
   \end{tikzcd}
   \end{center}

   #+BEGIN_proof
   Define \(\Phi:Sub(G;K)\to Sub(G/K)\) by \(\Phi:S\mapsto S/K\)

   To see that \Phi is injective, we begin by showing that if \(K\le S\le G\), then \(\pi^{-1}\pi(S)=S\).
   Apparently \(S\subseteq\pi^{-1}\pi(S)\). Let \(a\in\pi^{-1}\pi(S)\), so that \(\pi(a)=\pi(s)\) for some \(s\in S\). It
   follows taht \(as^{-1}\in\ker\pi=K\), so that \(a=sk\) for some \(k\in K\). But \(K\le S\), and
   so \(a=sk\in S\).

   Assume now that \(\pi(S)=\pi(S')\), where \(S\) and \(S'\) are subgroups of \(G\) containing \(K\).
   Then \(\pi^{-1}\pi(S)=\pi^{-1}\pi(S')\), and so \(S=S'\) as we have just proved in the preceding. Hence
   \Phi is injective.

   To see that \Phi is surjective, let \(U\) be a subgroup of \(G/K\). Now \(\pi^{-1}(U)\) is a subgroup
   of \(G\) containing \(K=\pi^{-1}(\{1\})\) and hence \(\pi(\pi^{-1}(U))=U\) as \pi is surjective

   \(T\le S\le G\) implies \(T/K=\pi(T)\le\pi(S)\le S/K\). Conversely, assume that \(T/K\le S/K\) . If \(t\in T\),
   then \(tK\in T/K\le S/K\) and so \(tK=sK\) for some \(s\in S\). Hence \(t=sk\) for some \(k\in K\le S\),
   and so \(t\in S\).

   For $[S:T]=[S^*:T^*]$, show there is a bijection between the family of all
   cosets of the form $sT$ and the family of all the cosets of the form
   $s^*T^*$.

   injective:
   \begin{align*}
   \pi(m)T^*=\pi(n)T^*&\Leftrightarrow \pi(m)\pi(n)^{-1}\in T^*\\
   &\Leftrightarrow mn^{-1}K\in T/K\\
   &\Rightarrow mn^{-1}t^{-1}\in K\\
   &\Rightarrow mn^{-1}=tk\in T\\
   &\Leftrightarrow mT=nT\\
   \end{align*}
   
   surjective:
   
   
   If $G$ is finite, then
   \begin{align*}
   [S^*:T^*]&=\abs{S^*}/\abs{T^*}\\
   &=\abs{S/K}/\abs{T/K}\\
   &=(\abs{S}/\abs{K})/(\abs{T}/\abs{K})\\
   &=\abs{S}/\abs{T}\\
   &=[S:T]
   \end{align*}

   If $T\tril S$, by third isomorphism theorem, $T/S\cong (T/K)/(S/K)=T^*/S^*$

   If $T^*\tril S^*$, 
   \begin{equation*}
   \pi(sts^{-1})\in \pi(s)T^*\pi(s)^{-1}=T^*
   \end{equation*}
   so that $sts^{-1}\in \pi^{-1}(T^*)=T$
   #+END_proof
   

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop2.78
   If $G$ is a finite abelian group and $d$ is a divisor of $\abs{G}$, then $G$
   contains a subgroup of order $d$
   #+END_proposition

   #+BEGIN_proof
   Abelian group's subgroup is normal and hence we can build quotient groups.
   p90 for proof. Use the correspondence theorem
   #+END_proof
   
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $H$ and $K$ are grops, then their *direct product*, denoted by 
   $H\times K$ 
   , is the set of all ordered pairs $(h,k)$ with the operation
   \begin{equation*}
   (h,k)(h',k')=(hh',kk')
   \end{equation*}
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let $G$ and $G'$ be groups and $K\tril G, K'\tril G'$. Then $K\times K'\tril
   G\times G'$ and
   \begin{equation*}
   (G\times G')/(K\times K')\cong (G/K)\times(G'/K')
   \end{equation*}
   #+END_proposition

   #+BEGIN_proof
   
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $G$ is a group containing normal subgroups $H$ and $K$ and $H\cap K=\{1\}$
   and $HK=G$, then $G\cong H\times K$
   #+END_proposition

   #+BEGIN_proof
   Note $\abs{HK}\abs{H\cap K}=\abs{H}\abs{K}$. Consider $\varphi:G\to H\times
   K$. Show it's homo and bijective.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm2.81
   If $m,n$ are relatively prime, then
   \begin{equation*}
   \I_{mn}\cong\I_m\times\I_n
   \end{equation*}
   #+END_theorem

   #+BEGIN_proof
   \begin{align*}
   f:&\Z\to\I_m\times\I_n\\
   &a\mapsto([a]_m,[a]_n)
   \end{align*}
   is a homo.
   $\Z/\la mn\ra\cong\I_m\times\I_n$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let $G$ be a group, and $a,b\in G$ be commuting elements of orders $m,n$. If
   $(m,n)=1$, then $ab$ has order $mn$
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $(m,n)=1$, then $\phi(mn)=\phi(m)\phi(n)$
   #+END_corollary

   #+BEGIN_proof
   Theorem ref:thm2.81 shows that $f:\I_{mn}\cong\I_m\times\I_n$. The result will
   follow if we prove that $f(U(\I_{mn}))=U(\I_m)\times U(\I_n)$, for then
   \begin{align*}
   \phi(mn)&=\abs{U(\I_{mn})}=\abs{f(U(\I_{mn}))}\\
   &=\abs{U(\I_m)\times U(\I_n)}=\abs{U(\I_m)}\cdot\abs{U(\I_n)}
   \end{align*}
   If $[a]\in U(\I_{mn})$, then $[a][b]=[1]$ for some $[b]\in\I_{mn}$ and
   \begin{align*}
   f([ab])&=([ab]_m,[ab]_n)=([a]_m[b]_m,[a]_n[b]_n)\\
   &=([a]_m,[a]_n)([b]_m,[b]_n)=([1]_m,[1]_n)
   \end{align*}
   Hence $f([a])=([a]_m,[a]_n)\in U(\I_m)\times U(\I_n)$

   For the reverse inclusion, if $f([c])=([c]_m,[c]_n)\in U(\I_m)\times
   U(\I_n)$, then we must show that $[c]\in U(\I_{mn})$. There is $[d]_m\in\I_m$
   with $[c]_m[d]_m=[1]_m$, and there is $[e]_n\I_n$ with $[c]_n[e]_n=[1]_n$.
   Since $f$ is surjective, there is $b\in\Z$ with
   $([b]_m,[b]_n)=([d]_m,[e]_n)$, so that
   \begin{equation*}
   f([1])=([1]_m,[1]_n)=([c]_m[b]_m,[c]_n[b]_n)=f([c][b])
   \end{equation*}
   Since $f$ is an injection, $[1]=[c][b]$ and $[c]\in U(\I_{mn})$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   1. If $p$ is a prime, then $\phi(p^e)=p^e-p^{e-1}=p^e(1-\frac{1}{p})$
   2. If $n=p_1^{e_1}\dots p_t^{e_t}$, then
      \begin{equation*}
      \phi(n)=n(1-\frac{1}{p_1})\dots(1-\frac{1}{p_t})
      \end{equation*}
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:nlemma1.92
   A cyclic group of order $n$ has a unique subgroup of order $d$, for each
   divisor $d$ of $n$, and this subgroup is cyclic.
   #+END_lemma

   Define an equivalence relation on a group $G$ by $x\equiv y$ if $\la x\ra=\la
   y\ra$. Denote the equivalence class containing $x$ by $\gen(C)$, where $C=\la
   x\ra$. Equivalence classes form a partition and we get
   \begin{equation*}
   G=\displaystyle\prod_{C}\gen(C)
   \end{equation*}
   where $C$ ranges over all cyclic subgroups of $G$. Note $\abs{\gen(C)}=\phi(n)$

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm2.86
   A group $G$ of order $n$ is cyclic if and only if for each divisor $d$ of
   $n$, there is at most one cyclic subgroup of order $d$
   #+END_theorem

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   If $G$ is an abelian group of order $n$ having at most one cyclic subgroup of
   order $p$ for each prime divisor $p$ of $n$, then $G$ is cyclic
   #+END_theorem

   Exercise:
   * 2.71 Suppose $H\le G, \abs{H}=\abs{K}$. Since $\abs{H}=[H:K]\abs{K}$,
     $[H:K]=1$. Hence $H=K$
   * 2.67 1. $\inn(S_3)\cong S_3/Z(S_3)\cong S_3$ and $\abs{\Aut(S_3)}\le 6$.
     Hence $\Aut(S_3)=\inn(S_3)$

     
   #+BEGIN_exercise
   label:ex2.69
   Prove that if $G$ is a group for which $G/Z(G)$ is cyclic, then $G$ is abelian
   #+END_exercise
   #+BEGIN_proof
   Suppose $G/Z(G)=\la a\ra$, let $g=a^kz^{-1},g'=a^{k'}z'^{-1}$, then 
   $gg'=a^kz^{-1}z^{k'}z'^{-1}=a^{k+k'}z'^{-1}z^{-1}=g'g$. Hence $G$ is abelian.
   #+END_proof
** Group Actions
   #+ATTR_LATEX: :options [Cayley]
   #+BEGIN_theorem
   Every group $G$ is isomorphic to a subgroup of the symmetric group $S_G$. In
   particular, if $\abs{G}=n$, then $G$ is isomorphic to a subgroup of $S_n$
   #+END_theorem

   #+BEGIN_proof
   For each $a\in G$, define $\tau_a(x)=ax$ for every $x\in G$. $\tau_a$ is a
   bijection for its inverse is $\tau_{a^{-1}}$
   \begin{equation*}
   \tau_a\tau_{a^{-1}}=\tau_1=\tau_{a^{-1}}\tau_a
   \end{equation*}
   #+END_proof

   #+ATTR_LATEX: :options [Representation on Cosets]
   #+BEGIN_theorem
   Let $G$ be a group and $H\le G$ having finite index $n$. Then there exists a
   homomorphism $\varphi:G\to S_n$ with $\ker\varphi\le H$
   #+END_theorem
  
   #+BEGIN_proof
   We still denote the family of all the cosets of $H$ in $G$ by $G/H$

   For each $a\in G$, define "translation" $\tau_a:G/H\to G/H$ by $\tau_a(xH)=axH$
   for every $x\in G$. For $a,b\in G$
   \begin{equation*}
   (\tau_a\circ\tau_b)(xH)=a(bxH)=(ab)xH
   \end{equation*}
   so that 
   \begin{equation*}
   \tau_a\tau_b=\tau_{ab}
   \end{equation*}
   It follows that each $\tau_a$ is a bijection and so $\tau_a\in S_{G/H}$. Define
   $\varphi:G\to S_{G/H}$ by $\varphi(a)=\tau_a$. Rewriting
   \begin{equation*}
   \varphi(a)\varphi(b)=\tau_a\tau_b=\tau_{ab}=\varphi(ab)
   \end{equation*}
   so that \varphi is a homomorphism. Finally if $a\in\ker\varphi$, then
   $\varphi(a)=1_{G/H}$, so that $\tau_a(xH)=xH$, in particular, when $x=1$, this
   gives $aH=H$ and $a\in H$. And $S_{G/H}\cong S_n$
   #+END_proof

   When $H=\{1\}$, this is the Cayley theorem.

   Four-group $\bV=\{1,(1 2)(3 4),(1 3)(2 4), (1 4)(2 3)\}$
   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Every group $G$ of order 4 is isomorphic to either $\I_4$ or the four-group
   $\bV$. And $\I_4\not\cong\bV$
   #+END_proposition

   #+BEGIN_proof
   By lagrange's theorem, every element in $G$ other than 1 has order 2 or 4. If
   4, then $G$ is cyclic.

   Suppose $x,y\neq 1$, then $xy\neq x,y$. Hence $G=\{1,x,y,xy\}$.
   #+END_proof
   

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $G$ is a group of order 6, then $G$ is isomorphic to either $\I_6$ or
   $S_3$. Moreover $\I_6\not\cong S_3$
   #+END_proposition

   #+BEGIN_proof

If $G$ is not cyclic, since $\abs{G}$ is even, it has some elements having
order 2, say $t$ by exercise ref:ex2.27

If $G$ is abelian. Suppose it has another different element $a$ with order 2.
Then $H=\{1,a,t,at\}$ is a subgroup which contradict. Hence it must contain
an element $b$ of order 3. Then $bt$ has order 6 and $G$ is cyclic.

If $G$ is not abelian. If $G$ doesn't have elements of order 3, then it's
abelian. Hence $G$ has an element $s$ of order 3.

Now $\abs{\la s\ra}=3$, so $[G:\la s\ra]=\abs{G}/\abs{\la s\ra}=2$ and $\la
s\ra$ is normal. 
Since $t=t^{-1}$, $tst\in\la s\ra$. If $tst=s^0=1$, $s=1$.
If $tst=s$, $\abs{\la st\ra}=6$. Therefore $tst=s^2=s^{-1}$.

Let $H=\la t\ra$, $\varphi:G\to S_{G/\la t\ra}$ given by
\begin{equation*}
\varphi(g):x\la t\ra\mapsto gx\la t\ra
\end{equation*}
By representation on cosets, $\ker\varphi\le\la t\ra$. Hence
$\ker\varphi=\{1\}$ or $\ker\varphi=\la t\ra$. Since
\begin{equation*}
\varphi(t)=
\begin{pmatrix}
\la t\ra&s\la t\ra&s^2\la t\ra\\
t\la t\ra&ts\la t\ra&ts^2\la t \ra
\end{pmatrix}
\end{equation*}
If $\varphi(t)$ is the identity permutation, then $ts\la t\ra=s\la t\ra$, so
that $s^{-1}ts\in\la t\ra=\{1,t\}$. But now $s^{-1}ts=t$. Therefore
$t\not\in\ker\varphi$ and $\ker\varphi=\{1\}$. Therefore $\varphi$ is
injective. Because $\abs{G}=\abs{S_3}$, $G\cong S_3$
   #+END_proof


   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $X$ is a set and $G$ is a group, then $G$ *acts* on $X$ if there is a
   function $G\times X\to X$, denoted by $(g,x)\to gx$ s.t.
   1. (gh)x=g(hx) for all $g,h\in G$ and $x\in X$
   2. $1x=x$ for all $x\in X$

      
   $X$ is a *\(G\)-set* if $G$ acts on $X$
   #+END_definition

   If a group $G$ acts on a set $X$, then fixing the first variable, say $g$,
   gives a function $\alpha_g:X\to X$, namely, $\alpha_g:x\mapsto gx$. This
   function is a permutation of $X$, for its inverse is $\alpha_{g^{-1}}$
   \begin{equation*}
   \alpha_g\alpha_{g^{-1}}=1=\alpha_{g^{-1}}\alpha_g
   \end{equation*}
   If's easy to see that $\alpha:G\to S_X$ defined by $\alpha:g\mapsto\alpha_g$
   is a homomorphism. Conversely, given any homomorphism $\varphi:G\to S_X$,
   define $gx=\varphi(g)(x)$. Thus an action of a group $G$ on a set $X$ is
   another way of viewing a homomorphism.

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $G$ acts on $X$ and $x\in X$, then the *orbit* of $x$, denoted by
   $\calo(x)$, is the subset of $X$
   \begin{equation*}
   \calo(x)=\{gx:g\in G\}\subseteq X
   \end{equation*}
   the *stabilizer* of $x$, denoted by $G_x$, is the subgroup
   \begin{equation*}
   G_x=\{g\in G:gx=x\}\le G
   \end{equation*}
   #+END_definition
   
   [[index:transitively]]
   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. Caylay's theorem says that $G$ acts on itself by translation:
      $\tau_g:a\mapsto ga$.  We say $G$ acts *transitively* on $X$ if there is
      only one orbit. 
   2. When $G$ acts on $G/H$ by translation $\tau_g:aH\mapsto gaH$, then the
      orbit $\calo(aH)=G/H$
   3. When a group $G$ acts on itself by conjugation, then the orbit  $\calo(x)$
      is 
      \begin{equation*}
      \{y\in G:y=axa^{-1}\text{ for some }a\in G\}
      \end{equation*}
      in this case, $\calo(x)$ is called the *conjugacy class* of $x$, and it is
      commonly denoted by $x^G$.

      *centralizer* $C_G(x)=\{g\in G:gxg^{-1}=x\}$
   4. Let $X=\{1,2,\dots,n\}$, let $\alpha\in S_n$ and regard the cyclic group
      $G=\la\alpha\ra$ as acting on $X$. If $i\in X$, then
      \begin{equation*}
      \calo(i)=\{\alpha^k(i):k\in\Z\}
      \end{equation*}
      Let the complete factorization of \alpha be $\alpha=\beta_1\dots\beta_{t(\alpha)}$,
      and let $i=i_1$ be moved by \alpha. If the cycle involving $i_1$ is
      $\beta_j=(i_1 i_2 \dots i_r)$,
      \begin{equation*}
      \calo(i)=\{i_1,\dots,i_r\}
      \end{equation*}
      where $i=i_1$. It follows that $\abs{\calo(i)}=r$. The stabilizer $G_l$ of
      a number $l$ is $G$ if \alpha fixes $l$
   #+END_examplle
   

   *Normalizer*
   \begin{equation*}
   N_G(H)=\{g\in G:gHg^{-1}=H\}
   \end{equation*}

   
   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $G$ acts on a set $X$, then $X$ is the disjoint union of the orbits. If
   $X$ is finite, then
   \begin{equation*}
   \abs{X}=\displaystyle\sum_i\abs{\calo(x_i)}
   \end{equation*}
   where $x_i$ is chosen from each orbit
   #+END_proposition

   #+BEGIN_proof
   $x\equiv y\Leftrightarrow$ there exists $g\in G$ with $y=gx$ is an
   equivalence relation
   #+END_proof


   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm2.98
   If $G$ acts on a set $X$ and $x\in X$ then
   \begin{equation*}
   \abs{\calo(x)}=[G:G_x]
   \end{equation*}
   #+END_theorem

   #+BEGIN_proof
   Let $G/G_x$ denote the family of cosets. Construct a bijection
   $\varphi:G/G_x\to \calo(x)$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   label:cor2.99
   If a finite group $G$ acts on a set X, then the number of elements in any
   orbit is a divisor of $\abs{G}$. 
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   label:cor2.100
   If $x$ lies in a finite group $G$, then the number of conjugates of $x$ is
   the index of its centralizer
   \begin{equation*}
   \abs{x^G}=[G:C_G(x)]
   \end{equation*}
   and hence it's a divisor of $G$
   #+END_corollary

   #+BEGIN_proof
   $x^G$ is the orbit, $C_G(x)$ is the stabilizer
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $H$ is a subgroup of a finite group $G$, then the number of conjugates of
   $H$ in $G$ is $[G:N_G(H)]$
   #+END_proposition

   #+BEGIN_proof
   Similar to theorem ref:thm2.98
   #+END_proof

   #+ATTR_LATEX: :options [Cauchy]
   #+BEGIN_theorem
   label:thmCauchy
   If $G$ is a finite group whose order is divisible by a prime $p$, then $G$
   contains an element of order $p$
   #+END_theorem

   #+BEGIN_proof
   Prove by induction on $m\ge 1$, where $\abs{G}=mp$. If $m=1$, it's obvious.

   If $x\in G$ , then $\abs{x^G}=[G:C_G(x)]$. If $x\not\in Z(G)$, then $x^G$
   has more than one element, so $\abs{C_G(x)}<\abs{G}$. If $p\mid \abs{C_G(x)}$, by
   inductive hypothesis, we are done. Else if $p\nmid \abs{C_G(x)}$ for all
   noncentral $x$ and $\abs{G}=[G:C_G(x)]\abs{C_G(x)}$, we have
   \begin{equation*}
   p\mid[G:C_G(x)]
   \end{equation*}
   $Z(G)$ consists of all those elements with $\abs{X^G}=1$, we have
   \begin{equation*}
   \abs{G}=\abs{Z(G)}+\displaystyle\sum_i[G:C_G(x_i)]
   \end{equation*}
   Hence $p\mid\abs{Z(G)}$ and by proposition ref:prop2.78
   #+END_proof
   
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   The *class equation* of a finite group $G$ is
   \begin{equation*}
   \abs{G}=\abs{Z(G)}+\displaystyle\sum_i[G:C_G(x_i)]
   \end{equation*}
   where each $x_i$ is selected from each conjugacy class having more than one element
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $p$ is a prime, then a finite group $G$ is called a *p-group* if
   $\abs{G}=p^n$ for some $n\ge 0$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm2.103
   If $p$ is a prime and $G$ is a p-group, then $Z(G)\neq\{1\}$
   #+END_theorem

   #+BEGIN_proof
   Consider 
   \begin{equation*}
   \abs{G}=\abs{Z(G)}+\displaystyle\sum_i[G:C_G(x_i)]
   \end{equation*}
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $p$ is a prime, then every group $G$ of order $p^2$ is abelian
   #+END_corollary
   #+BEGIN_proof
   If $G$ is not abelian, then $Z(G)$ has order $p$. The center is always normal, and
   so $G/Z(G)$ is defined; it has order $p$ and is cyclic by Lagrange's theorem. 
   This contradicts Exercise ref:ex2.69
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   Cauchy's theorem and Fermat's theorem are special cases of some common theorem.

   If $G$ is a finite group and $p$ is a prime, define
   \begin{equation*}
   X=\{(a_0,a_1,\dots,a_{p-1})\in G^p:a_0a_1\dots a_{p-1}=1\}
   \end{equation*}
   Note that $\abs{X}=\abs{G}^{p-1}$, for having chosen the last $p-1$ entries
   arbitrarily, the 0th entry must equal $(a_1a_2\dots a_{p-1})^{-1}$. Introduce an
   action of $\I_p$ on $X$ by defining, for $0\le i\le p-1$,
   \begin{equation*}
   [i](a_0,\dots,a_{p-1})=(a_{i+1},\dots,a_{p-1},a_0,\dots,a_i)
   \end{equation*}
   The product of the new \(p\)-tuple is a conjugate of $a_0a_1\dots a_{p-1}$
   \begin{equation*}
   a_{i+1}\dots a_{p-1}a_0\dots a_{i}=(a_0\dots a_i)^{-1}(a_0\dots a_{p-1})
   (a_0\dots a_i)
   \end{equation*}
   This conjugate is $1$ for $g^{-1}1g=1$, and so $[i](a_0,\dots,a_{p-1})\in X$. By
   Corollary ref:cor2.99, the size of every orbit of $X$ is a divisor of
   $\abs{\I_p}=p$. Now orbits with just one element consists of a \(p\)-tuple all
   of whose entries $a_i$ are equal, for all cyclic permutations of the \(p\)-tuple
   are the same. In other words, such an orbit corresponds to an element $a\in G$
   with $a^p=1$. Clearly $(1,1,\dots,1)$ is such an orbit; if it were the only such
   , then we would have
   \begin{equation*}
   \abs{G}^{p-1}=\abs{X}=1+kp
   \end{equation*}
   That is, $\abs{G}^{p-1}\equiv 1\mod p$. If $p$ is a divisor of $\abs{G}$, then
   we have a contradiction and thus proved Cauchy's theorem.
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $G$ is a group of order $\abs{G}=p^e$ then $G$ has a normal subgroup of order
   $p^k$ for every $k\le e$
   #+END_proposition

   #+BEGIN_proof
   We prove the result by induction on $e\ge 0$.

   By Theorem ref:thm2.103, $Z(G)\neq\{1\}$. Let $Z\le Z(G)$ be a subgroup of order
   $p$ and $Z$ is normal. If $k\le e$, then $p^{k-1}\le p^{e-1}=\abs{G/Z}$. By
   induction, $G/Z$ has a normal subgroup $H^*$ of order $p^{k-1}$. The
   correspondence theorem says there is a subgroup $H$ of $G$ containing $Z$ with
   $H^*=H/Z$; moreover $H^*\triangleleft G/Z$  implies $H\triangleleft G$. But
   $\abs{H/Z}=p^{k-1}$ implies $\abs{H}=p^k$ as desired.
   #+END_proof

   [[index: simple group]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A group $G\neq\{1\}$ is called *simple* if $G$ has no normal subgroups other than
   $\{1\}$ and  $G$ itself.
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   An abelian group $G$ is simple if and only if it is finite and of prime order
   #+END_proposition
   #+BEGIN_proof
   Assume $G$ is simple. Since $G$ is abelian, every subgroup is normal, and so $G$
   has no subgroups other than $\{1\}$ and $G$. Choose $x\in G$ with $x\neq 1$.
   Since $\la x\ra\le G$, we have $\la x\ra =G$. If $x$ has infinite order, then
   all the powers of $x$ are distinct, and so $\la x^2\ra<\la x\ra$ is a forbidden
   subgroup of $\la x\ra$, a contradiction. Therefore every $x\in G$ has finite
   order. If $x$ has order $m$ and if $m$ is composite, say $m=kl$, then 
   $\la x^k\ra$ is a proper subgroup of $\la x\ra$, a contradiction. Therefore
   $G=\la x\ra$ has prime order.
   #+END_proof

   Suppose that an element $x\in G$ has $k$ conjugates, that is 
   \begin{equation*}
   \abs{x^G}=\abs{\{gxg^{-1}:g\in G\}}=k
   \end{equation*}
   If there is a subgroup $H\le G$ with $x\in H\le G$, how many conjugates does $x
   $ have in $H$?

   Since
   \begin{equation*}
   x^H=\{hxh^{-1}:h\in H\}\subseteq x^G
   \end{equation*}
   we have $\abs{x^H}\le\abs{x^G}$. It is possible that there is a strict
   inequality $\abs{x^H}<\abs{x^G}$. For example, take $G=S_3,x=(1\; 2)$, and
   $H=\la x\ra$. Now let us consider this question, in particular, for
   $G=S_5,x=(1\;2\;3), H=A_5$

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   All 3-cycles are conjugate in $A_5$
   #+END_lemma

   #+BEGIN_proof
   Let $G=S_5,\alpha=(1\; 2\;3), H=A_5$. We know that $\abs{\alpha^{S_5}}=20$, for there
   are 20 3-cycles in $S_5$. Therefore, $20=\abs{S_5}/\abs{C_{S_5}(\alpha)}$ by
   Corollary ref:cor2.100 , so that $\abs{C_{S_5}(\alpha)}=6$. Here they are
   \begin{equation*}
   (1),\;(1\;2\;3),\;(1\;3\;2),\;(4\;5),\;(4\;5)(1\;2\;3),\;(4\;5\;)(1\;3\;2)
   \end{equation*}
   The last there of these are odd permutations, so that $\abs{C_{A_5}(\alpha)}=3$. We
   conclude that
   \begin{equation*}
   \abs{\alpha^{A_5}}=\abs{A_5}/\abs{C_{A_5}(\alpha)}=20
   \end{equation*}
   that is all 3-cycles are conjugate to \alpha in $A_5$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:lemma2.109
   If $n\ge 3$, every element in $A_n$ is a 3-cycle or a product of 3-cycles
   #+END_lemma
   #+BEGIN_proof
   Since each $\beta$ equals $\tau_1\dots\tau_{2q}$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   $A_5$ is a simple group
   #+END_theorem
   #+BEGIN_proof
   If $H\triangleleft A_5$ and $H\neq\{(1)\}$. Now if $H$ contains a 3-cycle, then
   normality forces $H$ to contain all its conjugates. Therefore it suffices to
   prove that $H$ contains 3-cycle.

   Since $\sigma\in H$, we may assume, after a harmless relabeling, that either
   $\sigma=(1\;\;2\;3),\sigma=(1\;2)(3\;4)$ or $\sigma=(1\;2\;3\;4\;5x)$

   If $\sigma=(1\;2)(3\;4)$, define $\tau=(1\;2)(3\;5)$. Now
   $(3\;5\;4)=(\tau\sigma\tau^{-1})\sigma^{-1}\in H$. If $\sigma=(1\;2\;3\;4\;5)$,
   define $\rho=(1\;3\;2)$ and $(1\;3\;4)=\rho\sigma\rho^{-1}\sigma^{-1}\in H$
   #+END_proof

   $A_4$ is not simple for $\bV\triangleleft A_4$.

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   $A_6$ is a simple group
   #+END_lemma

   #+BEGIN_proof
   Let $\{1\}\neq H\triangleleft A_6$; we must show that $H=A_6$. Assume that there
   is some $\alpha\in H$ with $\alpha\neq (1)$ that fixes some $i$, where $1\le
   i\le 6$. Define
   \begin{equation*}
   F=\{\sigma\in A_6:\sigma(i)=i\}
   \end{equation*}
   Note that $\alpha\in H\cap F$, so that $H\cap F\neq\{(1)\}$. The second
   isomorphism theorem gives $H\cap F\triangleleft F$. But $F$ is simple for
   $F\cong A_5$, we have $H\cap F=F$: that is $F\le H$. It follows that $H$
   contains a 3-cycle, and so $H=A_6$ by Exercise ref:ex2.91.

   If there is no $\alpha\in H$ with $\alpha\neq\{1\}$ that fixes some $i$ with
   $1\le i\le 6$. If we consider the cycle structures of permutations in $A_6$,
   however, any such \alpha must have cycle structure $(1\;2)(3\;4\;5\;6)$ or
   $(1\;2\;3)(4\;5\;6)$. In the first case $\alpha^2\in H$, $\alpha^2\in H$
   fixes 1. In the second case $\alpha(\beta\alpha^{-1}\beta^{-1})$ where
   $\beta=(2\;3\;4)$ fixes 1.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   $A_n$ is a simple group for all $n\ge 5$
   #+END_theorem

   #+BEGIN_proof
   If $H$ is a nontrivial normal subgroup of $A_n$, then we must show that $H=A_n$.
   By Exercise ref:ex2.91 it suffices to prove that $H$ contains a 3-cycle. If
   $\beta\in H$ is nontrivial, then there exists some $i$ that $\beta$ moves: say,
   $\beta(i)=j\neq i$. Choose a 3-cycle $\alpha$ that fixes $i$ and moves $j$. The
   permutations \alpha and \beta do not commute. It follows that
   $\gamma=(\alpha\beta\alpha^{-1})\beta^{-1}$ is a nontrivial element of $H$. But
   $\beta\alpha^{-1}\beta^{-1}$ is a 3-cycle, and so
   $\gamma=\alpha(\beta\alpha^{-1}\beta^{-1})$ is a product of two 3-cycles. Hence
   \gamma moves at most 6 symbols, say $i_1,\dots,i_6$. Define
   \begin{equation*}
   F=\{\sigma\in A_n:\sigma\text{ fixes all }i\neq i_1,\dots,i_6\}
   \end{equation*}
   Now $F\cong A_6$ and $\gamma\in H\cap F$. Hence $H\cap F\triangleleft F$. But
   $F$ is simple, and so $H\cap F=F$; that is $F\le H$. Therefore $H$ contains a
   3-cycle 
   #+END_proof

   #+ATTR_LATEX: :options [Burnside's Lemma]
   #+BEGIN_theorem
   Let $G$ act on a finite set $X$. If $N$ is the number of orbits, then
   \begin{equation*}
   N=\frac{1}{\abs{G}}\displaystyle\sum_{\tau\in G}Fix(\tau)
   \end{equation*}
   where $Fix(\tau)$ is the number of $x\in X$ fixed by \tau
   #+END_theorem
   #+BEGIN_proof
   List the elements of $X$ as follows: Choose $x_1\in X$ and then list all the
   elements $x_1,\dots,x_r$ in the orbit $\calo(x_1)$; then choose
   $x_{r+1}\not\in\calo(x_1)$, and so on until all the elements of $X$ are listed.
   Now list the elements $\tau_1,\dots,\tau_n$ of $G$ and form the following array,
   where
   \begin{equation*}
   f_{i,j}=
   \begin{cases}
   1&\text{ if }\tau_i\text{ fixes }x_j\\
   0&\text{ if }\tau_i\text{ moves }x_j
   \end{cases}
   \end{equation*}
   #+ATTR_LATEX: :mode math :align c|cccccc
   |          | $x_1$     | $x_2$     | $\dots$ | $x_{r+1}$   | $x_{r+2}$   | $\dots$ |
   |----------+-----------+-----------+---------+-------------+-------------+---------|
   | $\tau_1$ | $f_{1,1}$ | $f_{1,2}$ | $\dots$ | $f_{1,r+1}$ | $f_{1,r+2}$ | $\dots$ |
   | $\vdots$ |           |           |         |             |             |         |
   | $\tau_n$ | $f_{n,1}$ | $f_{n,2}$ | $\dots$ | $f_{n,r+1}$ | $f_{n,r+2}$ | $\dots$ |
   Now $Fix(\tau_i)$ is the number of 1's in the \(i\)th row. therefore 
   $\sum_{\tau\in G}Fix(\tau)$ is the total number of 1's in the array. The number
   of 1's in column 1 is $\abs{G_{x_1}}$. By Exercise ref:ex2.99
   $\abs{G_{x_1}}=\abs{G_{x_2}}$. By Theorem ref:thm2.98 the number of 1's in the
   $r$ columns labels by the $x_i\in\calo(x_i)$ is thus
   \begin{equation*}
   r\abs{G_{x_1}}=\abs{\calo(x_1)}\cdot\abs{G_{x_1}}=(\abs{G}/
   \abs{G_{x_1}})\abs{G_{x_1}}=\abs{G}
   \end{equation*}
   Therefore
   \begin{equation*}
   \displaystyle\sum_{\tau\in G}Fix(\tau)=N\abs{G}
   \end{equation*}
   #+END_proof

   We are going to use Burnside's lemma to solve problems of the following sort.
   How many striped flags are there having six stripes each of which can be colored
   red, white or blue?
   #+ATTR_LATEX: :align |c|c|c|c|c|c|
   |---+---+---+---+---+---|
   | r | w | b | r | w | b |
   |---+---+---+---+---+---|

   #+ATTR_LATEX: :align |c|c|c|c|c|c|
   |---+---+---+---+---+---|
   | b | w | r | b | w | r |
   |---+---+---+---+---+---|

   Let $X$ be the set of all 6-tuples of colors: if $x\in X$, then
   \begin{equation*}
   x=(c_1,c_2,c_3,c_4,c_5,c_6)
   \end{equation*}

   Let \tau be the permutation that reserves all the indices:
   #+ATTR_LATEX: :mode math :environment pmatrix :math-prefix \tau= :math-suffix =(1\;6)(2\;5)(3\;4)
   | 1 | 2 | 3 | 4 | 5 | 6 |
   | 6 | 5 | 4 | 3 | 2 | 1 |

   (thus $\tau$ turns over each 6-tuple $x$ of colored stripes). The cyclic group
   $G=\la\tau\ra$ acts on $X$; since $\abs{G}=2$, the orbit of any 6-tuple $x$
   consists of either 1 or 2 elements. Since a flag is unchanged by turning it
   over, it is reasonable to identify a flag with an orbit of 6-tuple. For example,
   the orbit consisting of the 6-tuples
   \begin{equation*}
   (r,w,b,r,w,b)\text{ and }(b,w,r,b,w,r)
   \end{equation*}
   above. The number of flags is thus the number $N$ of orbits; by Burnside's
   lemma, $N=\frac{1}{2}[Fix((1))+Fix(\tau)]$. The identity permutation $(1)$ fixes
   every $x\in X$, and so $Fix((1))=3^6$. Now \tau fixes a 6-tuple $x$ if it's a
   "palindrome". It follows that $Fix(x)=3^3$. The number of flags is thus
   \begin{equation*}
   N=\frac{1}{2}(3^6+3^3)=378
   \end{equation*}

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If a group $G$ acts on $X=\{1,\dots,n\}$ and if $\calc$ is a set of $q$ colors,
   then $G$ acts on the set $\calc^n$ of all \(n\)-tuples of colors by
   \begin{equation*}
   \tau(c_1,\dots,c_n)=(c_{\tau1},\dots,c_{\tau n})\text{ for all }\tau\in G
   \end{equation*}
   An orbit of $(c_1,\dots,c_n)\in\calc^n$ is called a *\((q,G)\)-coloring* of $X$.
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   Color each square in a $4\times 4$ grid red or black.

   If $X$ consists of the 16 squares in the grid and if $\calc$ consists of the two
   colors red and black, then the cyclic group $G=\la R\ra$ or order 4 acts on $X$,
   where $R$ is a clockwise rotation by $\ang{90}$; 

   \begin{figure}[h]
   \centering
   \begin{subfigure}[b]{0.4\textwidth}
   \begin{tabular}{|c|c|c|c|}
   \hline
   1 & 2 & 3 & 4\\
   \hline
   5 & 6 & 7 & 8\\
   \hline
   9 & 10 & 11 & 12\\
   \hline
   13 & 14 & 15 & 16\\
   \hline
   \end{tabular}
   \end{subfigure}
   \begin{subfigure}[b]{0.4\textwidth}
   \begin{tabular}{|c|c|c|c|}
   \hline
   13 & 9 & 5 & 1\\
   \hline
   14 & 10 & 6 & 2\\
   \hline
   15 & 11 & 7 & 3\\
   \hline
   16 & 12 & 8 & 4\\
   \hline
   \end{tabular}
   \end{subfigure}
   \label{fig2.10}
   \end{figure}

   Figure shows how $R$ acts: the right square is \(R)\)'s action on the left
   square. In cycle notation
   \begin{align*}
   &R=(1,\;4,\;16,\;13)(2,\;8,\;15,\;9)(3,\;12,\;14,\;5)(6,\;7,\;11,\;10)\\
   &R^2=(1,\;16)(4,\;13)(2,\;15)(8,\;9)(3,\;14)(12,\;5)(6,\;11)(7,\;10)\\
   &R^3=(1,\;13,\;16,\;4)(2,\;9,\;15,\;8)(3,\;5,\;14,\;12)(6,\;10,\;11,\;7)
   \end{align*}

   By Burnside's lemma, the number of chessboards is
   \begin{equation*}
   \frac{1}{4}[Fix((1))+Fix(R)+Fix(R^2)+Fix(R^3)]
   \end{equation*}

   #+END_examplle



   #+BEGIN_exercise
   Prove that if $p$ is a prime and $G$ is a finite group in which every element
   has order a power of $p$, then $G$ is a \(p\)-group. (A possibly infinite group
   $G$) is called a *\(p\)-group* if every element in $G$ has order a power of $p$
   #+END_exercise
   #+BEGIN_proof
   By Cauchy's theorem ref:thmCauchy
   #+END_proof

   #+BEGIN_exercise
   label:ex2.91
   1. For all $n\ge 5$, prove that all 3-cycles are conjugate in $A_n$
   2. Prove that if a normal subgroup $H\triangleleft A_n$ contains a 3-cycle,
      where $n\ge 5$, then $H=A_n$
   #+END_exercise
   #+BEGIN_proof
   1. If $(1\;2\;3)$ and $(i\; j\; k)$ are not disjoint. As Example ref:example2.8
      illustrated, $\alpha\in S_5$ 

      If they are disjoint, simple
   2. By lemma ref:lemma2.109
   #+END_proof

   #+BEGIN_exercise
   label:ex2.99
   1. Let a group $G$ act on a set $X$, and suppose that $x,y\in X$ lie in the same
      orbit: y=gx for some $g\in G$. Prove that $G_y=gG_xg^{-1}$
   2. Let $G$ be a finite group acting on a set $X$; prove that if $$x,y\in X lie
      in the same orbit, then $\abs{G_x}=\abs{G_y}$
   #+END_exercise

   #+BEGIN_proof
   1. If $f\in G_x$, then $gfg^{-1}(y)=gfg^{-1}gx=gx=y$
   2. There is a bijection.
   #+END_proof
* Commutative Rings \rom{1}
** First Properties
   [[index:commutative ring]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A *commutative ring* \(R\) is a set with two binary operations, addition and
   multiplication s.t.
   1. $R$ is an abelian group under addition
   2. (*commutativity*) \(ab=ba\) for all \(a,b\in R\)
   3. (*associativity*) \(a(bc)=(ab)c\) for every \(a,b,c\in R\)
   4. there is an element \(1\in R\) with \(1a=a\) for every \(a\in R\)
   5. (*distributivity*) \(a(b+c)=ab+ac\) for every \(a,b,c\in R\)
   #+END_definition

   The element 1 in a ring $R$ has several names: it is called *one*, the *unit* of
   \(R\), or the *identity* in \(R\)

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. \(\Z,\Q,\R\) and \(\C\) are commutative rings with the usual addition and
      multiplication
   2. Consider the set \(R\) of all real numbers $x$ of the form
      \begin{equation*}
      x=a+b\omega
      \end{equation*}
      where \(a,b\in\Q\) and \(\omega=\sqrt[3]{2}\). \(R\) is closed under ordinary
      addition. However, if \(R\) is closed under multiplication, then
      \(\omega^2\in R\) and there are rationals \(a\) and \(b\) with
      \begin{align*}
      &\omega^2=a+b\omega\\
      &2=a\omega+b\omega^2\\
      &b\omega^2=ab+b^2\omega
      \end{align*}
      Hence \(2-a\omega=ab+b^2\omega\) and so
      \begin{equation*}
      2-ab=(b^2+a)\omega
      \end{equation*}
      A contradiction.
   #+END_examplle

   
   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let $R$ be a commutative ring.
   1. \(0\cdot a=0\) for every \(a\in R\)
   2. If \(1=0\) then \(R\) consists of the single element 0. In this case \(R\)
      is called the *zero ring*
   3. If \(-a\) is the additive inverse of \(a\), then \((-1)(-a)=a\)
   4. \((-1)a=-a\) for every \(a\in R\)
   5. If $n\in\N$ and \(n1=0\), then \(na=0\) for all \(a\in R\)
   6. The binomial theorem holds: if \(a,b\in R\), then 
      \begin{equation*}
      (a+b)^n=\displaystyle\sum_{r=0}^n\binom{n}{r}a^rb^{n-r}
      \end{equation*}
   #+END_proposition

   #+BEGIN_proof
   6. [@6] \(\binom{n+1}{r}=\binom{n}{r-1}+\binom{n}{r}\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A subset $S$ of a commutative ring $R$ is a *subring* of \(R\) if
   1. \(1\in S\)
   2. if \(a,b\in S\) then \(a-b\in S\)
   3. if \(a,b\in S\), then \(ab\in S\)
   #+END_definition

   *Notation*. The tradition in ring theory is to write \(S\subseteq R\) for a
   subring

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   A subring \(S\) of a commutative ring \(R\) is itself a commutative ring.
   #+END_proposition

   [[index:domain]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A *domain* (often called an *integral domain*) is a commutative ring \(R\) that
   satisfies two extra axioms: first
   \begin{equation*}
   1\neq 0
   \end{equation*}
   second, the *cancellation law* for multiplication: for all \(a,b,c\in R\)
   \begin{equation*}
   \text{ if } ca=cb\text{ and }c\neq 0,\text{ then }a=b
   \end{equation*}
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   A nonzero commutative ring \(R\) is a domain if and only if the product of
   any two nonzero elements of \(R\) is nonzero
   #+END_proposition
   #+BEGIN_proof
   \(ab=ac\) if and only if \(a(b-c)=0\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop3.6
   The commutative ring \(\I_m\) is a domain if and only if \(m\) is a prime
   #+END_proposition
   #+BEGIN_proof
   If \(m=ab\), where \(1<a,b<m\), then \([a],[b]\neq[0]\) yet
   \([a][b]=[m]=[0]\)

   Conversely, if \(m\) is a prime and \([a][b]=[ab]=[0]\), then \(m\mid ab\)
   #+END_proof
   
   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. Let \(\calf(\R)\) be the set of all the function \(\R\to \R\) equipped
      with the operations of *point-wise addition* and *point-wise multiplication*:
      Given \(f,g\in\calf(\R)\), define functions \(f+g\) and \(fg\) by
      \begin{equation*}
      f+g:a\mapsto f(a)+f(b)\quad\text{ and }\quad fg:a\mapsto f(a)g(a)
      \end{equation*}
      We claim that \(\calf(\R)\) with these operations is a commutative ring.
      The zero element is the constant function \(z\) with value 0.
      \(\calf(\R)\) is not a domain by
      \begin{equation*}
      f(a)=
      \begin{cases}
      a&\text{ if }a\le 0\\
      0\\
      \end{cases}\e g(a)=
      \begin{cases}
      0&\text{ if }a\le 0\\
      a
      \end{cases}
      \end{equation*}
   #+END_examplle

   [[index:division]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let \(a\) and $b$ be elements of a commutative ring \(R\). Then \(a\) *divides*
   \(b\) *in* \(R\) (or \(a\) is a *divisor* of \(b\) or $b$ is a *multiple* of $a$),
   denoted by $a\mid b$, if there exists an element \(c\in R\) with \(b=ca\)
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   An element $u$ in a commutative ring $R$ is called a *unit* if \(u\mid 1\) in $R$.
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let \(R\) be a domain, and let \(a,b\in R\) be nonzero. Then \(a\mid b\) and
   \(b\mid a\) if and only if \(b=ua\) for some unit \(u\in R\)
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If \(a\) is an integer, then \([a]\) is a unit in \(\I_m\) if and only if
   \(a\) and \(m\) are relatively prime. 
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $p$ is a prime, then every nonzero $[a]$ in $\I_p$ is a unit.
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $R$ is a commutative ring, then the *group of units* of $R$ is
   \begin{equation*}
   U(R)=\{\text{all units in }R\}
   \end{equation*}
   #+END_definition

   [[index:field]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A *field* $F$ is a commutative ring in which $1\neq0$ and every nonzero element
   $a$ is a unit; that is, there is \(a^{-1}\in F\) with \(a^{-1}a=1\)
   #+END_definition
   A commutative ring $R$ is a field if and only if $U(R)=R^{\times}$, the
   nonzero elements of $R$.

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Every field $F$ is a domain
   #+END_proposition
   #+BEGIN_proof
   $ab=ac,b=a^{-1}ab=a^{-1}(ac)=c$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   The commutative ring $\I_m$ is a field if and only if $m$ is prime
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm3.13
   If $R$ is a domain then there is a field $F$ containing $R$ as a subring.
   Moreover, $F$ can be chosen so that for each $f\in  F$, there are \(a,b\in R\) with
   \(b\neq 0\) and \(f=ab^{-1}\) 
   #+END_theorem

   #+BEGIN_proof
   Let \(X=\{(a,b)\in R\times R:b\neq 0\}\) and define a relation $\equiv$ on
   $X$ by $(a,b)\equiv(c,d)$ if $ad=bc$. We claim that $\equiv$ is an
   equivalence relation. If $(a,b)\equiv(c,d)$ and $(c,d)\equiv(e,f)$, then
   $ad=bc,cf=de$ and \(adf=b(cf)=bde\), gives $af=be$

   Denote the equivalence class of (a,b) by $[a,b]$, define $F$ as the set of
   all equivalence classes $[a,b]$ and equip $F$ with the following addition and
   multiplication 
   \begin{align*}
   &[a,b]+[c,d]=[ad+bc,bd]\\
   &[a,b][c,d]=[ac,bd]
   \end{align*}
   Show addition and multiplication are well-defined.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   The field $F$ constructed from $R$ in Theorem ref:thm3.13 is called the
   *fraction field* of $R$, denoted by $\Frac(R)$, and we denote
   $[a,b]\in\Frac(R)$ by $a/b$
   #+END_definition

   Note that $\Frac(\Z)=\Q$
** Polynomials
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $R$ is a commutative ring, then a *sequence* \sigma in $R$ is
   \begin{equation*}
   \sigma=(s_0,s_1,\dots,s_i,\dots)
   \end{equation*}
   the entries \(s_i\in R\) for all \(i\ge 0\) are called the *coefficients* of \sigma
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A sequence \(\sigma=(s_0,\dots,s_i,\dots)\) in a commutative ring $R$ is
   called a *polynomial* if there is some integer \(m\ge 0\) with $s_i=0$ for all
   $i>m$; that is 
   \begin{equation*}
   \sigma=(s_0,\dots,s_m,0,\dots)
   \end{equation*}
   A polynomial has only finitely many nonzero coefficients. The *zero
   polynomial*, denoted by $\sigma=0$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $\sigma(s_0,\dots,s_n,0,\dots)\neq0$ is a polynomial, we call $s_n$ the
   *leading coefficient* of $\sigma$, we call $n$ the *degree* of \sigma, an we
   denote $n$ by $\deg(\sigma)$ 
   #+END_definition

   *Notation*. If $R$ is a commutative ring, then the set of all polynomials with
   coefficients in $R$ is denoted by $R[x]$


   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop3.14
   If $R$ is a commutative ring, then $R[x]$ is a commutative ring that contains
   $R$ as a subring
   #+END_proposition

   #+BEGIN_proof
   \(\sigma=(s_0,s_1,\dots),\tau=(t_0,t_1,\dots)\)
   \begin{align*}
   &\sigma+\tau=(s_0+t_0,s_1+t_1,\dots)\\
   &\sigma\tau=(c_0,c_1,\dots)
   \end{align*}
   where \(c_k=\sum_{i+j=k}s_it_j=\sum_{i=0}^ks_it_{k-i}\).
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:lemma3.15
   Let $R$ be a commutative ring and let \(\sigma,\tau\in R[x]\) be nonzero
   polynomials.
   1. Either \(\sigma\tau=0\) or \(\deg(\sigma\tau)\le\deg(\sigma)+\deg(\tau)\)
   2. If $R$ is a domain, then $\sigma\tau\neq0$ and 
      \begin{equation*}
      \deg(\sigma\tau)=\deg(\sigma)+\deg(\tau)
      \end{equation*}
   3. If $R$ is a domain, then $R[x]$ is a domain
   #+END_lemma

   #+BEGIN_proof
   \(\sigma=(s_0,s_1,\dots),\tau=(t_0,t_1,\dots)\) have degrees $m$ and $n$ respectively.
   1. if $k>m+n$, then each term in \(\sum_is_it_{k-i}\) is 0
   2. Each term in \(\sum_is_it_{m+n-i}\) is 0 with the possible exception of
      $s_mt_n$. Since $R$ is a domain, $s_m\neq 0$ and $t_n\neq0$ imply $s_mt_n\neq0$.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $R$ is a commutative ring, then $R[x]$ is called the *ring of polynomials
   over $R$*
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Define the element $x\in R[x]$ by
   \begin{equation*}
   x=(0,1,0,0,\dots)
   \end{equation*}
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   1. IF $\sigma=(s_0,\dots)$, then
      \begin{equation*}
      x\sigma=(0,s_0,s_1,\dots)
      \end{equation*}
   2. If $n\ge 1$, then $x^n$ is the polynomial having 0 everywhere except for 1
      in the \(n\)th coordinate
   3. If \(r\in R\), then
      \begin{equation*}
      (r,0,\dots)(s_0,s_1,\dots,s_j,\dots)=(rs_0,rs_1,\dots,rs_j,\dots)
      \end{equation*}
   #+END_lemma

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $\sigma=(s_0,\dots,s_n,0,\dots)$, then
   \begin{equation*}
   \sigma=s_0+s_1x+s_2x^2+\dots+s_nx^n
   \end{equation*}
   where each element \(s\in R\) is identified with the polynomial \((s,0,\dots)\)
   #+END_proposition

   As a customary, we shall write 
   \begin{equation*}
   f(x)=s_0+s_1x+\dots+s_nx^n
   \end{equation*}
   instead of \sigma. $s_0$ is called its *constant term*. If $s_n=1$ , then
   $f(x)$ is called *monic*.

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Polynomials $f(x)=s_0+\dots+s_nx^n$ and $g(x)=t_0+\dots+t_mx^m$ are equal if
   and only if $n=m$ and $s_i=t_i$ for all $i$.
   #+END_corollary

   If $R$ is a commutative ring, each polynomial $f(x)=s_0+\dots+s_nx^n$
   defines a *polynomial function* $f:R\to R$ by evaluation: If \(a\in R\), define
   \(f(a)=s_0+\dots+s_na^n\in R\).


   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let $k$ be a field. The fraction field of $k[x]$, denoted by $k(x)$, is
   called the *field of rational function* over $k$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $k$ is a field, then the elements of $k(x)$ have the form $f(x)/g(x)$
   where $f(x),g(x)\in k[x]$ and $g(x)\neq 0$
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $p$ is a prime, then the field of rational functions \(\I_p(x)\) is a n
   infinite field containing \(\I_p\) as a subfield.
   #+END_proposition

   #+BEGIN_proof
   By Lemma ref:lemma3.15 (3), \(\I_p[x]\) is an infinite domain for the powers
   $x^n$ for $n\in\N$ are distinct. Thus its fraction filed \(\I_p(x)\) is an
   infinite field containing \(\I_p[x]\) as a subring. But \(\I_p[x]\) contains
   \(\I_p\) as a subring, by Proposition ref:prop3.14.
   #+END_proof


   $R[x]$ is often called the ring of all *polynomials over $R$ in one variable*.
   If we write $A=R[x]$, then $A[y]$ is called the ring of all 
   *polynomials over $R$ in two variables $x$ and $y$*, and it is denoted by $R[x,y]$.

   #+BEGIN_exercise
   Show that if $R$ is a commutative ring, then $R[x]$ is never a field
   #+END_exercise

   #+BEGIN_proof
   If $R[x]$ is a field, then $x^{-1}\in R[x]$ and $x^{-1}=\sum_ic_ix^i$.
   However
   \begin{equation*}
   \deg(xx^{-1})=\deg(1)=1=\deg(x)+\deg(x^{-1})
   \end{equation*}
   A contradiction.
   #+END_proof

   #+BEGIN_exercise
   label:ex3.22
   Show that the polynomial function defined by $f(x)=x^p-x\in\I_p[x]$ is
   identically zero.
   #+END_exercise

   #+BEGIN_proof
   By Fermat's theorem ref:Fermat, \(a^p\equiv a\mod p\)
   #+END_proof

   #+BEGIN_exercise
   label:ex3.23
   If $R$ is a commutative ring and \(f(x)=\sum_{i=0}^n s_ix^i\in R[x]\) has
   degree \(n\ge1\), define its *derivative* \(f'(x)\in R[x]\) by
   \begin{equation*}
   f'(x)=s_1+2s_2x+3s_3x^2+\cdots+ns_nx^{n-1}
   \end{equation*}
   if \(f(x)\) is a constant polynomial, define its derivative to be the zero
   polynomial. Prove that the usual rules of calculus hold:
   \begin{align*}
   (f+g)'&=f'+g'\\
   (rf)'&=r(f)'\quad\text{if }r\in R\\
   (fg)'&=fg'+f'g\\
   (f^n)'&=nf^{n-1}f'\quad\text{for all }n\ge 1
   \end{align*}
   #+END_exercise

   #+BEGIN_exercise
   label:ex3.24
   Let $R$ be a commutative ring and let \(f(x)\in R[x]\)
   1. Prove that if \((x-a)^2\mid f(x)\), then \(x-a\mid f'(x)\) in $R[x]$
   2. Prove that if \(x-a\mid f(x)\) and \(x-a\mid f'(x)\), then \((x-a)^2\mid f(x)\)
   #+END_exercise
** Greatest Common Divisors
   #+ATTR_LATEX: :options [Division Algorithm]
   #+BEGIN_theorem
   Assume that $k$ is a field and that \(f(x),g(x)\in k[x]\) with $f(x)\neq 0$.
   Then there are unique polynomials \(q(x),r(x)\in k[x]\) with
   \begin{equation*}
   g(x)=q(x)f(x)+r(x)
   \end{equation*}
   and either $r(x)=0$ or $\deg(r)<\deg(f)$
   #+END_theorem

   #+BEGIN_proof
   We first prove the existence of such $q$ and $r$. If \(f\mid g\), then
   \(g=qf\) for some $q$; define the remainder $r=0$. If \(f\nmid g\), then
   consider all polynomials of the form \(g-qf\) as $q$ varies over $k[x]$. The
   least integer axiom provides a polynomial \(r=g-qf\) having least degree
   among all such polynomials. Since \(g=qf+r\), it suffices to show that
   \(\deg(r)<\deg(f)\). Write \(f(x)=s_nx^n+\dots+s_1x+s_0\) and
   \(r(x)=t^mx^m+\dots t_0\). Now $s_n\neq 0$ implies that $s_n$ is a unit
   because $k$ is a field and so $s_n^{-1}\in k$. If $\deg(r)\ge\deg(f)$, define
   \begin{equation*}
   h(x)=r(x)-t_ms_n^{-1}x^{m-n}f(x)
   \end{equation*}
   that is, if \(\LT(f)=s_nx^n\), where LT abbreviates *leading term*, then
   \begin{equation*}
   h=r-\frac{\LT(r)}{\LT(f)}f
   \end{equation*}
   note that $h=0$ or \(\deg(h)<\deg(r)\). If $h=0$, then \(r=[\LT(r)/\LT(f)]f\)
   and
   \begin{align*}
   g&=qf+r=qf+\frac{\LT(r)}{\LT(f)}f\\
   &=\left[q+\frac{\LT(r)}{\LT(f)}\right]f
   \end{align*}
   contradicting \(f\nmid g\). If \(h\neq0\), then $\deg(h)<\deg(r)$ and
   \begin{equation*}
   g-qf=r=h+\frac{\LT(r)}{\LT(f)}f
   \end{equation*}
   Thus $g-[q+\LT(r)/\LT(f)]f=h$, contradicting $r$ being a polynomial of least
   degree having this form. Therefore $\deg(r)<\deg(f)$

   To prove uniqueness of $q(x)$ and $r(x)$ assume that $g=q'f+r'$, where
   $\deg(r')<\deg(f)$. Then
   \begin{equation*}
   (q-q')f=r'-r
   \end{equation*}
   If $r'\neq r$, then each side has a degree. But
   \(\deg((q-q')f)=\deg(q-q')+\deg(f)\ge\deg(f)\), while
   \(\deg(r'-r)\le\max\{\deg(r'),\deg(r)\}<\deg(f)\), a contradiction. Hence
   $r'=r$ and $(q-q')f=0$. As $k[x]$ is a domain and $f\neq 0$, it follows that
   $q-q'=0$ and $q=q'$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $f(x)$ and $g(x)$ are polynomials in $k[x]$, where $k$ is a field, then
   the polynomials $q(x)$ and $r(x)$ occurring in the division algorithm are
   called the *quotient* and the *remainder* after dividing $g(x)$ by $f(x)$
   #+END_definition
   
   The hypothesis that $k$ is a filed is much too strong: long division can be
   carried out in $R[x]$ for every commutative ring $R$ as long as the leading
   coefficient of $f(x)$ is a unit in $R$; in particular, long division is
   always possible when $f(x)$ is monic.

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Let $R$ be a commutative ring and let $f(x)\in R[x]$ be a monic polynomial.
   If $g(x)\in R[x]$, then there exists $q(x),r(x)\in R[x]$ with
   \begin{equation*}
   g(x)=q(x)f(x)+r(x)
   \end{equation*}
   where either $r(x)=0$ or \(\deg(r)<\deg(f)\)
   #+END_corollary

   #+BEGIN_proof
   Note that \(\LT(r)/\LT(f)\in R\) because $f(x)$ is monic
   #+END_proof

   [[index:root]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $f(x)\in k[x]$, where $k$ is a field, then a *root* of $f(x)$ *in $k$* is an
   element $a\in k$ with $f(a)=0$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   Let $f(x)\in k[x]$, where $k$ is a field, and let $u\in k$. Then there is
   \(q(x)\in k[x]\) with
   \begin{equation*}
   f(x)=q(x)(x-u)+f(u)
   \end{equation*}
   #+END_lemma
   #+BEGIN_proof
   The division algorithm gives
   \begin{equation*}
   f(x)=q(x)(x-u)+r
   \end{equation*}
   Now evaluate
   \begin{equation*}
   f(u)=q(u)(u-u)+r
   \end{equation*}
   and so $r=f(u)$
   #+END_proof
   
   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop3.24
   If $f(x)\in k[x]$, where $k$ is a field, then $a$ is a root of $f(x)$ in $k$
   if and only if $x-a$ divides $f(x)$ in $k[x]$
   #+END_proposition

   #+BEGIN_proof
   If $a$ is a root of $f(x)$ in $k$, then $f(a)=0$ and the lemma gives
   $f(x)=q(x)(x-a)$. 
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm3.25
   Let $k$ be a field and let $f(x)\in k[x]$. If $f(x)$ has degree $n$, then
   $f(x)$ has at most $n$ roots in $k$
   #+END_theorem

   #+BEGIN_proof
   We prove the statement by induction on $n\ge 0$. If $n=0$, then $f(x)$ is a
   nonzero constant, and so the number of its roots in $k$ is zero. Now let
   $n>0$. If $f(x)$ has no roots in $k$, then we are done. Otherwise we may
   assume that there is $a\in k$ with $a$ a root of $f(x)$; hence by Proposition
   ref:prop3.24
   \begin{equation*}
   f(x)=q(x)(x-a)
   \end{equation*}
   moreover, \(q(x)\in k[x]\) has degree $n-1$. 
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   Theorem ref:thm3.25 is not true for polynomials with coefficients in an
   arbitrary commutative ring $R$. For example, if $R=\I_8$, then the quadratic
   polynomial $x^2-1$ has 4 roots: $[1],[3],[5],[7]$
   #+END_examplle
   

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Every \(n\)th root of unity in $\C$  is equal to
   \begin{equation*}
   e^{2\pi ik/n}=\cos\left(\frac{2\pi k}{n}\right)+i\sin\left(\frac{2\pi k}{n}\right)
   \end{equation*}
   where $k=0,1,\dots,n-1$
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Let $k$ be an infinite field and let $f(x)$ and $g(x)$ be polynomials in
   $k[x]$. If $f(x)$ and $g(x)$ determine the same polynomial function, then
   $f(x)=g(x)$ 
   #+END_corollary
   
   #+BEGIN_proof
   If $f(x)\neq g(x)$, then the polynomial $h(x)=f(x)-g(x)$ is nonzero, so that
   it has some degree, say $n$. Now every element of $k$ is a root of $h(x)$;
   since $k$ is infinite, $h(x)$ has more than $n$ roots, a contradiction.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:nthm2.46
   If $k$ is a field and $G$ is a finite subgroup of the multiplicative group
   $k^\times$., then $G$ is cyclic. In particular, if $k$ itself is finite, then
   $k^\times$ is cyclic.
   #+END_theorem

   #+BEGIN_proof
   Let $d$ be a divisor of $\abs{G}$. If there are two subgroups of $G$ of order
   $d$, say $S$ and $T$, then $\abs{S\cup T}>d$. But each $a\in S\cup T$
   satisfies $a^d=1$ and hence it's a root of $x^d-1$, a contradiction. Thus $G$
   is cyclic, by Theorem ref:thm2.86.
   #+END_proof

   [[index:primitive element]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $k$ is a finite field, a generator of the cyclic group $k^\times$ is
   called a *primitive element* of $k$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $f(x)$ and $g(x)$ are polynomials in $k[x]$, where $k$ is a field, then a
   *common divisor* is a polynomial \(c(x)\in k[x]\) with \(c(x)\mid f(x)\) and
   \(c(x)\mid g(x)\). If $f(x)$ and $g(x)$ in $k[x]$ are not both 0, define their
   *greatest common divisor*, abbreviated gcd, to be the monic common divisor
   having largest degree. If $f(x)=0=g(x)$, define their $\gcd=0$. The gcd of
   $f(x)$ and $g(x)$ is often denoted by $(f,g)$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm3.31
   If $k$ is a field and $f(x),g(x)\in k[x]$, then their \(\gcd d(x)\) is a
   nonlinear combination of $f(x)$ and $g(x)$; that is there are \(s(x),t(x)\in
   k[x]\) with
   \begin{equation*}
   d(x)=s(x)f(x)+t(x)g(x)
   \end{equation*}
   #+END_theorem

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Let $k$ be a field and let \(f(x),g(x)\in k[x]\). A monic common divisor
   $d(x)$ is the gcd if and only if $d(x)$ is divisible by every common divisor
   #+END_corollary

   [[index:irreducible]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   An element $p$ in a domain $R$ is *irreducible* if $p$ is neither 0 nor a unit
   and in any factorization $p=uv$ in $R$, either $u$ or $v$ is a unit. Elements
   \(a,b\in R\) are *associates* if there is a unit \(u\in R\) with $b=ua$
   #+END_definition

   For example, a prime $p$ is irreducible in $\Z$

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $k$ is a field, then a polynomial \(p(x)\in k[x]\) is irreducible if and
   only if \(\deg(p)=n\ge 1\) and there is no factorization in $k[x]$ of the
   form $p(x)=g(x)h(x)$ in which both factors have degree smaller than $n$
   #+END_proposition

   #+BEGIN_proof
   We show fist that $h(x)\in k[x]$ is a unit if and only if \(\deg(h)=0\). If
   $h(x)u(x)=1$, then $\deg(h)+\deg(u)=\deg(1)=0$, we have \(\deg(h)=0\).
   Conversely if $\deg(h)=0$, then $h(x)$ is a nonzero constant; that is, 
   $h\in k$; since $k$ is a field, $h$ has an inverse

   If $p(x)$ is irreducible, then its only factorization are of the form
   $p(x)=g(x)h(x)$ where $g(x)$ or $h(x)$ is a unit; that is, either $\deg(g)=0$
   or $\deg(h)=0$. 

   Conversely, if $p(x)$ is reducible, then it has factorization $p(x)=g(x)h(x)$
   where neither $g(x)$ nor $h(x)$ is a unit;
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   label:cor3.34
   Let $k$ be a field and let $f(x)\in k[x]$ be a quadratic or cubic polynomial.
   Then $f(x)$ is irreducible in $k[x]$ if and only if $f(x)$ does not have a root
   in $k$
   #+END_corollary

   #+BEGIN_proof
   If $f(x)=g(x)h(x)$, then $\deg(f)=\deg(g)+\deg(h)$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   label:example3.35
   1. We determine the irreducible polynomials in $\I_2[x]$ of small degree.

      As always, the linear polynomials $x$ and $x+1$ are irreducible

      There are four quadratics: $x^2,x^2+x,x^2+1,x^2+x+1$ . Since each of the
      first three has a root in \(\I_2\), there is only one irreducible
      quadratic

      There are eight cubics, of which four are reducible because their constant
      term is 0. The remaining polynomials are
      \begin{equation*}
      x^3+1;\quad x^3+x+1;\quad x^3+x^2+1;\quad x^3+x^2+x+1
      \end{equation*}
      Since 1 is a root of the first and fourth, the middle two are the only
      irreducible cubics.

      
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:lemma3.36
   Let $k$ be a field, let $p(x),f(x)\in k[x]$, and let $d(x)=(p,f)$. If $p(x)$
   is a monic irreducible polynomial, then
   \begin{equation*}
   d(x)=
   \begin{cases}
   1&\text{if }p(x)\nmid f(x)\\
   p(x)&\text{if }p(x)\mid f(x)
   \end{cases}
   \end{equation*}
   #+END_lemma

   #+ATTR_LATEX: :options [Euclid's Lemma]
   #+BEGIN_theorem
   Let $k$ be a field and let \(f(x),g(x)\in k[x]\). If $p(x)$ is an irreducible
   polynomial in $k[x]$, and \(p(x)\mid f(x)g(x)\), then either
   \begin{equation*}
   p(x)\mid f(x)\hspace{0.5cm}\text{or}\hspace{0.5cm}p(x)\mid g(x)
   \end{equation*}
   More generally, if \(p(x)\mid f_1(x)\dots f_n(x))\), then \(p(x)\mid f_i(x)\)
   for some $i$
   #+END_theorem

   #+BEGIN_proof
   Assume $p\mid fg$ but that \(p\nmid f\). Since $p$ is irreducible, $(p,f)=1$,
   and so $1=sp+tf$ for some polynomials $s$ and $t$. Therefore
   \begin{equation*}
   g=spg+tfg
   \end{equation*}
   and so \(p\mid g\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Two polynomials \(f(x),g(x)\in k[x]\) where $k$ is a field, are called
   *relatively prime* if their gcd is 1
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Let \(f(x),g(x),h(x)\in k[x]\), where $k$ is a field and let $h(x)$ and
   $f(x)$ be relatively prime. If \(h(x)\mid f(x)g(x)\), then \(h(x)\mid g(x)\)
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $k$ is a field, then a rational function \(f(x)/g(x)\in k(x)\) is in
   *lowest terms* if $f(x)$ and $g(x)$ are relatively prime
   #+END_definition


   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $k$ is a field, every nonzero \(f(x)/g(x)\in k(x)\) can be put in lowest terms
   #+END_proposition


   #+ATTR_LATEX: :options [Euclidean Algorithm]
   #+BEGIN_theorem
   If $k$ is a field and \(f(x),g(x)\in k[x]\), then there are algorithms for
   computing \(\gcd(f,g)\) as well as for finding a pair of polynomials $s(x)$
   and $t(x)$ with 
   \begin{equation*}
   (f,g)=s(x)f(x)+t(x)g(x)
   \end{equation*}
   #+END_theorem

   #+BEGIN_proof
   \begin{gather*}
   g=q_1f+r_1\\
   f=q_2r_1+r_2\\
   r_1=q_3r_2+r_3\\
   \vdots\\
   r_{n-4}=q_{n-2}r_{n-3}+r_{n-2}\\
   r_{n-3}=q_{n-1}r_{n-2}+r_{n-1}\\
   r_{n-2}=q_nr_{n-1}+r_n\\
   r_{n-1}=q_{n+1}r_n
   \end{gather*}
   Since the degrees of the remainders are strictly decreasing, this procedure
   must stop after a finite number of steps. The claim is that $d=r_n$ is the
   gcd. If $c$ is any common divisor of $f$ and $g$, then $c\mid r_i$  for every
   $i$. Also
   \begin{align*}
   r_n&=r_{n-2}-q_nr_{n-1}\\
   &=r_{n-2}-q_n(r_{n-3}-q_{n-1}r_{n-2})\\
   &=(1+q_{n-1})r_{n-2}-q_nr_{n-3}\\
   &=(1+q_{n-1})(r_{n-4}-q_{n-2}r_{n-3})-q_nr_{n-3}\\
   &=(1+q_{n-1})r_{n-4}-[(1+q_{n-1})q_{n-2}+q_n]r_{n-3}\\
   &\vdots\\
   &=sf+tg
   \end{align*}
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Let $k$ be a subfield of a field $K$, so that $k[x]$ is a subring of $K[x]$.
   If \(f(x),g(x)\in k[x]\), then their gcd in $k[x]$ is equal to their gcd in $K[x]$
   #+END_corollary

   #+BEGIN_proof
   The division algorithm in $K[x]$ gives
   \begin{equation*}
   g(x)=Q(x)f(x)+R(x)
   \end{equation*}
   $k[x]$ gives
   \begin{equation*}
   g(x)=q(x)f(x)+r(x)
   \end{equation*}
   and this also holds in $K[x]$. So that uniqueness of quotient and remainder
   gives $Q(x)=q(x),R(x)=r(x)$.
   #+END_proof

   #+ATTR_LATEX: :options [Unique Factorization]
   #+BEGIN_theorem
   If $k$ is a field, then every polynomial \(f(x)\in k[x]\) of degree $\ge1$ is
   a product of a nonzero constant and monic irreducibles. Moreover, if $f(x)$
   has two such factorizations
   \begin{equation*}
   f(x)=ap_1(x)\dots p_m(x)\hspace{0.5cm}\text{and}\hspace{0.5cm}
   f(x)=bq_1(x)\dots q_n(x)
   \end{equation*}
   then \(a=b,m=n\) and the \(q\)'s may be reindexed so that \(q_i=p_i\) for all $i$
   #+END_theorem

   #+BEGIN_proof
   We prove the existence of a factorization for a polynomial $f(x)$ by
   induction on $\deg(f)\ge1$. If \(\deg(f)=1=\), then $f(x)=ax+c=a(x+a^{-1}c)$.
   As every linear polynomial, $x+a^{-1}c$ is irreducible.

   Assume now that $\deg(f)\ge1$. If $f(x)$ is irreducible and its leading
   coefficient is $a$, write $f(x)=a(a^{-1}f(x))$; we are done. If $f(x)$ is not
   irreducible, then $f(x)=g(x)h(x)$, where \(\deg(g)<\deg(f)\) and
   \(\deg(h)<\deg(f)\). By the inductive hypothesis, 
   \(g(x)=bp_1(x)\dots p_m(x)\) and \(h(x)=cq_1(x)\dots q_n(x)\). It follows
   that 
   \begin{equation*}
   f(x)=(bc)p_1(x)\dots p_m(x)q_x(x)\dots q_n(x)
   \end{equation*}

   We now prove by induction on \(M=\max\{m,n\}\ge1\) if there is an
   equation
   \begin{equation*}
   ap_1(x)\dots p_m(x)=bq_1(x)\dots q_n(x)
   \end{equation*}
   where $a$ and $b$ are nonzero constants and the \(p\)'s and \(q\)'s are monic
   irreducibles. For the inductive step, \(p_m(x)\mid q_1(x)\dots q_n(x)\). By
   Euclid's lemma, there is $i$ with \(p_m(x)\mid q_i(x)\). But $q_i(x)$ are
   monic irreducible, so that $q_i(x)=p_m(x)$. Canceling this factor we will use
   inductive hypothesis
   #+END_proof

   Let $k$ be a field and assume that there are $a,r_1,\dots,r_n\in k$ with
   \begin{equation*}
   f(x)=a \displaystyle\prod_{i=1}^n(x-r_i)
   \end{equation*}
   
   If \(r_1,\dots,r_s\) where \(s\le n\) are the distinct roots of $f(x)$, then
   collecting terms gives
   \begin{equation*}
   f(x)=a(x-r_1)^{e_1}\dots (x-r_s)^{e_s}
   \end{equation*}
   where $r_j$ are distinct and $e_j\ge1$. We call $e_j$ the *multiplicity* of the
   root $r_j$. 

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm3.43
   Let \(f(x)=a_0+a_1x+\dots+a_nx^n\in\Z[x]\subseteq\Q[x]\). Every rational root
   $r$ of $f(x)$ has the form $b/c$, where \(b\mid a_0\) and \(c\mid a_n\)
   #+END_theorem

   #+BEGIN_proof
   We may assume that $r=b/c$ is in lowest form. 
   \begin{gather*}
   0=f(b/c)=a_0+a_1(b/c)+\dots+a_n(b/c)^n\\
   0=a_0c^n+a_1bc^{n-1}+\dots+a_nb^n
   \end{gather*}
   Hence \(a_0c^n=b(-a_1c^{n-1}-\dots-a_nb^{n-1})\), that is \(b\mid a_0c^n\).
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A complex number \alpha is called an *algebraic integer* if \alpha is a root of a monic
   \(f(x)\in\Z[x]\)
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   label:cor3.44
   A rational number $z$ that is an algebraic integer must lie in $\Z$. More
   precisely, if \(f(x)\in\Z[x]\subseteq\Q[x]\) is a monic polynomial, then
   every rational root of $f(x)$ is an integer that divides the constant term
   #+END_corollary

   #+BEGIN_proof
   $a_n=1$ in Theorem ref:thm3.43
   #+END_proof

   For example, consider \(f(x)=x^3+4x^2-2x-1\in\Q[x]\). By Corollary
   ref:cor3.34, this cubic is irreducible if and only if it has no rational
   root. As $f(x)$ is monic, the candidates for rational roots are \(\pm 1\),
   for these are the only divisor of -1 in $\Z$. Thus $f(x)$ has no roots in
   $\Q$ and hence $f(x)$ is irreducible in $\Q[x]$

   #+BEGIN_exercise
   label:ex3.37
   1. Let \(f(x)=(x-a_1)\cdots(x-a_n)\in k[x]\) where $k$ is a field. Show that
      \(f(x)\) has *no repeated roots* if and only if \(\gcd(f,f')=1\) where
      \(f'(x)\) is the derivative of $f$
   2. Prove that if \(p(x)\in\Q[x]\) is an irreducible polynomial, then \(p(x)\)
      has no repeated ro！ots in \(\C\)
   #+END_exercise
   
   #+BEGIN_proof
   1. \(f'(x)=\sum_{i=1}^n(x-a_1)\cdots(x-a_{i-1})(x-a_{i+1})\cdots(x-a_n)\)
   #+END_proof
** Homomorphisms
   [[index:ring homomorphism]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $A$ and $R$ are (commutative) rings, a *(ring) homomorphism* is a function
   \(f:A\to R\) s.t.
   1. \(f(1)=1\)
   2. \(f(a+a')=f(a)+f(a')\)
   3. \(f(aa')=f(a)f(a')\)
   #+END_definition


   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. Let $R$ be a domain and let \(F=\Frac(R)\). 
      \(R'=\{[a,1]:a\in R\}\subseteq F\), then the function \(f:R\to R'\) given by \(f(a) =[a,1]\), is an isomorphism
   2. Complex conjugation \(z=a+ib\mapsto\overline{z}=a-ib\) is an isomorphism
      \(\C\to \C\).
   3. Let $R$ be a commutative ring, and let \(a\in R\). Define the *evaluation
      homomorphism* \(e_a:R[x]\to R\) by \(e_a(f(x))=f(a)\).
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   If \(f:A\to R\) is a ring homomorphism, then for all \(a\in A\)
   1. \(f(a^n)=f(a)^n\)
   2. if \(a\) is a unit, then \(f(a)\) is a unit and \(f(a^{-1})=f(a)^{-1}\)
   3. if \(f:A\to R\) is a ring homomorphism, then
      \begin{equation*}
      f(U(A))\le U(R)
      \end{equation*}
      where \(U(A)\) is the group of units of $A$; if $f$ is an isomorphism,
      then
      \begin{equation*}
      U(A)\cong U(R)
      \end{equation*}
   #+END_lemma

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop3.48
   If $R$ and $S$ are commutative rings and \(\varphi:R\to S\) is a ring
   homomorphism, then there is a ring homomorphism \(\varphi^*:R[x]\to S[x]\)
   given by
   \begin{equation*}
   \varphi^*:r_0+r_1x+r_2x^2+\dots\mapsto\varphi(r_0)+\varphi(r_1)x+
   \varphi(r_2)x^2+\dots
   \end{equation*}
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If \(f:A\to R\) is a ring homomorphism, then its *kernel* is
   \begin{equation*}
   \ker f=\{a\in A:f(a)=0\}
   \end{equation*}
   and its *image* is 
   \begin{equation*}
   \im f=\{r\in R:\exists a\in R\e r=f(a)\}
   \end{equation*}
   #+END_definition

   The kernel of a group homomorphism is not merely a subgroup; it is a *normal*
   subgroup. Similarly, the kernel of a ring homomorphism is almost a subring
   (\(1\not\in\ker f\))
   and is closed under multiplication.

   [[index:ideal]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   An *ideal* in a commutative ring $R$ is a subset $I$ of $R$ s.t. 
   1. \(0\in I\)
   2. if \(a,b\in I\), then \(a+b\in I\)
   3. if \(a\in I\) and \(r\in R\), then \(ra\in I\)
   #+END_definition

   An ideal \(I\neq R\) is called a *proper ideal*

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   If \(b_1,\dots,b_n\in R\), then the set of all linear combinations
   \begin{equation*}
   I=\{r_1b_1+\dots+r_nb_n:r_i\in R\}
   \end{equation*}
   is an ideal in $R$. We write \(I=(b_1,\dots,b_n)\) in this case and we call
   $I$ the *ideal generated by* \(b_1,\dots,b_n\). In particular, if $n=1$, then
   \begin{equation*}
   I=(b)=\{rb:r\in R\}
   \end{equation*}
   is an ideal in $R$; $(b)$ consists of all the multiplies of $b$ and it is
   called the *principal ideal* generated by $b$. Notice that $R$ and $\{0\}$ are
   always principal ideals: \(R=(1),\{0\}=(0)\)
   #+END_examplle


   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If \(f:A\to R\) is a ring homomorphism, then \(\ker f\) is an ideal in $A$
   and \(\im f\) is a subring of $R$. Moreover, if $A$ and $R$ are not zero rings,
   then \(\ker f\) is a proper ideal.
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. If an ideal $I$ in a commutative ring $R$ contains 1, then $I=R$
   2. it follows from 1 that if $R$ is a field, then the only ideals are $\{0\}$
      and $R$
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   A ring homomorphism \(f:A \to R\) is an injection if and only if \(\ker f=\{0\}\)
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   label:ncor2.32
   If \(f:k\to R\) is a ring homomorphism, where $k$ is a field and $R$ is not the
   zero ring, then $f$ is an injection
   #+END_corollary

   #+BEGIN_proof
   the only proper ideal in $k$ is $\{0\}$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm3.54
   If $k$ is a field, then every ideal $I$ in $k[x]$ is a principal ideal.
   Moreover, if \(I\neq\{0\}\), there is a monic polynomial that generates $I$
   #+END_theorem

   #+BEGIN_proof
   If $k$ is a field, then $k[x]$ is an example of a *euclidean ring*. Follows
   Theorem ref:thm3.60
   #+END_proof

   [[index:principal ideal domain]]
   
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A domain $R$ is a *principal ideal domain* (PID) if every ideal in $R$ is a principal
   ideal. 
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. The ring of integers is a PID
   2. Every field is a PID
   3. If $k$ is a field, then the polynomial ring $k[x]$ is a PID
   4. There are rings other than \(\Z\) and \(k[x]\) where $k$ is a field that
      have a division algorithm; they are called *euclidean rings*.
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   Let \(R=\Z[x]\). The set of all polynomials with even constant term is an
   ideal in \(\Z[x]\). We show that \(I\) is not a principal ideal.

   Suppose there is \(d(x)\in\Z[x]\) with \(I=(d(x))\). The constant \(2\in I\),
   so that there is \(f(x)\in\Z[x]\) with \(2=d(x)f(x)\). We have
   \(0=\deg(2)=\deg(d)+\deg(f)\). The candidates for $d(x)$ are $\pm 1$ and
   $\pm2$. Suppose $d(x)=\pm2$; since \(x\in I\), there is \(g(x)\in\Z[x]\) with
   \(x=d(x)g(x)=\pm2g(x)\). But every coefficients on the right side is even.
   This contradiction gives \(d(x)=\pm1\). Hence $I=\Z[x]$, another
   contradiction. Therefore $I$ is not a principal ideal.
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   An element \delta in a commutative ring $R$ is a *greatest commmon divisor*, gcd, of
   elements \(\alpha,\beta \in R\) if
   1. \delta is a common divisor of \alpha and \beta 
   2. if \gamma is any common divisor of \alpha and \beta, then \(\gamma\mid\delta\)
   #+END_definition

   #+BEGIN_remark
   Let $R$ be a PID and let \(\pi,\alpha\in R\) with \pi irreducible. A gcd \delta of
   \pi and \alpha is a divisor of \pi. Hence \(\pi=\delta\epsilon\). And
   irreducibility of \pi forces either \delta or \epsilon to be a unit. Now
   \(\alpha=\delta\beta\). If \delta is not a unit, then \epsilon is a unit and so 
   \begin{equation*}
   \alpha=\delta\beta=\pi\epsilon^{-1}\beta
   \end{equation*}
   that is \(\pi\mid\alpha\). We conclude that if \(\pi\nmid\alpha\) then \delta is a
   unit; that is 1 is a gcd of \pi and \alpha
   #+END_remark

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm3.57
   Let $R$ be a PID
   1. Every \(\alpha,\beta\in R\) has a gcd, \delta, which is a linear combination of
      \alpha and \beta
      \begin{equation*}
      \delta=\sigma\alpha+\tau\beta
      \end{equation*}
   2. If an irreducible element \(\pi\in R\) divides a product \(\alpha\beta\), then
      either \(\pi\mid\alpha\) or \(\pi\mid\beta\)
   #+END_theorem
   #+BEGIN_proof
   1. We may assume that at least one of \alpha and \beta is not zero. Consider the set $I$
      of all the linear combinations
      \begin{equation*}
      I=\{\sigma\alpha+\tau\beta:\sigma,\tau\in R\}
      \end{equation*}
      $I$ is an ideal and so there is \(\delta\in I\) with \(I=(\delta)\); we claim
      that \delta is gcd of \alpha and \beta. Note that \(\alpha,\beta,\delta \in I\)
   2. If \(\pi\nmid\alpha\), then the remark says that 1 is a gcd of \pi and \alpha.
      Thus \(1=\sigma\pi+\tau\alpha\) and so
      \begin{equation*}
      \beta=\sigma\pi\beta+\tau\alpha\beta
      \end{equation*}
      Since \(\pi\mid\alpha\beta\), it follows that \(\pi\mid\beta\)
   #+END_proof
   
   [[index:least common multiple]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $f$ and $g$ are elements in a commutative ring $R$, then a *common multiple*
   is an element \(m\in R\) with \(f\mid m\) and \(g\mid m\). If $f$ and $g$ in
   $R$ are not both 0, define their *least common multiple*, abbreviated lcm.
   #+END_definition

   #+BEGIN_exercise
   label:ex3.34
   If $k$ is a field, prove that \(\sqrt{1-x^2}\not\in k(x)\), where \(k(x)\) is
   the field of rational functions
   #+END_exercise

   #+BEGIN_proof
   If \(\sqrt{1-x^2}\in k(x)\), then \(1-x^2=p^2(x)/q^2(x)\) and 
   \(q^2(x)=p^2(x)+x^2q^2(x)\). However \(\deg(x^2q^2(x)>\deg(q^2(x)))\)
   #+END_proof

   #+BEGIN_exercise
   label:ex3.45
   1. Show that every element \(a\in\I_p\) has a \(p\)th root
   2. Let $k$ be a field that contains \(\I_p\) as a subfield. For every
      positive integer $n$, show that the function
      \(\varphi_n:k\to k\), given by \(\varphi(a)=a^{p^n}\) is a ring homomorphism
   #+END_exercise

   #+BEGIN_proof
   1. \(a^p\equiv a\mod p\)
   2. [[https://math.stackexchange.com/questions/565135/finite-fields-existence-of-field-of-order-pn-proof-help][stackexchange]]
   #+END_proof

   #+BEGIN_exercise
   label:ex3.47
   1. If $A$ and $R$ are domains and \(\varphi:A\to R\) is a ring homomorphism,
      prove that 
      \begin{equation*}
      [a,b]\to [\varphi(a),\varphi(b)]
      \end{equation*}
      is a ring homomorphism \(\Frac(A)\to\Frac(R)\)
   2. Prove that if a field $k$ contains an isomorphic copy of \(\Z\) as a
      subring, then $k$ must contain an isomorphic copy of \(\Q\)
   3. Let $R$ be a domain and let \(\varphi:R\to k\) be an injective ring
      homomorphism, where $k$ is a field. Prove that there exists a unique ring
      homomorphism \(\Phi:\Frac(R)\to k\) extending \varphi ; that is, \(\Phi|R=\varphi\)
   #+END_exercise

   #+BEGIN_proof
   1. 
      \begin{align*}
      f([1,1])&=[1,1]\\
      f([a,b]+[c,d])&=f([ad+bc,bd])=[\varphi(ad+bc),\varphi(bd)]\\
      &=[\varphi(a)\varphi(d)+\varphi(b)\varphi(c),\varphi(b)\varphi(d)]\\
      &=[\varphi(a),\varphi(b)]+[\varphi(c),\varphi(d)]\\
      &=f([a,b])+f([c,d])\\
      f([a,b][c,d])&=f([ac,bd])=[\varphi(ac),\varphi(bd)]=[\varphi(a)\varphi(c),
      \varphi(b)\varphi(d)]\\&=f([a,b])f([c,d])
      \end{align*}
   2. Suppose \(k'\le k\) and \(k'\cong\Z\), then \(\Frac(k')\cong\Frac(\Z)\).
      Obviously.
   3. $k$ is a field and has inverse.
   #+END_proof
   
** Euclidean Rings
   [[index:degree function]]
   [[index:euclidean ring]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A *euclidean ring* is a domain that is equipped with a function
   \begin{equation*}
   \partial:R-\{0\}\to\N
   \end{equation*}
   called a *degree function*, s.t.
   1. \(\partial(f)\le\partial(fg)\) for all \(f,g\in R\) with \(f,g\neq0\)
   2. for all \(f,g\in R\) with \(f\neq0\), there exists \(q,r\in R\) with
      \begin{equation*}
      g=qf+r
      \end{equation*}
      where either \(r=0\) or \(\partial(r)<\partial(f)\)
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. The integers \(\Z\) is a euclidean ring with the degree function
      \(\partial(m)=\abs{m}\). In \(\Z\) we have
      \begin{equation*}
      \partial(mn)=\abs{mn}=\abs{m}\abs{n}=\partial(m)\partial(n)
      \end{equation*}
   2. when $k$ is a field, the domain $k[x]$ is a euclidean ring with degree
      function the usual degree of a nonzero polynomial. In $k[x]$, we have
      \begin{align*}
      \partial(fg)=\deg(fg)=\deg(f)+\deg(g)=\partial(f)+\partial(g)
      \end{align*}
      If a degree function is multiplicative, then \partial is called a *norm*
   3. The Gaussian integers \(\Z[i]\) form a euclidean ring whose degree
      function 
      \begin{equation*}
      \partial(a+bi)=a^2+b^2
      \end{equation*}
      is a norm. One reason to show that \(\Z[i]\) is a euclidean ring is that
      it is a PID, and hence it has unique factorization of its elements of into
      products of irreducibles.

      \(\partial\) is a multiplicative degree function for
      \begin{equation*}
      \partial(\alpha\beta)=\alpha\beta\overline{\alpha\beta}=\alpha\beta\overline{\alpha}
      \overline{\beta}=\alpha\overline{\alpha}\beta\overline{\beta}= \partial(\alpha)\partial(\beta)
      \end{equation*}

      Let us show that \partial satisfies the second desired property. Given
      \(\alpha,\beta\in\Z[i]\) with \(\beta\neq0\), regard \(\alpha/\beta\) as an
      element of \(\C\). Rationalizing the denominator gives
      \(\alpha/\beta=\alpha\overline{\beta}/\beta\overline{\beta}=\alpha\overline{\beta}/\partial{\beta}\), so that
      \begin{equation*}
      a/\beta=x+yi
      \end{equation*}
      where \(x,y\in\Q\). Write \(x=a+u\) and \(y=b+v\), where \(a,b\in\Z\)are
      integers closest to $x$ and $y$, respectively; thus
      \(\abs{u},\abs{v}\le1/2\). It follows that
      \begin{equation*}
      \alpha=\beta(a+bi)+\beta(u+vi)
      \end{equation*}
      Notice that \(\beta(u+vi)\in\Z[i]\). Finally we have
      \begin{equation*}
      \partial(\beta(u+vi))=\partial(\beta)\partial(u+vi)<\partial(\beta)
      \end{equation*}
      And so \(\Z[i]\) is a euclidean ring whose degree function is a norm

      Note that quotients and remainders are not unique because of the choice
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm3.60
   Every euclidean ring $R$ is a PID
   #+END_theorem

   #+BEGIN_proof
   Let $I$ be an ideal in $R$. If $I\neq\{0\}$, by the least integer axiom, the
   set of all degrees of nonzero elements in $I$ has a smallest element, say
   $n$; choose $d\in I$ with \(\partial(d)=n\). Clearly \((d)\subseteq I\). For any
   \(a\in I\), then there are \(q,r\in R\) with \(a=qd+r\), where either $r=0$
   or \(\partial(r)<\partial(a)\). But \(r=a-qd\in I\) and so $d$ having the least degree
   implies that $r=0$. Hence $a=qd\in(d)$.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   The ring of Gaussian integers $\Z[i]$ is a PID
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   An element $u$ in a domain $R$ is a *universal side divisor* if $u$ is not a unit
   and for every $x\in R$, either \(u\mid x\) or there is a unit \(z\in R\) with \(u\mid(x+z)\)
   #+END_definition


   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $R$ is a euclidean ring but not a field, then $R$ has a universal side divisor
   #+END_proposition

   #+BEGIN_proof
   Define 
   \begin{equation*}
   S=\{\partial(v):v\neq 0\text{ and }v\text{ is not a unit}\}
   \end{equation*}
   where \partial is the degree function on $R$. Since $R$ is not a field, $S$ is a
   nonempty subset of the natural number. By the least integer axiom, $S$ has a
   smallest element, say, $\partial(u)$. We claim that $u$ is a universal side
   divisor. If \(x\in R\), then there are \(q,r\) with \(x=qu+r\).
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop3.64
   1. Let $R$ be a euclidean ring $R$ that is not a field. If the degree function
      \partial is a norm, then \alpha is a unit if and only if \(\partial(\alpha)=1\)
   2. Let $R$ be a euclidean ring $R$ that is not a field. If the degree function
      \partial is a norm and if \(\partial(a)=p\), where $p$ is a prime, then \alpha is not
      irreducible
   3. The only units in the ring \(\Z[i]\) of Gaussian integers are \(\pm1\) and
      \(\pm i\)
   #+END_proposition

   #+BEGIN_proof
   1. Since \(1^2=1\), we have \(\partial(1)^2=\partial(1)\), so that \(\partial(1)=0\) or
      \(\partial(1)=1\). If \(\partial(1)=0\), then \(\partial(a)=\partial(1a)=0\). But $R$ is not a
      field, and so \partial is not identically zero. We conclude that \(\partial(1)=1\)

      If \(a\in R\) is a unit, then there is \(\beta\in R\) with
      \(\alpha\beta=1\). Therefore \(\partial(\alpha)\partial(\beta)=1\) and hence \(\partial(\alpha)=1\)

      For the converse, we begin by showing that there is no element \(\beta\in
      R\) with 
      \(\partial(\beta)=0\). If such an element exists, the division algorithms gives 
      \(1=q\beta+r\) and so \(\partial(r)=0\). That is \beta is a unit, then \(\partial(\beta)=1\), a
      contradiction

      Assume now that \(\partial(\alpha)=1\). The division algorithm gives
      \begin{equation*}
      \alpha=q\alpha^2+r
      \end{equation*}
      As \(\partial(\alpha^2)=\partial(\alpha)^2=1\), \(r=0\) or \(\partial(r)=0\), which would not occur.
      Hence $r=0$ and \(\alpha=q\alpha^2\). It follows that \(1=q\alpha\), and
      so \alpha is a unit
   2. If on the contrary, \(\alpha=\beta\gamma\), where neither \(\beta\) or
      \gamma is a unit, then \(p=\partial(\alpha)=\partial(\beta)\partial(\gamma)\). 
   3. If \(\alpha=a+bi\in\Z[i]\) is a unit, then \(1=\partial(\alpha)=a^2+b^2\).
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   If $p$ is a prime and \(p\equiv 1\mod 4\), then there is an integer $m$ with
   \begin{equation*}
    m^2\equiv -1\mod p
   \end{equation*}
   #+END_lemma
   #+BEGIN_proof
   If \(G=(\I_p)^\times\) is the multiplicative group of nonzero elements in
   \(\I_p\), then \(\abs{G}=p-1\equiv 0\mod 4\). By Proposition ref:prop2.78,
   $G$ contains a subgroup $S$ of order $4$. By Exercise ref:ex2.36 either $S$
   is cyclic or \(a^2=1\) for all \(a\in S\). Since \(\I_p\) is a field,
   however, it cannot contain four roots of the quadratic \(x^2-1\). Therefore,
   $S$ is cyclic, say \(S=\la[m]\ra\) where \([m]\) is the congruence class of
   $m$ mod $p$. Since \([m]\) has order 4, we have \([m^4]=[1],[m^2]\neq1\), and
   so \([m^2]=[-1]\) for \([-1]\)  is the unique element in \(S\) of order 2.
   Therefore, \(m^2\equiv -1\mod p\)
   #+END_proof

   #+ATTR_LATEX: :options [Fermat's Two-Squares Theorem]
   #+BEGIN_theorem
   label:thm3.66
   An odd prime $p$ is a sum of two squares,
   \begin{equation*}
   p=a^2+b^2
   \end{equation*}
   where $a$ and $b$ are integers if and only if \(p\equiv 1\mod 4\)
   #+END_theorem

   #+BEGIN_proof
   Assume that \(p=a^2+b^2\). Since $p$ is odd, $a$ and $b$ have different
   parity; say, $a$ is even and $b$ is odd. Hence \(a=2m\) and \(b=2n+1\) and
   \begin{equation*}
   p=a^2+b^2=4m^2+4n^2+4n+1\equiv1\mod 4
   \end{equation*}
   Conversely, assume that \(p\equiv1\mod4\). By the lemma, there is an integer
   $m$ s.t.
   \begin{equation*}
   p\mid(m^2+1)
   \end{equation*}
   In \(\Z[i]\), there is a factorization \(m^2+1=(m+i)(m-i)\) and so 
   \begin{equation*}
   p\mid(m+i)(m-i) \text{ in }\Z[i]
   \end{equation*}
   If \(p\mid(m\pm i)\) in \(\Z[i]\), then there are integers $u$ and $v$ with
   \(m\pm i=p(u+iv)\). Comparing the imaginary parts gives \(pv=1\), a
   contradiction. We conclude that $p$ does not satisfy the analog of Euclid's
   lemma in Theorem ref:thm3.57; it follows from Exercise ref:ex3.62 that  $p$
   is not irreducible. Hence there is a factorization
   \begin{equation*}
   p=\alpha\beta\in\Z[i]
   \end{equation*}
   Therefore, taking norms gives an equation in \(\Z\)
   \begin{align*}
   p^2&=\partial(p)=\partial(\alpha\beta)\\
   &=\partial(\alpha)\partial(\beta)=(a^2+b^2)(c^2+d^2)
   \end{align*}
   By Proposition ref:prop3.64, the only units in \(\Z[i]\) are \(\pm1\) and
   \(\pm i\), so that any nonzero Gaussian integers that is not a unit has a norm
   $>1$; therefore \(a^2+b^2\neq1\) and \(c^2+d^2\neq1\). Euclid's lemma now
   gives \(p\mid a^2+b^2\)  or \(p\mid c^2+d^2\); then fundamental theorem of arithmetic
   gives \(p=a^2+b^2\).
   
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   [[label:lemma3.67]]
   If \(\alpha\in\Z[i]\) is irreducible, then there is a unique prime number $p$
   with \(a\mid p\) in \(\Z[i]\)
   #+END_lemma

   #+BEGIN_proof
   Since \(\partial(\alpha)=\alpha\overline{\alpha}\), we have \(\alpha\mid\partial(\alpha)\). Now
   \(\partial(\alpha)=p_1\dots p_n\). If \(\alpha\mid q\) for some prime \(q\neq p_i\), then 
   \(\alpha\mid(q,p_i)=1\), forcing \alpha to be unit. A contradiction
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let \(\alpha=a+bi\in\Z[i]\) be neither 0 nor a unit. Then \alpha is irreducible if
   and only if
   1. \alpha is an associate of a prime $p$ in \(\Z\) of the form \(p=4m+3\); or
   2. \alpha is an associate of $1+i$ or its conjugate; or
   3. \(\partial(\alpha)=a^2+b^2\) is a prime in \(\Z\) of the form \(4m+1\)
   #+END_proposition

   #+BEGIN_proof
   By Lemma ref:lemma3.67 there is a unique prime number $p$ divides by \alpha in
   \(\Z[i]\). Since \(\alpha\mid p\), we have \(\partial(\alpha)\mid\partial(p)=p^2\) in
   \(\Z\), so that \(\partial(\alpha)=p\) or \(\partial(\alpha)=p^2\).
   1. \(p\equiv 3\mod4\)

      By Theorem ref:thm3.66 \(p^2=a^2+b^2\). We have \(\alpha\beta=p\) and
      \(\partial(\alpha)\partial(\beta)=\partial(p)\). Therefore, \(p^2\partial(\beta)=p^2\) and
      \(\partial(\beta)=1\). Thus \beta is a unit by Proposition ref:prop3.64 and $p$ is
      irreducible.
   2. \(p\equiv 2\mod 4\)

      \(a^2+b^2=2\)
   3. \(p\equiv1\mod4\)
      If \(\partial(\alpha)=p^2\), \beta is a unit as case 1. Now 
      \(\alpha\overline{\alpha}=p^2=(\alpha\beta)^2\), so that
      \(\overline{\alpha}=\alpha\beta^2\) but \(\beta^2=\pm1\) by Proposition ref:prop3.64

   
   #+END_proof

   #+BEGIN_exercise
   label:ex3.62
   If $R$ is a euclidean ring and \(\pi\in R\) is irreducible, prove that 
   \(\pi\mid\alpha\beta\) implies \(\pi\mid\alpha\) or \(\pi\mid\beta\)
   #+END_exercise


   #+BEGIN_proof
   $R$ is PID and follow Theorem ref:thm3.57.
   #+END_proof
** Linear Algebra
*** Vector Spaces
    [[index:vector space]] [[index:vector]]
    
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If $k$ is a field, then a *vector space over* $k$ is an (additive) abelian
    group $V$ equipped with a *scalar multiplication*; there is a function 
    \(k\times V\to V\), denoted by \((a,v)\mapsto av\) s.t. for all \(a,b,1\in
    k\) and all 
    \(u,v\in V\)
    1. \(a(u+v)=au+av\)
    2. \((a+b)v=av+bv\)
    3. \((abv)=a(bv)\)
    4. \(1v=v\)


    The elements of \(V\) are called *vectors* and the elements of $k$ are called *scalars*
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    1. Euclidean space \(V=\R^n\) is a vector space over $\R$
    2. If $R$ is a commutative ring and $k$ is a subring that is a field, then
       $R$ is a vector space over $k$

       For example, if $k$ is a field, then the polynomial ring \(R=k[x]\) is a
       vector space over $k$.
    #+END_examplle

    [[index:subspace]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If $V$ is a vector space over a field $k$, then a *subspace* of $V$ is a
    subset $U$ of $V$ s.t.
    1. \(0\in U\)
    2. \(u,u'\in U\) imply \(u+u'\in U\)
    3. \(u\in U\) and \(a\in k\) imply \(au\in U\)
    #+END_definition

    
    [[index:k-linear combination]] [[index:span]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    Let $V$ be a vector space over a field $k$. A *\(k\)-linear combination* of a
    list \(v_1,\dots,v_n\) in $V$ is a vector of $v$ of the form
    \begin{equation*}
    v=a_1v_1+\dots+a_nv_n
    \end{equation*}
    where \(a_i\in k\) for all \(i\)
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If \(X=v_1,\dots,v_m\) is a list in a vector space $V$, then
    \begin{equation*}
    \la v_1,\dots,v_m\ra
    \end{equation*}
    the set of all the \(k\)-linear combinations of \(v_1,\dots,v_m\) is called
    the *subspace spanned by \(X\)*. We also say that \(v_1,\dots,v_m\) *spans*
    \(\la v_1,\dots,v_m\ra\)
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_lemma
    Let $V$ be a vector space over a field $k$
    1. Every intersection of subspaces of $V$ is itself a subspace
    2. If \(X=v_1,\dots,v_m\) is a list in $V$, then the intersection of all the
       subspaces of $V$ containing $X$ is \(\la v_1,\dots,v_m\ra\), and so 
       \(\la v_1,\dots,v_m\ra\) is the *smallest subspace*
    #+END_lemma

    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    Let \(V=\R^2\), let \(e_1=(1,0)\) and let \(e_2=(0,1)\). then \(V=\la e_1,e_2\ra\)
    #+END_examplle

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    A vector space $V$ is called *finite-dimensional* if it is spanned by a finite
    list; otherwise $V$ is called *infinite-dimensional*
    #+END_definition

    *Notation*. If \(v_1,\dots,v_m\) is a list, then
    \(v_1,\dots,\hatv_i,\dots,v_m\) is the shorter list with \(v_i\) deleted

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    label:prop3.73
    If $V$ is a vector space, then the following conditions on a list
    \(X=v_1,\dots,v_m\) spanning $V$ are equivalent
    1. $X$ is not a shortest spanning list
    2. some \(v_i\) is in the subspace spanned by the others; that is
       \begin{equation*}
       v_i\in\la v_1,\dots,\hatv_i,\dots,v_m\ra
       \end{equation*}
    3. there are scalars \(a_1,\dots,a_m\) not all zero with
       \begin{equation*}
       \displaystyle\sum_{l=1}^ma_lv_l=0
       \end{equation*}
    #+END_proposition
    

    [[index:linearly dependent]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    A list \(X=v_1,\dots,v_m\) in a vector space $V$ is *linearly dependent* if
    there are scalars \(a_1,\dots,a_m\) not all zero, with
    \(\sum_{l=1}^ma_lv_l=0\); otherwise $X$ is called *linearly independent*
    #+END_definition


    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    If \(X=v_1,\dots,v_m\) is a list spanning a vector space $V$, then $X$ is a
    shortest spanning list if and only if $X$ is linearly independent
    #+END_corollary


    [[index:basis]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    A *basis* of a vector space $V$ is a linearly independent list that spans $V$
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    Let \(X=v_1,\dots,v_n\) be a list in a vector space $V$ over a field $k$.
    Then $X$ is a basis if and only if each vector in $V$ has a unique
    expression as a \(k\)-linear combination of vectors in $X$
    #+END_proposition

    #+BEGIN_proof
    If a vector \(v=\sum a_iv_i=\sum b_iv_i\), then \(\sum(a_i-b_i)=0\)
    #+END_proof

    [[index:coordinate set]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If \(X=v_1,\dots,v_n\) is a basis of a vector space $V$ and if \(v\in V\),
    then there are unique scalars \(a_1,\dots,a_n\) with
    \(v=\sum_{i=1}^na_iv_i\). The \(n\)-tuple \((a_1,\dots,a_n)\) is called the
    *coordinate set* of a vector \(v\in V\) relative to the basis $X$
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_theorem
    Every finite-dimensional vector space $V$ has a basis
    #+END_theorem

    #+BEGIN_proof
    A finite spanning list \(X\) exists, since $V$ is finite-dimensional. If it
    is  linearly independent, it is a basis; if not, $X$ can be shortened to a
    spanning list \(X'\) by Proposition ref:prop3.73
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_lemma
    Let \(u_1,\dots,u_n\) be elements in a vector space $V$, and let
    \(v_1,\dots,v_m\in\la u_1,\dots,u_n\ra\). If \(m>n\), then \(v_1,\dots,v_m\)
    is a linearly dependent list
    #+END_lemma

    #+BEGIN_proof
    Induction on \(n\ge 1\)
    
    /Base step/. If \(n=1\)
    
    /Inductive step/. For \(i=1,\dots,m\)
    \begin{equation*}
    v_i=a_{i1}u_1+\dots+a_{in}u_n
    \end{equation*}
    We may assume that some \(a_{i1}\neq0\) otherwise 
    \(v_1,\dots,v_m\in\la u_2,\dots,u_n\ra\) , and the inductive hypothesis
    applies. Changing notation 
    if necessary we may assume \(a_{11}\neq 0\). For each \(i\ge 2\), define
    \begin{equation*}
    v_i'=v_i-a_{i1}a_{11}^{-1}v_1\in\la u_2,\dots,u_n\ra
    \end{equation*}

    Since \(m-1>n-1\)
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    A homogeneous system of linear equations, over a field $k$, with more
    unknowns than equations has a nontrivial solution.
    #+END_corollary
    
    #+BEGIN_proof
    An \(n\)-tuple \((\beta_1,\dots,\beta_n)\) is a solution of a system
    \begin{gather*}
    \alpha_{11}x_1+\dots+\alpha_{1n}x_n=0\\
    \vdots\quad\vdots\quad\vdots\\
    \alpha_{m1}x_1+\dots+\alpha_{mn}x_n=0
    \end{gather*}
    if \(\alpha_{i1}\beta_1+\dots+\alpha_{in}\beta_n=0\) for all $i$. In other
    words, if \(c_1,\dots,c_n\) are the columns of the \(m\times n\) coefficient
    matrix \(A=[\alpha_{ij}]\), then 
    \begin{equation*}
    \beta_1c_1+\dots+\beta_nc_n=0
    \end{equation*}
    Note that \(c_i\in k^m\). Now \(k^m\) can be spanned by \(m\) vectors. Since
    \(n>m\), \(c_1,\dots,c_n\) is linearly dependent
    #+END_proof

    #+ATTR_LATEX: :options [Invariance of Dimension]
    #+BEGIN_theorem
    If \(X=x_1,\dots,x_n\) and \(Y=y_1,\dots,y_m\) are bases of a vector space
    $V$, then \(m=n\)
    #+END_theorem

    #+BEGIN_proof
    Otherwise \(n<m\) or \(m<n\)
    #+END_proof

    [[index:dimension]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If $V$ is a finite-dimensional vector space over a field $k$, then its
    *dimension* denoted by \(\dim_k(V)\) or \(\dim(V)\), is the number of elements
    in a basis of $V$
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    Let \(X=\{x_1,\dots,x_n\}\) be a finite set. Define
    \begin{equation*}
    k^X=\{\text{functions }f:X\to k\}
    \end{equation*}
    Now \(k^X\) is a vector space if we define addition
    \begin{equation*}
    f+f':x\mapsto f(x)+f'(x)
    \end{equation*}
    and scalar multiplication for \(a\in k\)
    \begin{equation*}
    af:x\mapsto af(x)
    \end{equation*}

    It's easy to check that the set of \(n\) functions of the form \(f_x\),
    where \(x\in X\) defined by
    \begin{equation*}
    f_x(y)=
    \begin{cases}
    1&\text{if }y=x\\
    0
    \end{cases}
    \end{equation*}
    form a basis.

    An \(n\)-tuple \((a_1,\dots,a_n)\) is really a function \(f:\{1,\dots,n\}\to
    k\) with \(f(i)=a_i\)
    #+END_examplle

    #+ATTR_LATEX: :options []
    #+BEGIN_lemma
    label:lemma3.73
    If \(X=v_1,\dots,v_n\) is a linearly dependent list of vectors in a vector
    space $V$, then there exists \(v_r\) with \(r\ge1\) with \(v_r\in\la v_1,\dots,v_{r-1}\ra\)
    #+END_lemma

    #+ATTR_LATEX: :options [Exchange Lemma]
    #+BEGIN_lemma
    label:lemma3.74
    If \(X=x_1,\dots,x_m\) is a basis of a vector space $V$ and
    \(y_1,\dots,y_n\) is a linearly independent subset of $V$, then \(n\le m\)
    #+END_lemma

    #+BEGIN_proof
    We begin by showing that one of the \(x\)'s in \(X\) can be replaced by
    \(y_n\) so that the new list still spans \(V\). Now \(y_n\in\la X\ra\), so
    that the list
    \begin{equation*}
    y_n,x_1,\dots,x_m
    \end{equation*}
    is linearly dependent. By Lemma ref:lemma3.73 there is some \(i\) with 
    \(x_i=ay_n+\sum_{j<i}a_jx_j\). Throwing out \(x_i\) and replacing it by
    \(y_n\) gives a spanning list 
    \begin{equation*}
    X'=y_n,x_1,\dots,\hatx_i,\dots,x_m
    \end{equation*}

    Now repeat this argument for the spanning list 
    \(y_{n-1},y_n,x_1,\dots,\hatx_i,\dots,x_m\). It follows that the
    disposable vector must be one of the remaining \(x\)'s, say \(x_l\). After
    throwing out \(x_l\), we have a new spanning list \(X''\). If \(n>m\), then
    this procedure ends with a spanning list consisting of \(m\) \(y\)'s and no
    \(x\)'. Thus a proper sublist of \(Y=y_1,\dots,y_n\) spans $V$, a contradiction
    #+END_proof

    #+ATTR_LATEX: :options [Invariance of Dimension]
    #+BEGIN_theorem
    If \(X=x_1,\dots,x_n\) and \(Y=y_1,\dots,y_m\) are bases of a vector space
    $V$, then \(m=n\)
    #+END_theorem

    #+BEGIN_proof
    By Lemma ref:lemma3.74, \(n\le m\) and \(m\le n\)
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    A *longest* (or a *maximal*) linearly independent list \(u_1,\dots,u_m\) is a
    linearly independent list for which there is no vector \(v\in V\) s.t. 
    \(u_1,\dots,u_m,v\) is linearly independent
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_lemma
    If $V$ is a finite-dimensional vector space, then a longest linearly
    independent list \(v_1,\dots,v_n\) is a basis of $V$
    #+END_lemma

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    Let \(Z=u_1,\dots,u_m\) be a linearly independent list in an
    \(n\)-dimensional vector space $V$. Then \(Z\) can be extended to a basis
    #+END_proposition

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    If \(\dim(V)=n\), then any list of \(n+1\) or more vectors is linearly dependent
    #+END_corollary

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    Let $V$ be a vector space with \(\dim(V)=n\)
    1. A list of \(n\) vectors that spans $V$ must be linearly independent
    2. Any linearly independent list of \(n\) vectors must span $V$
    #+END_corollary

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    Let $U$ be a subspace of a vector space $V$ of dimension $n$
    1. $U$ is finite-dimensional and \(\dim(U)\le\dim(V)\)
    2. If \(\dim(U)=\dim(V)\), then \(U=V\)
    #+END_corollary

    #+BEGIN_exercise
    label:nex2.77
    If \(V\) is a finite-dimensional vector space and \(U\) is a subspace, prove
    that
    \begin{equation*}
    \dim(U)+\dim(V/U)=\dim(V)
    \end{equation*}
    #+END_exercise

    #+BEGIN_proof
    If \(v_1+U,\dots,v_r+U\) is a basis of \(V/U\), then \(v_1,\dots,v_r\) are linearly independent.
    #+END_proof
*** Linear Tranformations
    [[index:linear transformation]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If \(V\) and $W$ are vector spaces over a field $k$, then a function
    \(T:V\to W\) is a *linear transformation* if for all vectors \(u,v\in V\), and
    all scalars \(a\in k\)
    1. \(T(u+v)=T(u)+T(v)\)
    2. \(T(av)=aT(v)\)
    #+END_definition

    [[index:singular]]
    We say that a linear transformation $T$ is *nonsingular* (or is an
    *isomorphism*) if $T$ is a bijection.
    
    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    1. If \theta is an angle, then the rotation about the origin by \theta is a linear
       transformation \(R_\theta:\R^2\to \R^2\)
    2. If \(V\) and $W$ are vector spaces over a field $k$, write
       \(\Hom_k(V,W)\) for the set of all linear transformations \(V\to W\).
       It's a vector space
    
    #+END_examplle

    [[index:general linear group]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If $V$ is a vector space over a field $k$, then the *general linear group*,
    denoted by \(\GL(V)\), is the set of all nonsingular linear transformations 
    \(V\to V\)
    #+END_definition

    A composite $ST$ of linear transformation $S$ and $T$ is again a linear
    transformation 

    #+ATTR_LATEX: :options []
    #+BEGIN_theorem
    label:thm3.92
    Let \(v_1,\dots,v_n\) be a basis of a vector space $V$ over a field $k$. If
    $W$ is a vector space over $k$ and \(u_1,\dots,u_n\) is a list in $W$, then
    there exists a unique linear transformation \(T:V\to W\) with \(T(v_i)=u_i\)
    for all $i$
    #+END_theorem

    #+BEGIN_proof
    Each \(v\in V\) has a unique expression of the form \(v=\sum_ia_iv_i\) and
    so \(T:V\to W\) given by \(T(v)=\sum a_iu_i\) is a well-defined function

    To prove the uniqueness of $T$, assume that \(S:V\to W\) is a linear
    transformation with 
    \begin{equation*}
    S(v_i)=u_i=T(v_i)
    \end{equation*}
    Then 
    \begin{align*}
    S(v)&=S(\sum a_iv_i)=\sum S(a_iv_i)\\
    &=\sum a_iS(v_i)=\sum a_iT(v_i)=T(v)
    \end{align*}
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    If two linear transformations \(S,T:V\to W\) agree on a basis, then \(S=T\)
    #+END_corollary

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    If \(T:k^n\to k^m\) is a linear transformation, then there exists an
    \(m\times n\) matrix $A$ s.t.
    \begin{equation*}
    T(y)=Ay
    \end{equation*}
    for all \(y\in k^n\) (here \(y\) is an \(n\times 1\) column matrix) 
    #+END_proposition

    #+BEGIN_proof
    If \(e_1,\dots,e_n\) is the standard basis of \(k^n\) and
    \(e_1',\dots,e_m'\) is the standard basis of \(k^m\), define \(A=[a_{ij}]\)
    to be the matrix whose \(j\)th column is the coordinate set of \(T(e_j)\).
    If \(S:k^n\to k^m\) is defined by \(S(y)=Ay\), then \(S=T\) since they agree
    on a basis: \(T(e_j)=\sum_ia_{ij}e_i'=Ae_j\)
    #+END_proof

    [[index:matrix]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    Let \(X=v_1,\dots,v_n\) be a basis of $V$ and let \(Y=w_1,\dots,w_m\) be a
    basis of $W$. If \(T:V\to W\) is a linear transformation, then the *matrix
    of $T$* is the \(m\times n\) matrix \(A=[a_{ij}]\), whose \(j\)th column
    \(a_{1j},a_{2j},\dots,a_{mj}\) is the coordinate set of \(T(v_j)\)
    determined by \(w\)'s: \(T(v_j)=\sum_{i=1}^ma_{ij}w_j\). The matrix $A$ does
    depend on the choice of bases $X$ and $Y$: we will write
    \begin{equation*}
    A={}_Y[T]_X
    \end{equation*}

    In case \(V=W\), we often let the basis \(X=v_1,\dots,v_n\) and
    \(w_1,\dots,w_m\) coincide. If \(1_V:V\to V\), given by \(v\mapsto v\) is
    the identity linear transformation, then \({}_X[1_V]_X\) is the
    \(n\times n\)  *identity matrix $I_n$*, defined by
    \begin{equation*}
    I=[\delta_{ij}]
    \end{equation*}
    where \(\delta_{ij}\) is the Kronecker delta. A matrix is *nonsingular* if it
    has inverse.
    #+END_definition
    
    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    Let \(T:V\to W\) be a linear transformation, and let \(X=v_1,\dots,v_n\) and
    \(Y=w_1,\dots,w_n\) be bases of $V$ and $W$ ,respectively. The matrix for
    $T$ is set up from the equation
    \begin{equation*}
    T(v_j)=a_{1j}w_1+\dots+a_{mj}w_m
    \end{equation*}
    #+END_examplle

    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    1. Let \(T:\R^2\to \R^2\) be rotation by \(\ang{90}\). The matrix of $T$
       related to the standard basis \(X=(1,0),(0,1)\) is 
       #+ATTR_LATEX: :mode math :environment bmatrix :math-prefix {}_X[T]_X=
       | 0 | -1 |
       | 1 | 0  |

       However if \(Y=(0,1)(1,0)\), then
       #+ATTR_LATEX: :mode math :environment bmatrix :math-prefix {}_Y[T]_Y=
       |  0 | 1 |
       | -1 | 0 |
    2. Let $k$ be a field, let \(T:V\to V\) be a linear transformation on a
       two-dimensional vector space, and assume that there is some vector
       \(v\in V\) with \(T(v)\) not a scalar multiple of \(v\). The assumption on
       $v$ says that the list \(X=v,T(v)\) is linearly independent, and hence
       it's a basis of $V$. Write \(v_1=v,v_2=Tv\).

       We compute \({}_X[T]_X\)
       \begin{equation*}
       T(v_1)=v_2\quad\text{ and }\quad T(v_2)=av_1+bv_2
       \end{equation*}
       for some \(a,b\in k\). We conclude that
       #+ATTR_LATEX: :mode math :environment bmatrix :math-prefix {}_X[T]_X=
       | 0 | a |
       | 1 | b |
    #+END_examplle

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    label:prop3.97
    Let \(V\) and \(W\) be a vector spaces over a field $k$, and let
    \(X=v_1,\dots,v_n\) and \(Y=w_1,\dots,w_m\) be bases of \(V\) and \(W\),
    respectively. If \(\Hom_k(V,W)\) denotes the set of all linear
    transformations \(T:V\to W\) and \(\Mat_{m\times n}k\) denotes the set of
    all \(m\times n\) matrices with entries in $k$, then the function 
    \(T\mapsto{}_Y[T]_X\) is a bijection \(\Hom_k(V,W)\to\Mat_{m\times n}(k)\) 
    #+END_proposition

    #+BEGIN_proof
    Given a matrix \(A\), its columns define vectors in \(W\); in more detail,
    if the \(j\)th column of $A$ is \(a_{1j},\dots,a_{mj}\), define
    \(z_j=\sum_{i=1}^ma_{ij}w_i\). By Theorem ref:thm3.92, there exists a linear
    transformation \(T:V\to W\) with \(T(v_j)=z_j\) and \({}_Y[T]_X=A\).
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    label:prop3.98
    Let \(T:V\to W\) and \(S:W\to U\) be linear transformations. Choose bases 
    \(X=x_1,\dots,x_n\) of \(V\), \(Y=y_1,\dots,y_m\) of \(W\), and
    \(Z=z_1,\dots,z_l\) of $U$, then
    \begin{equation*}
    {}_Z[S\circ T]_X=({}_Z[S]_Y)({}_Y[T]_X)
    \end{equation*}
    #+END_proposition

    #+BEGIN_proof
    Let \({}_Y[T]_X=[a_{ij}]\), so that \(T(x_j)=\sum_pa_{pj}y_p\), and let
    \({}_Z[S]_Y=[b_{qp}]\), so that \(S(y_p)=\sum_qb_{qp}z_q\). Then
    \begin{align*}
    ST(x_j)=S(T(x_j))&=S(\displaystyle\sum_{p}a_{pj}y_p)\\
    &=\displaystyle\sum_{p}a_{pj}S(y_p)=\displaystyle\sum_{p}
    \displaystyle\sum_qa_{pj}b_{qp}z_q=\displaystyle\sum_{q}c_{qj}z_q
    \end{align*}
    where \(c_{qj}=\sum_{p}b_{qp}a_{pj}\). Therefore
    \begin{equation*}
    {}_Z[ST]_X=[c_{qj}]={}_Z[S]_Y{}_Y[T]_X
    \end{equation*}
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    Matrix multiplication is associative
    #+END_corollary

    #+BEGIN_proof
    Let $A$ be an \(m\times n\) matrix, let \(B\) be an \(n\times p\) matrix,
    and let \(C\) be a \(p\times q\) matrix. By Theorem ref:thm3.92, there are
    linear transformations
    \begin{equation*}
    k^q\xrightarrow{T}k^p\xrightarrow{S}k^n\xrightarrow{R}k^m
    \end{equation*}
    with \(C=[T],B=[S],A=[R]\)
    
    Then 
    \begin{equation*}
    [R\circ(S\circ T)]=[R][S\circ T]=[R]([S][T])=A(BC)
    \end{equation*}
    On the other hand
    \begin{equation*}
    [(R\circ S)\circ T]=[R\circ S][T]=([R][S])[T]=(AB)C
    \end{equation*}
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    label:cor3.100
    Let \(T:V\to W\) be a linear transformation of vector space \(V\) over a
    field $k$, and let \(X\) and \(Y\) be bases of \(V\) and \(W\),
    respectively. If \(T\) is nonsingular, then the matrix of \(T^{-1}\) is the
    inverse of the matrix of $T$
    \begin{equation*}
    {}_X[T^{-1}]_Y=({}_Y[T]_X)^{-1}
    \end{equation*}
    #+END_corollary
    
    #+BEGIN_proof
    \(I={}_Y[1_W]_Y={}_Y[T]_{XX}[T^{-1}]_Y\) and \(I={}_X[1_V]_X={}_X[T^{-1}]_{YY}[T]_X\)
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    label:cor3.101
    Let \(T:V\to V\) be a linear transformation on a vector space $V$ over a
    field $k$. If \(X\) and \(Y\) are bases of $V$, then there is a nonsingular
    matrix $P$ with entries in $k$ so that
    \begin{equation*}
    {}_Y[T]_Y=P(_X[T]_X)P^{-1}
    \end{equation*}
    Conversely, if \(B=PAP^{-1}\), where \(B,A,P\) are \(n\times n\) matrices
    with entries in $k$ and $P$ is nonsingular, then there is a linear
    transformation \(T:k^n\to k^n\) and bases $X$ and $Y$ of \(k^n\) s.t.
    \(B=_Y[T]_Y,A=_X[T]_X\)
    #+END_corollary

    #+BEGIN_proof
    The first statement follows from Proposition ref:prop3.98 and associativity
    \begin{equation*}
    {}_Y[T]_Y=_Y[1_VT1_V]_Y=(_Y[1_V]_X)(_X[T]_X)(_X[1_V]_Y)
    \end{equation*}
    Set \(P=_Y[1_V]_X\)

    For the converse, let \(E=e_1,\dots,e_n\) be the standard basis of \(k^n\),
    and define \(T:k^n\to k^n\) be \(T(e_j)=Ae_j\). If follows that
    \(A=_E[T]_E\). Now define a basis \(Y=y_1,\dots,y_n\) by \(y_j=P^{-1}e_j\).
    $Y$ is a basis because $P^{-1}$ is nonsingular. It suffices to prove that 
    \(B=_Y[T]_Y\); that is \(T(y_j)=\sum_ib_{ij}y_i\), where \(B=[b_{ij}]\)
    \begin{align*}
    T(y_j)&=Ay_y=AP^{-1}e_j=P^{-1}Be_j\\
    &=P^{-1}\displaystyle\sum_{i}b_{ij}e_i
    =\displaystyle\sum_{i}b_{ij}P^{-1}e_i\\
    &=\displaystyle\sum_{i}b_{ij}y_i
    \end{align*}
    #+END_proof

    [[index:similar]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    Two \(n\times n\) matrices $B$  and \(A\) with entries in field $k$ are
    *similar* if there is a nonsingular matrix \(P\) with entries in $k$ with
    \(B=PAP^{-1}\) 
    #+END_definition



    Corollary ref:cor3.101 says the two matrices arise from the same linear
    transformation on a vector space $V$ if and only if they are similar

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If \(T:V\to W\) is a linear transformation, then the *kernel* (or the *null
    space*) of $T$ is
    \begin{equation*}
    \ker T=\{v\in V:T(v)=0\}
    \end{equation*}
    and the *image* of $T$ is
    \begin{equation*}
    \im T=\{w\in W:w=T(v)\text{ for some }v\in V\}
    \end{equation*}
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    Let \(T:V\to W\) be a linear transformation
    1. \(\ker T\) is a subspace of $V$ and \(\im T\) is a subspace of $W$
    2. $T$ is injective if and only if \(\ker T=\{0\}\)
    #+END_proposition

    #+ATTR_LATEX: :options []
    #+BEGIN_lemma
    label:lemma3.103
    Let \(T:V\to W\) be a linear transformation
    1. If $T$ is nonsingular, then for every basis \(X=v_1,\dots,v_n\) of $V$,
       we have \(T(X)=T(v_1),\dots,T(v_n)\) a basis of $W$
    2. Conversely, if there exists some basis \(X=v_1,\dots,v_n\) of $V$ for
       which \(T(X)\) is a basis of $W$, then $T$ is nonsingular
    #+END_lemma
    #+BEGIN_proof
    1. If \(\sum c_iT(v_i)=0\), then \(T(\sum c_iv_i)=0\) and so 
       \(\sum c_iv_i\in\ker T=\{0\}\). Hence each \(c_i=0\) because $X$ is linearly
       independent. If \(w\in W\), then the surjectivity of $T$ provides \(v\in V\)
       with \(w=T(v)\). But \(v=\sum a_iv_i\), and so 
       \(w=T(v)=T(\sum a_iv_i)=\sum a_iT(v_i)\). Therefore $T(X)$ is a basis of
       $W$
    2. Let \(w\in W\). Since \(T(X)\) is a basis of $W$, we have 
       \(w=\sum c_iT(v_i)=T(\sum c_iv_i)\). Add so $T$ is surjective. If 
       \(\sum c_iv_i\in\ker T\), then \(\sum c_iT(v_i)=0\) and so linear
       independence gives all \(c_i=0\); hence \(\ker T=\{0\}\). Therefore $T$
       is nonsingular
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_theorem
    If $V$ is an \(n\)-dimensional vector space over a field $k$, then \(V\) is
    isomorphic to \(k^n\)
    #+END_theorem

    #+BEGIN_proof
    Choose a basis \(v_1,\dots,v_n\) of $V$. If \(e_1,\dots,e_n\) is the
    standard basis of \(k^n\), then Theorem ref:thm3.92 says that there is a
    linear transformation \(T:V\to k^n\) with \(T(v_i)=e_i\); by Lemma
    ref:lemma3.103 $T$ is nonsingular
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    Two finite-dimensional vector space $V$ and $W$ over a field $k$ are
    isomorphic if and only if \(\dim(V)=\dim(W)\)
    #+END_corollary

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    label:prop3.106
    Let $V$ be a finite-dimensional vector space with \(\dim(V)=n\), and let 
    \(T:V\to V\) be a linear transformation. The following statements are
    equivalent
    1. $T$ is an isomorphism
    2. $T$ is surjective
    3. $T$ is injective
    #+END_proposition

    #+BEGIN_proof
    \(2\to 3\). Let \(v_1,\dots,v_n\) be the basis of $V$. Since $T$ is
    surjective, there are vectors \(u_1,\dots,u_n\) with \(Tu_i=v_i\). We claim
    that \(u_1,\dots,u_n\) are linearly independent. To show that $T$ is
    injective, it suffices to show that \(\ker T=\{0\}\)

    \(3\to 1\). Let \(v_1,\dots,v_n\) be a basis of $V$. If \(c_1,\dots,c_n\)
    are scalars not all 0, then \(\sum c_iv_i\neq0\). Since $T$ is injective, it
    follows that \(\sum c_iT(v_i)\neq0\) and so \(Tv_1,\dots,Tv_n\) are linearly
    independent. Therefore Lemma ref:lemma3.103 shows that $T$ is an isomorphism

    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    If \(A\) and $B$ are \(n\times n\) matrices with \(AB=I\), then \(BA=I\).
    Therefore $A$ is nonsingular with inverse $B$
    #+END_corollary
    #+BEGIN_proof
    There are linear transformations \(T,S:k^n\to k^n\) with \([T]=A,[S]=B\),
    and \(AB=I\) gives
    \begin{equation*}
    [TS]=[T][S]=[1_{k^n}]
    \end{equation*}
    Since \(T\mapsto[T]\) is a bijection, by Proposition ref:prop3.97, it
    follows that \(TS=1_{k^n}\). Hence $T$ is a surjection and \(S\) is an
    injection by Proposition ref:prop1.47 But Proposition ref:prop3.106 says taht $T,S$ are both
    isomorphism, so that \(S=T^{-1}\) and \(TS=1_{k^n}=ST\)
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    The set of all nonsingular \(n\times n\) matrices with entries in $k$ is
    denoted by \(\GL(n,k)\)
    #+END_definition

    It's easy to prove that \(\GL(n,k)\) is a group

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    Let $V$ be an \(n\)-dimensional vector space over a field $k$, and let
    \(X=v_1,\dots,v_n\) be a basis of $V$. Then \(\mu:\GL(V)\to\GL(n,k)\) defined
    by \(T\mapsto[T]={}_X[T]_X\) is an isomorphism
    #+END_proposition

    #+BEGIN_proof
    By Proposition ref:prop3.97 the function \(\mu':T\mapsto[T]\) is a bijection
    \begin{equation*}
    \Hom_k(V,V)\to\Mat_n(k)
    \end{equation*}

    If \(T\in\GL(V)\), then \([T]\) is a nonsingular matrix by Corollary
    ref:cor3.100; that is, if \mu is the restriction of \(\mu'\), then 
    \(\mu:\GL(V)\to\GL(n,k)\) is an injective homomorphism.

    If \(A\in\GL(n,k)\), then \(A=[T]\) for some \(T:V\to V\). It suffices to
    show that \(T\) is an isomorphism; that is, \(T\in\GL(V)\). Since \([T]\) is
    a nonsingular matrix, there is a matrix $B$ with \([T]B=I\). Now \(B=[S]\)
    for some \(S:V\to V\) and 
    \begin{equation*}
    [TS]=[T][S]=I=[1_V]
    \end{equation*}
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    A linear transformation $T:V\to V$ is a *scalar transformation* if there is 
    \(c\in k\) with \(T(v)=cv\) for all \(v\in V\); that is \(T=c1_V\). A
    *scalar matrix* is a matrix of the form \(cI\)
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    1. The center of the group \(\GL(V)\) consists of all the nonsingular scalar
       transformations
    2. The center of the group \(\GL(n,k)\) consists of all the nonsingular
       scalar matrices
    #+END_corollary
** Quotient Rings and Finite Fields
   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm3.110
   If $I$ is an ideal in a commutative ring $R$, then the additive abelian group
   \(R/I\) can be made into a commutative ring in such a way that the natural
   map \(\pi:R\to R/I\) is a surjective ring homomorphism
   #+END_theorem

   #+BEGIN_proof
   Define multiplication on the additive abelian group \(R/I\) by
   \begin{equation*}
   (a+I)(b+I)=ab+I
   \end{equation*}
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   The commutative ring \(R/I\) constructed in Theorem ref:thm3.110 is called
   the *quotient ring* of $R$ modulo $I$
   #+END_definition


   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $I$ is an ideal in a commutative ring $R$, then there are a commutative
   ring $A$ and a ring homomorphism \(\pi:R\to A\) with \(I=\ker\pi\)
   #+END_corollary

   #+BEGIN_proof
   Natural map \(\pi:R\to R/I\)
   #+END_proof

   #+ATTR_LATEX: :options [First Isomorphism Theorem]
   #+BEGIN_theorem
   If \(f:R\to A\) is a homomorphism of rings, then \(\ker f\) is an ideal in
   $R$, \(\im f\) is a subring of $A$, and
   \begin{equation*}
   R/\ker f\cong\im f
   \end{equation*}
   #+END_theorem

   [[index:prime field]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $k$ is a field, the intersection of all the subfields of $k$ is called the
   *prime field* of $k$
   #+END_definition


   Every subfield of \(\C\) contains $\Q$ and so the prime field of \(\C\) and
   of $\R$ is $\Q$.


   *Notation*. From now on, we will denote \(\I_p\) by \(\F_p\) when we are
   regarding it as a field.

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $k$ is a field, then its prime field is isomorphic to $\Q$ or to $\F_p$
   for some prime $p$
   #+END_proposition

   #+BEGIN_proof
   Consider the ring homomorphism \(\chi:\Z\to k\) defined by \(\chi(n)=n\epsilon\),
   where we denote the *one* in $k$ by \epsilon. Since every ideal in \(\Z\) is
   principal, there is an integer $m$ with \(\ker\chi=(m)\). If \(m=0\), then
   \chi is an injection, and so there is an isomorphism copy of \(\Z\) that is a
   subring of $k$. By Exercise ref:ex3.47 , there is a field
   \(Q\cong\Frac(\Z)=\Q\). with \(\im\chi\subseteq Q\subseteq k\). Now \(Q\)
   is the prime ideal of $k$, for it is the subfield generated by by \epsilon. 
   . If \(m\neq0\), the first isomorphism theorem gives
   \(\I_m=\Z/(m)\cong\im\chi\subseteq k\). Since $k$ is a field, \(\im\chi\) is
   a domain, and so Proposition ref:prop3.6 gives $m$ prime. If we now write $p$
   instead of $m$, then \(\im\chi=\{0,\epsilon,2\epsilon,\dots,(p-1)\epsilon\}\) is a
   subfield of $k$ isomorphic to \(\F_p\)
   #+END_proof

   [[index:characteristic]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A field $k$ has *characteristic 0* if its prime field is isomorphic to \(\Q\);
   a field $k$ has *characteristic $p$* if its prime field is isomorphic to
   \(\F_p\) for some prime $p$
   #+END_definition

   The fields \(\Q,\R,\C\) have characteristic 0

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $k$ is a field of characteristic \(p>0\), then \(pa=0\) for all \(a\in k\)
   #+END_proposition

   #+BEGIN_proof
   \(p\cdot 1=0\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If \(k\) is a finite field, then \(\abs{k}=p^n\) for some prime $p$ and some \(n\ge1\)
   #+END_proposition

   #+BEGIN_proof
   The prime field \(P\) of $k$ cannot be the infinite field $\Q$, and so
   \(P\cong\F_p\) for some $p$. Now $k$ is a  vector space over $P$, and so it is
   a vector space over \(\F_p\). Clearly, $k$ is finite-dimensional, and if
   \(\dim_{\F_p}(k)=n\), then \(\abs{k}=p^n\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop3.116
   If $k$ is a field and \(I=(p(x))\), where \(p(x)\) is a nonzero polynomial in
   $k[x]$, then the following are equivalent: \(p(x)\) is irreducible;
   \(k[x]/I\) is a field; \(k[x]/I\) is a domain
   #+END_proposition

   #+BEGIN_proof
   Assume \(p(x)\) is irreducible. Note that \(I=(p(x))\) is a proper ideal so
   that the /one/ in \(k[x]/I\), namely, \(1+I\) is not zero. If 
   \(f(x)+I\in k[x]/I\) is nonzero, then \(f(x)\not\in I\). Since \(p(x)\) is
   irreducible, by Lemma ref:lemma3.36, $p$ and $f$ are relatively prime. By 
   Theorem ref:thm3.31, there are polynomials \(s\) and $t$ s.t. \(1=sp+tf\).
   Thus \(tf-1\in I\) and so \(1+I=tf+I=(t+I)(f+I)\). Therefore every nonzero
   element of \(k[x]/I\) has an inverse and so \(k[x]/I\) is a field.

   If \(k[x]/I\) is a domain. If \(p(x)\) is not an irreducible polynomial. Then 
   \(p(x)=g(x)f(x)\) with \(\deg(p)> \deg(g)\), \(\deg(p)>\deg(f)\). It follows
   that \(f+I\) and \(g+I\) is nonzero but \((f+I)(g+I)=p+I\) is zero,
   contradicting to the fact that \(k[x]/I\) is a domain.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop3.117
   label:nprop2.141
   Let $k$ be a field, let \(p(x)\in k[x]\) be a monic irreducible polynomial of
   degree $d$, let \(K=k[x]/I\), where \(I=(p(x))\), and let \(\beta=x+I\in K\)
   1. $K$ is a field and \(k'=\{a+I;a\in k\}\) is a subfield of $K$ isomorphic
      to $k$
   2. \beta is a root of \(p(x)\) in $K$
   3. if \(g(x)\in k[x]\) and \beta is a root of $g(x)$, then \(p(x)\mid g(x)\) in 
      $k[x]$
   4. \(p(x)\) is the unique monic irreducible polynomial in $k[x]$ having \beta as
      a root
   5. The list \(1,\beta,\beta^2,\dots,\beta^{d-1}\) is a basis of $K$ as a vector space
      over $k$, and so \(\dim_k(K)=d\)
   #+END_proposition

   #+BEGIN_proof
   2. [@2] Note that \beta is a root in $K$, suppose 
      \(p(x)=a_0+a_1x+\dots+a_{d-1}x^{d-1}+x^d\), hence
      \begin{align*}
      p(\beta)&=(a_0+I)+(a_1+I)\beta+\dots+(a_{d-1}+I)\beta^{d-1}+(1+I)\beta^{d}\\
      &=(a_0+I)+(a_1+I)(x+I)+\dots+(a_{d-1}+I)(x+I)^{d-1}+(1+I)(x+I)^d\\
      &=a_0+a_1x+\dots+x^d+I=p(x)+I=I
      \end{align*}
   3. If \(p(x)\nmid g(x)\), then \(p(x)\) and \(g(x)\) are relatively prime
      since $p(x)$ is irreducible. Hence \(1=p(x)s(x)+g(x)t(x)\). Since 
      \(k[x]\subseteq K[x]\), we may regard this equation in $K[x]$. Hence
      \(1=0\), a contradiction
   5. [@5] Every element of $K$ has the form \(f(x)+I\), where \(f(x)\in k[x]\).
      By the division algorithm, there are \(f(x)=g(x)p(x)+r(x)\) with either 
      \(r(x)=0\) or \(\deg(r)< d\). We know that \(r(\beta)=r+I\), hence 
      \(1,\beta,\dots,\beta^{d-1}\) spans $K$

      To prove uniqueness, suppose
      \begin{equation*}
      b_0+b_1\beta+\dots+b_{d-1}\beta^{d-1}=c_0+c_1\beta+\dots+c_{d-1}\beta^{d-1}
      \end{equation*}
      Define \(g(x)=\sum_{i=0}^{d-1}(c_i-b_i)x^i\); if \(g(x)=0\) we are done.
      Otherwise, then \(\deg(g)\) is defined and \(\deg(g)<d\). On the other
      hand, \beta is a root of $g(x)$ and hence \(p(x)\mid g(x)\), a contradiction
   #+END_proof

   [[index:extension]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $K$ is a field containing $k$ as a subfield, then $K$ is called a (field) 
   *extension* of $k$, and we write "$K/k$ is a field extension"

   An extension field $K$ of a field $k$ is a *finite extension* of $k$ if $K$ is
   a finite-dimensional vector space over $k$. The dimension of $K$, denoted by
   $[K:k]$, is called the *degree* of $K/k$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   The polynomial \(x^2+1\in\R[x]\) is irreducible, and so \(K=\R[x]/(x^2+1)\)
   is a field extension \(K/\R\) of degree 2. If \beta is a root of \(x^2+1\), then 
   \(\beta^2=-1\); moreover, every element of $K$ has a unique expression of the
   form \(a+b\beta\), where \(a,b\in\R\). Clearly this is another construction
   of \(\C\).

   Consider the evaluation map \(\varphi:\R[x]\to\C\) given by 
   \(\varphi:f(x)\mapsto f(i)\). First \varphi is surjective, for
   \(a+ib=\varphi(a+bx)\in\im\varphi\). Second,
   \(\ker\varphi=\{f(x\in\R[x]:f(i)=0)\}\). We know that
   \(x^2+1\in\ker\varphi\), so that \((x^2+1)\subseteq\ker\varphi\). For the
   reverse inclusion, take \(g(x)\in\ker\varphi\). Now $i$ is a root of $g(x)$,
   and so \(\gcd(g,x^2+1)\neq1\) in \(\C[x]\); therefore \(\gcd(g,x^2+1)\neq1\)
   in \(\R[x]\). Irreducibility of \(x^2+1\) in $\R[x]$ gives \(x^2+1\mid g(x)\)
   and so \(g(x)\in(x^2+1)\). Therefore \(\ker\varphi=(x^2+1)\). The first
   isomorphism theorem now gives \(\R[x]/(x^2+1)\cong\C\)
   #+END_examplle

   [[index:algebraic]]
   [[index:transcendental]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let \(K/k\) be a field extension. An element \(\alpha\in K\) is *algebraic*
   over $k$ if there is some nonzero polynomial \(f(x)\in k[x]\) having \alpha as
   root; otherwise \alpha is *transcendental* over $k$. An extension \(K/k\) is
   *algebraic* if every \(\alpha\in K\) is algebraic over $k$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If \(K/k\) is a finite field extension, then \(K/k\) is an algebraic
   extension.
   #+END_proposition

   #+BEGIN_proof
   \([K:k]=n<\infty\). Hence the list of \(1,\alpha,\dots,\alpha^n\) is dependent. Thus
   there are \(c_0,\dots,c_n\in k\), not all 0, with \(\sum c_i\alpha^i=0\). Thus
   the polynomial \(f(x)=\sum c_ix^i\) is not the zero polynomial, and \alpha is a root
   of \(f(x)\).
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If \(K/k\) is an extension and \(\alpha\in K\), then \(k(\alpha)\) is the
   intersection of all those subfields of $K$ that contain $k$ and \alpha; we call
   \(k(\alpha)\) the subfield of $K$ obtained by *adjoining* \alpha to $k$

   More generally, if $A$ is a (possibly infinite) subset of $K$, define $k(A)$
   to be the intersection of all subfields of $K$ contain \(k\cup A\)
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:thm3.120
   label:nthm2.144
   1. If \(K/k\) is an extension and \(\alpha\in K\) is algebraic over $k$, then
      there is a unique monic irreducible polynomial \(p(x)\in k[x]\) having
      \alpha as a root. Moreover, if \(I=(p(x))\), then \(k[x]/I\cong k(\alpha)\); indeed,
      there exists an isomorphism
      \begin{equation*}
      \varphi:k[x]/I\to k(\alpha)
      \end{equation*}
      with \(\varphi(x+I)=\alpha\) and \(\varphi(c+I)=c\) for all \(c\in k\)
   2. If \(\alpha'\in K\) is another root of \(p(x)\), then there is an
      isomorphism
      \begin{equation*}
      \theta:k(\alpha)\to k(\alpha')
      \end{equation*}
      with \(\theta(\alpha)=\alpha'\) and \(\theta(c)=c\) for all \(c\in k\)
   #+END_theorem

   #+BEGIN_proof
   1. Consider the evaluation, the ring homomorphism \(\varphi:k[x]\to K\) defined by
      \begin{equation*}
      \varphi:f(x)\mapsto f(\alpha)
      \end{equation*}
      Now \(\ker\varphi\) is the ideal in \(k[x]\). Since k[x] is an Euclidean
      ring and hence a PID, we have \(\ker\varphi=(p(x))\) for some monic
      polynomial \(p(x)\in k[x]\). But \(k[x]/(p(x))\cong\im\varphi\), which is
      a domain, and so $p(x)$ is irreducible by Proposition ref:prop3.116. The
      same proposition says that \(k[x]/p(x)\) is a field, and so \(\im\varphi\)
      is a subfield of $K$ containing $K$ and \alpha.Since every subfield of $K$ that
      contains $k$ and \alpha must contain  \(\im\varphi\), we have \(\im\varphi=k(\alpha)\).
   2. \(k[x]/I\cong k(\alpha)\) and \(k[x]/I\cong k(\alpha')\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $K/k$ is a field extension and \(\alpha\in K\) is algebraic over $k$, then
   the unique monic irreducible polynomial \(p(x)\in k[x]\) having \alpha as a root
   is called the *minimal polynomial* of \alpha over $k$, and it is denoted by
   \begin{equation*}
   \irr(\alpha,k)=p(x)
   \end{equation*}
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   Let \(k\subseteq E\subseteq K\) be fields with $E$ a finite extension of $k$
   and $K$ a finite extension of $E$. Then $K$ is a finite extension of $k$, and
   \begin{equation*}
   [K:k]=[K:E][E:k]
   \end{equation*}
   #+END_theorem

   #+BEGIN_proof
   If \(A=a_1,\dots,a_n\) is a basis of $E$ over $k$ and if \(B=b_1,\dots,b_m\)
   is a basis of $K$ over $E$, then it suffices to prove that a list $X$ of all
   \(a_ib_j\) is a basis of $K$ over $k$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   Let \(f(x)=x^4-10x^2+1\in\Q[x]\), then the root are
   \begin{equation*}
   \sqrt{2}+\sqrt{3},\quad 
   \sqrt{2}-\sqrt{3},\quad 
   -\sqrt{2}-\sqrt{3},\quad 
   -\sqrt{2}+\sqrt{3}
   \end{equation*}

   We claim that \(f(x)\) is irreducible in \(\Q[x]\). If \(g(x)\) is a
   quadratic factor of \(f(x)\) in \(\Q[x]\), then
   \begin{equation*}
   g(x)=(x-a\sqrt{2}-b\sqrt{3})(x-c\sqrt{2}-d\sqrt{3})
   \end{equation*}
   where \(a,b,c,d\in\{1,-1\}\). Multiplying
   \begin{equation*}
   g(x)=x^2-((a+c)\sqrt{2}+(b+d)\sqrt{3})x+2ac+3bd+(ad+bc)\sqrt{6}\not\in\Q[x]
   \end{equation*}
   Therefore \(f(x)\) is irreducible in \(\Q[x]\). If
   \(\beta=\sqrt{2}+\sqrt{3}\), then \(f(x)=\irr(\beta,\Q)\)

   Consider the field \(E=\Q(\beta)\). There is a tower of fields 
   \(\Q\subseteq E\subseteq F\), where \(F=\Q(\sqrt{2},\sqrt{3})\), and so 
   \begin{equation*}
   [F:\Q]=[F:E][E:\Q]
   \end{equation*}
   Since \(f(x)\) is a monic irreducible polynomial, \([E:\Q]=4\). On the other
   hand
   \begin{equation*}
   [F:\Q]=[F:\Q(\sqrt{2})][\Q(\sqrt{2}):\Q]
   \end{equation*}
   Now \([\Q(\sqrt{2}):\Q]=2\) because \(\sqrt{2}\) is root of \(x^2-2\). We
   claim that \([F:\Q(\sqrt{2})]\le2\). The field $F$ arises by adjoining
   \(\sqrt{3}\) to \(\Q(\sqrt{2})\); either \(\sqrt{3}\in\Q(\sqrt{2})\), in
   which case the degree is 1, or \(x^2-3\) is irreducible in
   \(\Q(\sqrt{2})[x]\), in which the degree is 2. It follows that \([F:\Q]\le4\)
   and so \([F:E]=1\); that is, \(F=E\)
   #+END_examplle

   #+ATTR_LATEX: :options [Kronecker]
   #+BEGIN_theorem
   If $k$ is a field and \(f(x)\in k[x]\), then there exists a field $K$
   containing $k$ as a subfield and with \(f(x)\) a product of linear
   polynomials in \(K[x]\)
   #+END_theorem
   
   #+BEGIN_proof
   Induction on $\deg(f)$. If \(\deg(f)=1\), then \(f(x)\) is linear and we can
   choose \(K=k\). If \(\deg(f)>1\), write \(f(x)=p(x)g(x)\) where \(p(x)\) is
   irreducible. Now Proposition ref:prop3.117 provides a field $F$ containing
   $k$ and root \(z\) of \(p(x)\). Hence in \(F[x]\), we have \(p(x)=(x-z)h(x)\)
   and \(f(x)=(x-z)h(x)g(x)\). By induction, there is a field $K$ containing $F$
   so that \(h(x)g(x)\), and hence \(f(x)\) is a product of linear factors in \(K[x]\)
   #+END_proof

   [[index:split]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let $k$ be a subfield of a field $K$, and let \(f(x)\in k[x]\). We say that
   \(f(x)\) *splits over $K$* if
   \begin{equation*}
   f(x)=a(x-z_1)\cdots(x-z_n)
   \end{equation*}
   where \(z_1,\dots,z_n\in K\) and \(a\in k\) is nonzero

   If \(f(x)\in k[x]\) is a polynomial, then a field extension \(E/k\) is called
   a *splitting field* of \(f(x)\) *over* $k$ if \(f(x)\) splits over $E$, but
   \(f(x)\) does not split over any proper subfield of $E$
   #+END_definition

   \(f(x)=x^2+1\in\Q[x]\). \(f(x)\) splits over \(\C\). \(\Q(i)\) is a splitting
   field of \(f(x)\) over \(\Q\). \(\R(i)=\C\) is a splitting field of \(f(x)\)
   over \(\R\) 


   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Let $k$ be a field, and let \(f(x)\in k[x]\). Then a splitting field of
   $f(x)$ over $k$ exists.
   #+END_corollary

   #+BEGIN_proof
   By Kronecker theorem
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   label:nexample2.151
   Let \(f(x)=x^n-1\in k[x]\) for some field $k$ ,and let $E/k$ be a splitting
   field. In Theorem ref:nthm2.46, we saw that the group \(\Gamma_n\) of all
   \(n\)th roots of unity in $E$ is a cyclic group: \(\Gamma_n=\la\omega\ra\),
   where \omega is a primitive \(n\)th root of unity.
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop3.126
   Let $p$ be a prime, and let $k$ be a field. If \(f(x)=x^p-c\in k[x]\) and
   \alpha is a \(p\)th root of $c$ (in some splitting field), then either \(f(x)\) is
   irreducible in \(k[x]\) or $c$ has a \(p\)th root in $k$. In either case, if
   $k$ contains the \(p\)th roots of unity, then \(k(\alpha)\) is a splitting field
   of $f(x)$
   #+END_proposition

   #+BEGIN_proof
   By Kronecker's theorem, there exists a field extension \(K/k\) that contains
   all the roots of \(f(x)\); that is, $K$ contains all the \(p\)th roots of
   $c$. If \(\alpha^p=c\), then every such root has the form \(\omega\alpha\), where
   \omega is a \(p\)th root of unity.

   If \(f(x)\) is not irreducible in \(k[x]\). Then there is a factorization
   \(f(x)=g(x)h(x)\). Now the constant term $b$ of $g(x)$ is, up to sign, the
   product of some of the roots of $f(x)$:
   \begin{equation*}
   \pm b=\alpha^d\omega
   \end{equation*}
   where \omega, which is a product of $d$ \(p\)th roots of unity, is itself a
   \(p\)th root of unity. It follows that 
   \begin{equation*}
   (\pm b)^p=\alpha^{dp}=c^d
   \end{equation*}
   _But $p$ being prime and \(d<p\) forces \((d,p)=1\)_. hence \(1=sd+tp\),
   therefore
   \begin{equation*}
   c=c^{sd+tp}=c^{sd}c^{tp}=(\pm b)^{ps}c^{tp}=[(\pm b)^sc^t]^p
   \end{equation*}
   therefore $c$ has a \(p\)th roots of unity.
   #+END_proof

   #+ATTR_LATEX: :options [Galois]
   #+BEGIN_theorem
   label:thm3.127
   If $p$ is a prime and $n$ is a positive integer, then there is a field having
   exactly $p^n$ elements.
   #+END_theorem

   #+BEGIN_proof
   Write \(q=p^n\), and consider the polynomial
   \begin{equation*}
   g(x)=x^q-x\in\F_p[x]
   \end{equation*}
   By Kronecker's theorem, there is a field $K$ containing $\F_p$ s.t. $g(x)$ is
   a product of linear factors in $K[x]$. Define
   \begin{equation*}
   E=\{\alpha\in K:g(\alpha)=0\}
   \end{equation*}
   Since the derivative \(g'(x)=qx^{q-1}-1=p^nx^{q-1}-1=-1\) (Exercise
   ref:ex3.23), it follows that the \(\gcd(g,g')\) is 1. By Exercise ref:ex3.37,
   all the roots of \(g(x)\) are distinct

   We claim that \(E\) is a subfield of $K$, and this will complete the proof.
   If \(a,b\in E\) then \(a^q=a\) and \(b^q=b\). Therefore \((ab)^q=ab\) and
   \(ab\in E\). By Exercise ref:ex3.45, \((a-b)^q=a^q-b^q=a-b\), so that
   \(a-b\in E\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   label:ncor2.154
   For every prime $p$ and every integer \(n\ge1\), there exists an irreducible
   polynomial \(g(x)\in\F_p[x]\) of degree $n$. In fact, if \alpha is a primitive
   element of \(\F_{p^n}\), then its minimal polynomial \(g(x)=\irr(\alpha,\F_p)\)
   #+END_corollary

   #+BEGIN_proof
   Let \(E/\F_p\) be an extension field with \(p^n\) elements, and let
   \(\alpha\in E\) be a primitive element. Clearly, \(\F_p(\alpha)=E\). By Theorem 
   ref:nthm2.144, \(g(x)=\irr(\alpha,\F_p)\in\F_p[x]\) is an irreducible polynomial
   having \alpha as a root. If \(\deg(g)=d\), then Proposition ref:nprop2.141 gives 
   \([\F_p[x]/(g(x)):\F_p]=d\); but \(\F_p[x]/(g(x))\cong\F_p(\alpha)=E\) by Theorem
   ref:nthm2.144, so that \([E:\F_p]=n\). Therefore \(n=d\), and so \(g(x)\) is
   an irreducible polynomial of degree of $n$
   #+END_proof

   This corollary can also be proved by counting. If \(m=p_1^{e_1}\dots
   p_n^{e_n}\), define the *Möbius function* by
   \begin{equation*}
   \mu(m)=
   \begin{cases}
   1&\text{if }m=1\\
   0&\text{if any }e_i>1\\
   (-1)^n&\text{if }1=e_1=e_2=\dots=e_n
   \end{cases}
   \end{equation*}
   If \(N_n\) is the number of irreducible polynomials in \(\F_p[x]\) of degree
   $n$, then
   \begin{equation*}
   N_n=\frac{1}{n}\sum_{d\mid n}\mu(d)p^{n/d}
   \end{equation*}

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. Consider a field with four elements:
      \begin{equation*}
      \F_4=\left\{\begin{bsmallmatrix}a&b\\b&a+b\end{bsmallmatrix}
      :a,b\in\F_2
      \right\}
      \end{equation*}
      On the other hand, we may construct a field of order 4 as the quotient
      \(F=\F_2[x]/(q(x))\), where \(q(x)\in\F_2[x]\) is the irreducible
      polynomial \(x^2+x+1\). By Proposition ref:nprop2.141, $F$ is a field
      consisting of all \(a+b\beta\), where \(\beta=x+(q(x))\) is a root of
      \(q(x)\) and \(a,b\in\F_2\). Since \(\beta^2+\beta+1=0\), we have
      \(\beta^2=-\beta-1=\beta+1\); moreover,
      \(\beta^3=\beta\beta^2=\beta(\beta+1)=\beta^2+\beta=1\). There is a ring
      isomorphism
      \(\varphi:\F_4\to F\) with 
      \(\varphi(\begin{bsmallmatrix}a&b\\b&a+b\end{bsmallmatrix})=a+b\beta\)
   2. There are three monic irreducible quadratics in \(\F_3[x]\), namely,
      \begin{equation*}
      p(x)=x^2+1,\quad q(x)=x^2+x-1,\quad r(x)=x^2-x-1
      \end{equation*}
      each give rise to a field with \(9=3^2\) elements. Let us look at the
      first two in more detail. Proposition ref:nprop2.141 says that
      \(E=\F_3[x]/(p(x))\) is given by
      \begin{equation*}
      E=\{a+b\alpha:\text{where }\alpha^2+1=0\}
      \end{equation*}
      Similarly, if \(F=\F_3[x]/(q(x))\), then
      \begin{equation*}
      F=\{a+b\beta:\text{where }\beta^2+\beta-1=0\}
      \end{equation*}
      There two fields are isomorphic. The map \(\varphi:E\to F\), defined by 
      \(\varphi(a+b\alpha)=a+b(1-\beta)\)
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:nlemma2.156
   Let \(\varphi:k\to k'\) be an isomorphism of fields, and let \(\varphi^*:k[x]\to k'[x]\)
   be the ring isomorphism: 
   \(\varphi^*:g(x)=a_0+a_1x+\dots+a_nx^n\mapsto
   g^*(x)=\varphi(a_0)+\dots+\varphi(a_n)x^n\). 
   Let \(f(x)\in k[x]\) and \(f^*(x)=\varphi^*(x)\in k'[x]\). If $E$ is a
   splitting field of \(f(x)\) over $k$ and \(E'\) is a splitting field of
   \(f^*(x)\) over $k'$, then there is an isomorphism \(\Phi:E\to E'\) extending \varphi:
   \begin{tikzcd}
   E\arrow[r,dashed,"\Phi"]\arrow[d,no head]&E'\\
   k\arrow[r,"\varphi"]&k'\arrow[u,no head]
   \end{tikzcd}
   #+END_lemma

   #+BEGIN_proof
   Induction on \(d=[E:k]\). If \(d=1\), then \(f(x)\) is a product of linear
   polynomials in \(k[x]\). \(\Phi=\varphi\)
   
   Choose a root \(z\) of \(f(x)\) in $E$ that is not in $k$, and let
   \(p(x)=\irr(z,k)\). Now \(\deg(p)>1\) and \([k(z):k]=\deg(p)\). Let \(z'\) be
   a root of \(p^*(x)\) in \(E'\), and let \(p^*(x)=\irr(z',k')\) be the
   corresponding monic irreducible polynomial in \(k'[x]\).

   By a generalization of Proposition ref:nthm2.144, there is an isomorphism
   \(\widetilde{\varphi}:k(z)\to k'(z')\) extending \varphi with 
   \(\widetilde{\varphi}z\mapsto z\). We may regard \(f(x)\) as a polynomial with
   coefficients in \(k(z)\), for \(k\subseteq k(z)\) implies 
   \(k[x]\subseteq k(z)[x]\). We claim that \(E\) is a splitting field of
   \(f(x)\) over \(k(z)\); that is
   \begin{equation*}
   E=k(z)(z_1,\dots,z_n)
   \end{equation*}
   where \(z_1,\dots,z_n\) are the roots of \(f(x)/(x-z)\). Similarly, \(E'\) is
   a splitting field of \(f^*(x)\) over \(k'(z')\). But \([E:k(z)]<[E:k]\), so
   that the inductive hypothesis gives an isomorphism \(\Phi:E\to E'\) that extends
   \(\widetilde{\varphi}\) 
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   If $k$ is a field and \(f(x)\in k[x]\), then any two splitting fields of
   \(f(x)\) over $k$ are isomorphic via an isomorphism that fixes \(k\) pointwise
   #+END_theorem

   #+ATTR_LATEX: :options [Moore]
   #+BEGIN_corollary
   Any two finite fields having exactly \(p^n\) elements are isomorphic
   #+END_corollary

   #+BEGIN_proof
   If $E$ is a field with \(q=p^n\) elements, then Lagrange's Theorem applied to
   the multiplicative group \(E^\times\) shows that \(a^{q-1}=1\)  for every
   \(a\in E^\times\). It follows that every element of $E$ is a root of
   \(f(x)=x^q-x\in\F_p[x]\), and so $E$ is a splitting field of \(f(x)\) over \(\F_p\)
   #+END_proof

   Finite fields are often called *Galois fields*. We speak of the field with
   \(q\) elements, where \(q=p^n\) is a power of a prime $p$, and we denote it
   by
   \begin{equation*}
   \F_q
   \end{equation*}


   #+BEGIN_exercise
   label:ex3.83
   For every commutative ring $R$, prove that \(R[x]/(x)\cong R\)
   #+END_exercise
* Fields
  :PROPERTIES:
  :PP:       212
  :END:
** Insolvability of the Quintic
   :PROPERTIES:
   :PP:       217
   :END:

   By Kronecker's theorem, for each monic \(f(x)\in k[x]\), where $k$ is a
   field, there is an extension field \(K/k\) and roots \(z_1,\dots,z_n\in K\)
   with
   \begin{equation*}
   f(x)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0=(x-z_1)\dots(x-z_n)
   \end{equation*}
   Hence we have
   \begin{equation*}
   \begin{cases}
   a_{n-1}=-\displaystyle\sum_{i}z_i\\
   a_{n-2}=\displaystyle\sum_{i<j}z_iz_j\\
   a_{n-3}=-\displaystyle\sum_{i<j<k}z_iz_jz_k\\
   \vdots\\
   a_0=(-1)^nz_1\dots z_n
   \end{cases}
   \end{equation*}

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   The *elementary symmetric functions* of $n$ variables are the polynomials, for
   \(j=1,\dots,n\)
   \begin{equation*}
   e_j(x_1,\dots,x_n)=\displaystyle\sum_{i_1<\cdots<i_j}x_{i_1}\dots x_{i_j}
   \end{equation*}
   #+END_definition

   We have 
   \begin{equation*}
   e_j(z_1,\dots,z_n)=(-1)^ja_{n-j}
   \end{equation*}

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let $E$ be a field containing a subfield $k$. An *automorphism* of $E$ is an
   isomorphism \(\sigma:E\to E\); we say that \sigma *fixes* $k$ if \(\sigma(a)=a\) for every 
   \(a\in k\)
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop4.1
   Let $k$ be a subfield of a field $K$, let 
   \begin{equation*}
   f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in k[x]
   \end{equation*}
   and let \(E=k(z_1,\dots,z_n)\subseteq K\) be a splitting field. If 
   \(\sigma:E\to E\) is an automorphism fixing $k$, then \sigma permutes the set of root 
   \(\{z_1,\dots,z_n\}\) of \(f(x)\)
   #+END_proposition

   #+BEGIN_proof
   If \(r\) is a root of \(f(x)\), then 
   \begin{equation*}
   0=f(r)=r^n+a_{n-1}r^{n-1}+\cdots+a_1r+a_0
   \end{equation*}
   
   Applying \sigma to this equation gives
   \begin{align*}
   0&=\sigma(r)^n+\sigma(a_{n-1})\sigma(r)^{n-1}+\cdots+\sigma(a_0)\\
   &=\sigma(r)^n+a_{n-1}\sigma(r)^{n-1}+\cdots+a_0\\
   &=f(\sigma(r))
   \end{align*}
   Therefore \(\sigma(r)\) is a root of \(f(x)\)
   #+END_proof

   [[index:Galois group]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let $k$ be a subfield of a field $E$. The *Galois group* of $E$ over $k$,
   denoted by \(\Gal(E/k)\), is the set of all those automorphisms of $E$ that
   fix $k$. If \(f(x)\in k[x]\), and if \(E=k(z_1,\dots,z_n)\) is a splitting
   field, then the *Galois group* of $f(x)$ over $k$ is defined to be \(\Gal(E/k)\)
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:lemma4.2
   label:nlemma3.2
   Let \(E=k(z_1,\dots,z_n)\). If \(\sigma:E\to E\) is an automorphism fixing $k$,
   that is, if \(\sigma\in\Gal(E/k)\), and if \(\sigma(z_i)=z_i\) for all $i$, then
   \sigma is the identity \(1_E\)
   #+END_lemma

   #+BEGIN_proof
   Induction on \(n\ge1\). If \(n=1\), then each \(u\in E\) has the form 
   \(u=f(z_1)/g(z_1)\), where \(f(x),g(x)\in k[x]\) and \(g(z_1)\neq0\). Hence
   \sigma fixes all \(u\in E\). For the inductive steop, write
   \(K=k(z_1,\dots,z_{n-1})\) and note that \(E=K(z_n)\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:nthm3.3
   If \(f(x)\in k[x]\) has degree $n$, then its Galois group \(\Gal(E/k)\) is
   isomorphic to a subgroup of \(S_n\)
   #+END_theorem

   #+BEGIN_proof
   Let \(X=\{z_1,\dots,z_n\}\). If \(\sigma\in\Gal(E/k)\), then Proposition
   ref:prop4.1 shows that its restriction \(\sigma|X\) is a permutation of $X$.
   Define \(\varphi:\Gal(E/k)\to S_X\) by \(\varphi:\sigma\mapsto\sigma|X\). This is a
   homomorphism.

   \(\im\varphi\le S_X\cong S_n\). Since \varphi fixes $k$, \(\ker\varphi=\{1\}\).
   Therefore \varphi is injective.
   #+END_proof

   If \(f(x)=x^2+1\in\Q[x]\), then complex conjugate \sigma is an automorphism of its
   splitting field \(\Q(i)\) which fixes \(\Q\). Since \(\Gal(\Q(i)/\Q)\) is a
   subgroup of the symmetric group \(S_2\), which has order 2, it follows that 
   \(\Gal(\Q(i)/\Q)=\la\sigma\ra\cong\I_2\)

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:lemma4.4
   If $k$ is a field of characteristic 0, then every irreducible polynomial 
   \(p(x)\in k[x]\) has no repeated roots
   #+END_lemma

   #+BEGIN_proof
   Note Exercise ref:ex3.37. Either \(p'(x)=0\) or \(\deg(p')<\deg(p)\).
   Since $p(x)$ is irreducible, it not a constant and so it has some nonzero
   monomial \(a_ix^i\) where \(i\ge1\). Therefore \(ia_ix^{i-1}\) is a nonzero
   monomial in \(p'(x)\), _because $k$ has characteristic 0, and so_ 
   _\(p'(x)\neq 0\)_. Finally, since \(p(x)\) is irreducible, its only divisors
   are constant and associates; as \(p'(x)\) has smaller degree, \(\gcd(p',p)=1\)
   #+END_proof

   [[index:separable]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let \(E/k\) be an algebraic extension. An /irreducible/ polynomial \(p(x)\) is
   *separable* if it has no repeated roots. An arbitrary polynomial \(f(x)\) is
   *separable* if each of its irreducible factor has no repeated roots.

   An element \(\alpha\in E\) is called *separable* if either \alpha is transcendental
   over $k$ or if \alpha is algebraic over $k$ and its minimal polynomial
   \(\irr(\alpha,k)\) has no repeated roots

   A field extension \(E/k\) is called a *separable extension* if each of its
   elements is separable
   #+END_definition

   Lemma ref:lemma4.4 shows that every extension of a field of characteristic 0
   is a separable extension. If $E$ is a finite field with $p^n$ elements, then
   Lagrange's theorem (for the multiplicative group $E^\times$) shows that every
   element of $E$ is a root of \(x^{p^n}-x\). We saw in the proof of Theorem
   ref:thm3.127 that \(x^{p^n}-x\) has no repeated roots. It follows that 
   if \(k\subseteq E\), then \(E/k\) is a separable extension, for if
   \(\alpha\in E\), then \(\irr(\alpha,k)\) is a divisor of \(x^{p^n}-x\)

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   Example of an inseparable extension. Let \(k=\F_p(t)=\Frac(\F_p[t])\), and
   let \(E=k(\alpha)\), where \alpha is a root of \(f(x)=x^p-t\). In \(E[x]\) we have
   \begin{equation*}
   f(x)=x^p-t=x^p-\alpha^p=(x-\alpha)^p
   \end{equation*}
   If we show that \(\alpha\not\in k\), then \(f(x)\) is irreducible, by
   Proposition ref:prop3.126, and so \(f(x)=\irr(\alpha,k)\) is an inseparable
   polynomial. Therefore \(E/k\) is an inseparable extension

   If \(\alpha\in k\), then \(\alpha=g(t)/h(t)\). Hence \(g=\alpha h\) and 
   \(g^p=\alpha^p h^p=th^p\), so that
   \begin{equation*}
   \deg(g^p)=\deg(th^p)=1+\deg(h^p)
   \end{equation*}
   But \(p\mid\deg(g^p)\) and \(p\mid\deg(h^p)\)
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:nthm3.7
   1. Let \(E/k\) be a splitting field of a separable polynomial
      \(f(x)\in k[x]\), let \(\varphi:k\to k'\) be a field isomorphism, and let 
      \(E'/k'\) be a splitting field of \(f^*(x)\in k'[x]\) (where \(f^*(x)\) is
      obtained from \(f(x)\) by applying \varphi to its coefficients)
      \begin{center}
      \begin{tikzcd}
      E\arrow[r,densely dotted,"\Phi"] \arrow[d,dash] &E'\arrow[d,dash]\\
      k\arrow[r,"\varphi"]&k'
      \end{tikzcd}
      \end{center}
      Then there are exactly \([E:k]\) isomorphisms \(\Phi:E\to E'\) that extend \varphi
   2. If \(E/k\) is a splitting field of a separable \(f(x)\in k[x]\), then
      \begin{equation*}
      \abs{\Gal(E/k)}=[E:k]
      \end{equation*}
   #+END_theorem


   #+BEGIN_proof
   1. Induction on \([E:k]\). If \([E:k]=1\), then \(E=k\) and there is only one
      extension. If \([E:k]>1\), let \(f(x)=p(x)g(x)\), where \(p(x)\) is an
      irreducible factor of largest degree, say $d$. We may assume \(d>1\),
      otherwise \(f(x)\) splits over $k$ and \([E:k]=1\). Choose a root \alpha of
      \(p(x)\). If \(\widetilde{\varphi}:k(\alpha)\to E'\) is any extension of \varphi, then
      \(\varphi(\alpha)\) is a root \(\alpha'\) of \(p^*(x)\), by Proposition ref:prop4.1.
      Since \(f^*(x)\) is separable, \(p^*(x)\) has exactly $d$ roots
      \(\alpha'\in E'\); by Lemma ref:lemma4.2 and Theorem ref:thm3.120, there
      are exactly $d$ isomorphisms \(\widetilde{\varphi}:k(\alpha)\to k'(\alpha')\)
      extending \varphi, for each \(\alpha'\). Now $E$ is also a splitting field of
      \(f(x)\) over $k(\alpha)$, because adjoining all the roots of \(f(x)\) to
      \(k(\alpha)\) ; similarly, \(E^*\) is a splitting field of \(f^*(x)\) over
      \(k'(\alpha)\). Now \([E:k(\alpha)]<[E:k]\) because \([E:k(\alpha)]=[E:k]/d\)
      (\(1,\alpha,\dots,\alpha^{d-1}\) is an basis by Proposition ref:prop3.117), so that
      induction shows that each of the $d$ isomorphisms \(\widehat{\varphi}\) has
      exactly \([E:k]/d\) extensions \(\Phi:E\to E^*\). Thus we have constructed
      \([E:k]\) isomorphisms extending \varphi. But there are no others, because every
      \tau extending \varphi has \(\tau|k(\alpha)=\widehat{\varphi}\) for some 
      \(\widehat{\varphi}:k(\alpha)\to k'(\alpha^*)\)
   2. take \(k=k',E=E^*,\varphi=1_k\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   label:ncor3.9
   Let \(E/k\) be a splitting field of a separable polynomial \(f(x)\in k[x]\)
   of degree $n$. If \(f(x)\) is irreducible, then \(n\mid\abs{\Gal(E/k)}\)
   #+END_corollary

   #+BEGIN_proof
   By Theorem ref:nthm3.7, \(\abs{\Gal(E/k)}=[E:k]\). Let \(\alpha\in E\) be a
   root of \(f(x)\). Since \(f(x)\) is irreducible, \([k(\alpha):k]=n\), and 
   \begin{equation*}
   [E:k]=[E:k(\alpha)][k(\alpha):k]=n[E:k(\alpha)]
   \end{equation*}
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   The polynomial \(f(x)=x^4+1\) is irreducible in \(\Q[x]\), yet it factors in 
   \(\F_p[x]\) for every prime $p$
   #+END_proposition

   #+BEGIN_proof
   To see that \(f(x)\) is irreducible, it suffices to show that \(f(x-1)\) is
   irreducible by Lemma ref:nlemma2.74. But \(f(x-1)=x^4-4x^3+6x^2-4x+2\) is
   irreducible, by Eisenstein's Criterion.

   We now show for all prime $p$, that \(x^4+1\) factors in \(\F_p[x]\). If
   \(p=2\), then \(x^4+1=(x+1)^4\) and so we may assume that $p$ is an odd
   prime. It is easy to check that every square \(n^2\) is congruent to \(0,1\)
   or \(4\mod 8\); since $p$ is odd, we must have \(p^2\equiv1\mod8\).
   Therefore, \(\abs{(\F_{p^2})^\times}=p^2-1\) is divisible by 8. But 
   \((\F_{p^2})^\times\) is a cyclic group, by Theorem ref:nthm2.46, and so it
   has a (cyclic) subgroup of order 8, by Lemma ref:nlemma1.92. It follows that 
   \((\F_{p^2})\) contains all the 8th roots of unity; in particular,
   \(\F_{p^2}\) contains all the roots of \(x^4+1\). Hence, the splitting field
   \(E_p\) of \(x^4+1\) over \(\F_{p}\) is \(\F_{p^2}\), and so \([E_p:\F_p]=2\)
   (\(qp+r\)). Since \(x^4+1\) is irreducible in \(\F_p[x]\), then 
   \(4\mid[E_p:\F_p]\), by Corollary label:ncor3.9. Therefore, \(x^4+1\) factors
   in \(\F_p[x]\) for every prime $p$.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. Let \(f(x)=x^3-1\in\Q[x]\). Now \(f(x)=(x-1)(x^2+x+1)\). Suppose roots of
      \(x^2+x+1\) is \(\omega\) and \(\bar{\omega}\). The splitting field of
      \(f(x)\) is \(\Q(\omega)\), for \(\omega^2=\bar{\omega}\), and so \([\Q(\omega),\Q]=2\).
      Therefore \(\abs{\Gal(\Q(\omega)/\Q)}=2\). It's nontrivial element is complex conjugation.
   2. Let \(f(x)=x^2-2\in\Q[x]\). Now \(f(x)\) is irreducible with roots
      \(\pm\sqrt{2}\), so that \(E=\Q(\sqrt{2})\) is a splitting field.
      \(\abs{\Gal(E/\Q)}=2\). Now every element of $E$ has a unique expression
      of the form \(a+b\sqrt{2}\).
   3. Let \(g(x)=x^3-2\in\Q[x]\). The roots of \(g(x)\) are
      \(\beta,\omega\beta,\omega^2\beta\), where \(\beta=\sqrt[3]{2}\) and \omega is a
      primitive cube root of unity. The splitting field of \(g(x)\) is
      \(E=\Q(\beta,\omega)\). Now that 
      \begin{equation*}
      [E:\Q]=[E:\Q(\beta)][\Q(\beta):\Q]=3[E:\Q(\beta)]
      \end{equation*}
      for \(g(x)\) is irreducible over $\Q$. Now \(E\neq\Q(\beta)\) for every
      element in \(\Q(\beta)\) is real. Therefore \([E:\Q]=\abs{\Gal(E/\Q)}>3\). On the
      other hand, we know that \(\Gal(E/\Q)\) is isomorphic to \(S_3\), and so
      we must have \(\Gal(E/\Q)\cong S_3\)
   4. We examined \(f(x)=x^4-10x^2+1\in\Q[x]\) which is irreducible; in fact,
      \(f(x)=\irr(\beta,\Q)\) , where \(\beta=\sqrt{2}+\sqrt{3}\). If \(E=\Q(\beta)\),
      then \([E:\Q]=4\); moreover $E$ is a splitting field of $f(x)$. It follows
      that \(\abs{G}=\abs{\Gal(E/\Q)}=4\); hence either \(G\cong\I_4\) or
      \(G\cong\bV\). Actually \(G\cong\bV\)
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:nprop3.12
   If $m$ is a positive integer, $k$ is a field, and $E$ is a splitting field of
   \(x^m-1\) over $k$, then \(\Gal(E/k)\) is abelian; in fact, \(\Gal(E/k)\) is
   isomorphic to a subgroup of the group of units \(U(\I_m)=\{[i]\in\I_m:(i,m=1\}\}\)q
   #+END_proposition

   #+BEGIN_proof
   By Example ref:nexample2.151, \(E=k(\omega)\), where \omega is a primitive \(m\)th root
   of unity. The group \(\Gamma_m\) of all roots of \(x^m-1\) in $E$ is cyclic
   and if \(\sigma\in\Gal(E/k)\), then its restriction to (\Gamma_m) is an
   automorphism of \(\Gamma_m\). Hence \(\sigma(\omega)=\omega^i\) must also be a
   generator of \(\Gamma_m\); that is, \((i,m)=1\), by Theorem ref:nthm1.39. $i$
   is uniquely determined mod $m$, so that the function 
   \(\theta:\Gal(k(\omega)/k)\to U(\I_m)\) is well-defined. \theta is homomorphism.

   Therefore, Lemma ref:nlemma3.2 shows that \theta is injective
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   If $p$ is prime, then
   \begin{equation*}
   \Gal(\F_{p^n}/\F_p)\cong\I_n
   \end{equation*}
   and a generator is the *Frobenius automorphism*  \(\Fr:u\mapsto u^p\)
   #+END_theorem

   #+BEGIN_proof
   Let \(q=p^n\) and \(G=\Gal(\F_q/\F_p)\). Since \(\F_q\) has characteristic
   $p$, we have \((a+b)^p=a^p+b^p\), and so the Frobenius Fr is a homomorphism
   of fields. As any homomorphism of fields, Fr is injection; as \(\F_q\) is
   finite, Fr must be an automorphism, by the Pigeonhole principle; that is,
   \(\Fr\in G\) (Fr fixes \(\F_p\), by Fermat's Theorem)

   If \(\pi\in\F_q\) is a primitive element, then \(d(x)=\irr(\pi,\F_p)\) has
   degree $n$, by Corollary ref:ncor2.154. It suffices to prove that the order
   \(j\) of \(\Fr\) is not less than $n$. But if \(\Fr^j=1_{\F_q}\) for \(j<n\),
   then \(u^{p^j}=u\) for all of the \(q=p^n\) elements \(u\in\F_q\), giving too
   many roots of the polynomial \(x^{p^j}-x\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let $k$ be a field, let \(f(x)\in k[x]\), and let \(E/k\) be a splitting
   field of \(f(x)\). If \(f(x)\) has no repeated roots, then \(f(x)\) is
   irreducible if and only if \(\Gal(E/k)\) acts transitively on the roots of
   \(f(x)\). 
   #+END_proposition

   #+BEGIN_proof
   Assume that \(f(x)\) is irreducible, and let \(\alpha,\beta\in E\) be roots of
   \(f(x)\). By Theorem ref:nthm2.144, there is an isomorphism 
   \(\varphi:k(\alpha)\to k(\beta)\) with \(\varphi(\alpha)=\beta\) which fixes $k$. Lemma ref:nlemma2.156
   shows that \varphi extends to an automorphism \Phi of $E$ that fixes $k$; that is,
   \(\Phi\in\Gal(E/k)\). Now \(\Phi(\alpha)=\varphi(\alpha)=\beta\), and so \(\Gal(E/k)\)
   acts transitively on the roots.

   Assume that \(\Gal(E/k)\) acts transitively on the roots of \(f(x)\). Let 
   \(f(x)=p_1(x)\dots p_t(x)\) be a factorization into irreducibles in \(k[x]\),
   where \(t\ge2\). Choose a root \(\alpha\in E\) of \(p_1(x)\) and a root
   \(\beta\in E\) of \(p_2(x)\). By hypothesis, there is \(\sigma\in\Gal(E/k)\)
   with \(\sigma(\alpha)=\beta\). Now \sigma permutes the roots of \(p_1(x)\), contradicting \beta
   not being a root of \(p_1(x)\)
   #+END_proof

   The analogy
   \begin{center}
   Polygon \(\Omega\)\dotfill polynomial \(f(x)\in k[x]\)\\
   Regular polygon\dotfill irreducible polynomial\\
   Vertices of \(\Omega\)\dotfill roots of \(f(x)\)\\
   Plain\dotfill splitting field $E$ of \(f(x)\)\\
   Motion\dotfill automorphism fixing $k$\\
   Symmetry group \(\Sigma(\Omega)\)\dotfill Galois group \(\Gal(E/k)\)
   \end{center}
*** Classical Formulas and Solvability by Radicals
    [[index:pure extension]]
    [[index:radical extension]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    A *pure extension* of *type* $m$ is an extension field \(k(u)/k\), where
    \(u^m\in k\) for some \(m\ge1\). An extension field \(K/k\) is a *radical
    extension* if there is a tower of intermediate fields
    \begin{equation*}
    k=K_0\subseteq K_1\subseteq\cdots\subseteq K_t=K
    \end{equation*}
    in which each \(K_{i+1}/K_i\) is a pure extension
    #+END_definition

    Every pure extension is a radical extension. If \(u^m=a\in k\), then
    \(k(u)\) arises from $k$ by adjoining an \(m\)th root of \(a\). If
    \(k\subseteq\C\), there are $m$ different \(m\)th roots of $a$, namely,
    \(u,\omega u,\dots,\omega^{m-1}u\), where \(\omega=e^{2\pi i/m}\) is a primitive
    \(m\)th root.

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    Let \(f(x)\in k[x]\) have a splitting field $E$. We say that \(f(x)\) is
    *solvable by radicals* if there is a radical extension
    \begin{equation*}
    k=K_0\subseteq K_1\subseteq\cdots\subseteq K_t
    \end{equation*}
    where \(E\subseteq K_t\)
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    label:nexample3.15
    1. For every field $k$ and every \(n\ge1\), we show that 
       \(f(x)=x^n-1\in k[x]\) is solvable by radicals. A splitting field of
       \(x^n-1\) is \(E=k(\omega)\)
    2. Let $p$ be a prime and let $k$ contain all \(p\)th roots of unity. If
       \(k(u)/k\)is a pure extension of type $p$, then we claim that \(k(u)\) is
       a splitting field of \(f(x)=x^p-u^p\). If \(k\) has characteristic $p$,
       then \(x^p-u^p=(x-u)^p\), and \(f(x)\) splits over \(k(u)\); otherwise,
       $k$ contains a primitive \(p\)th root of unity, \omega, and
       \(f(x)=\prod_i(x-\omega^iu)\). Note that \(f(x)\) is separable if
       characteristic \(k\neq p\)
    #+END_examplle
**** Quadratics
     If \(f(x)=x^b+bx+c\), then the quadratic formula gives its roots as
     \begin{equation*}
     \frac{1}{2}(-b\pm\sqrt{b^2-4c})
     \end{equation*}
     Let \(k=\Q(b,c)\). Define \(K_1=k(u)\), where \(u=\sqrt{b^2-c}\). Then
     \(K_1\) is a radical extension of $k$, for \(u^2\in k\). Moreover, the
     quadratic formula implies that \(K_1\) is the splitting field of \(f(x)\),
     and so \(f(x)\) is solvable by radicals
**** Cubics
     Let \(f(X)=X^3+bX^2+cX+d\), and let \(k=\Q(b,c,d)\). Change of variable
     \(X=x-b/3\) yields a new polynomial: \(\widetilde{f}(x)=x^3+qx+r\in k[x]\)
     having the same splitting field $E$; it follows that \(\widetilde{f}(x)\)
     is solvable by radicals if and only if \(f(x)\) is. The cubic formula gives
     the roots of \(\widetilde{f}(x)\) as
     \begin{equation*}
     g+h,\quad,\omega g+\omega^2h,\quad w^2g+\omega h
     \end{equation*}
     where \(g^3=\frac{1}{2}(-r+\sqrt{R})\), \(h=-q/3g\),
     \(R=r^2+\frac{4}{27}q^3\), and \omega is a primitive cube root of unity.

     Define \(K_1=k(\sqrt{R})\), \(K_2=K_1(\alpha)\), where
     \(\alpha^3=\frac{1}{2}(-r+\sqrt{R})\). \(K_2\) contains the root
     \(\alpha+\beta\) of \(\widetilde{f}(x)\), where \(\beta=-q/3\alpha\).
     Finally define \(K_3=K_2(\omega)\)x
*** Translation into Group Theory
    if \(k\subseteq k(u)\) is of type $m$, then factor \(m=p_1\dots p_q\), where
    the \(p\)'s are primes, and replace \(k\subseteq k(u)\) by
    \begin{equation*}
    k\subseteq k(u^{m/p_1})\subseteq k(u^{m/p_1p_2})\subseteq\dots\subseteq
    k(u)
    \end{equation*}

    [[index:normal extension]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    An extension field \(E/k\) is called *normal* if it is the splitting field of
    a polynomial in \(k[x]\)
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_theorem
    label:nthm3.17
    Let \(k\subseteq B\subseteq E\) be a tower of fields. If \(B/k\) and \(E/k\)
    are normal extensions, then \(\sigma(B)=B\) for all \(\sigma\in\Gal(E/k)\),
    \(\Gal(E/B)\tril\Gal(E/k)\) and
    \begin{equation*}
    \Gal(E/k)/\Gal(E/B)\cong\Gal(B/k)
    \end{equation*}
    #+END_theorem

    #+BEGIN_proof
    Since \(B/k\) is a normal extension, it is a splitting field of some
    \(f(x)\) in \(k[x]\); that is, \(B=k(z_1,\dots,z_t)\subseteq E\), where
    \(z_1,\dots,z_t\) are the roots of \(f(x)\). If \(\sigma\in\Gal(E/k)\), the
    restriction of \sigma to $B$ permutes the roots and hence \(\sigma(B)=B\).

    Define \(\rho:\Gal(E/k)\to\Gal(B/k)\) by \(\sigma\mapsto\sigma|B\). \rho is a
    homomorphism and \(\ker\rho=\Gal(E/B)\); thus, \(\Gal(E/B)\tril\Gal(E/k)\).
    But \rho is surjective: if \(\tau\in\Gal(B/k)\), then Lemma ref:nlemma2.156
    applies to show that there is \(\sigma\in\Gal(E/k)\) extending \tau
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_lemma
    label:nlemma3.18
    1. If \(B=k(u_1,\dots,u_t)/k\) is a finite extension field, then there is a
       normal extension \(E/k\) containing $B$; that is, $E$ is a splitting
       field of some \(f(x)\in k[x]\). If each \(u_i\) is separable over $k$,
       then \(f(x)\) is a separable polynomial and, if \(G=\Gal(E/k)\), then
       \begin{equation*}
       E=k(\sigma(u_1),\dots,\sigma(u_t):\sigma\in G)
       \end{equation*}
    2. If \(B/k\) is a radical extension, then the normal extension \(E/k\) is a
       radical extension
    #+END_lemma

    #+BEGIN_proof
    1. By Theorem ref:nthm2.144, there are irreducible polynomials 
       \(p_i(x)=\irr(u_i,k)\in k[x]\). Define $E$ to be a splitting field of 
       \(f(x)=p_1(x)\dots p_t(x)\) over $k$. Since \(u_i\in E\) for all $i$, we
       have \(B=k(u_1,\dots,u_t)\subseteq E\). If each \(u_i\) is separable over
       $k$, then each \(p_i(x)\) is a separable polynomial, and hence \(f(x)\)
       is a separable polynomial.

       For each pair of roots \(u\) and \(u'\) of any \(p_i(x)\), Theorem
       ref:nthm2.144 gives an isomorphism \(\gamma:k(u)\to k(u')\). By Lemma
       ref:nlemma2.156, each such \gamma extends to an automorphism
       \(\sigma\in G=\Gal(E/k)\). Thus \(f(x)\) splits over 
       \(k(\sigma(u_1),\dots,\sigma(u_t):\sigma\in G)\). But \(E/k\) is a splitting field
       of \(f(x)\) over \(k\) and \(k(\sigma(u_1),\dots,\sigma(u_t):\sigma\in G)\subseteq E\); hence,
       \begin{equation*}
       E=k(\sigma(u_1),\dots,\sigma(u_t):\sigma\in G)
       \end{equation*}
       because a splitting field is the smallest field over which \(f(x)\) splits

    2. Assume now that \(B/k\)is a radical extension; say,
       \(B=k(v_1,\dots,v_s)\), where 
       \begin{equation*}
       k\subseteq k(v_1)\subseteq k(v_1,v_2)\subseteq\dots\subseteq k(v_1,\dots,v_s)=B
       \end{equation*}
       and each \(k(v_1,\dots,v_{i+1})/k(v_1,\dots,v_i)\) is a pure extension;
       of course, \(\sigma(B)=k(\sigma(v_1),\dots,\sigma(v_s))\) is a radical extension of $k$
       for every \(\sigma\in G\). Define 
       \begin{equation*}
       B_1=k(\sigma(v_1):\sigma\in G)
       \end{equation*}
       Now if \(G=\{1,\sigma,\tau,\dots\}\), then the tower
       \begin{equation*}
       k\subseteq k(v_1)\subseteq k(v_1,\sigma(v_1))\subseteq
       k(v_1,\sigma(v_1),\tau(v_1))\subseteq\dots\subseteq B_1
       \end{equation*}
       displays $B_1$ as a radical extension of $k$. If \(v_1^m\in k\), 
       \(\tau(v_1)^m=\tau(v_1^m)\) lies in \(\tau(k)=k\). Define
       \begin{equation*}
       B_{i+1}=B_i(\sigma(v_{i+1}):\sigma\in G)
       \end{equation*}
       Assume, by induction, that \(B_i/k\) is a radical extension and that 
       \(\sigma(B_i)\subseteq B_i\) for all \(\sigma\in G\). Now \(B_{i+1}/B_i\) is a
       radical extension
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_lemma
    label:nlemma3.19
    Let 
    \begin{equation*}
    k=K_0\subseteq K_1\subseteq K_2\subseteq\cdots\subseteq K_t
    \end{equation*}
    be a tower with each \(K_i/K_{i-1}\)  a pure extension of prime type
    \(p_i\). If \(K_t/k\) is a normal extension and $k$ contains all the
    \(p_i\)th roots of unity, for \(i=1,\dots,t\), then there is a sequence of
    subgroups
    \begin{equation*}
    \Gal(K_t/k)=G_0\supseteq G_1\supseteq\cdots\supseteq G_t=\{1\}
    \end{equation*}
    with each \(G_{i+1}\tril G_i\), and \(G_i/G_{i+1}\) cyclic of prime order
    \(p_{i+1}\) or \(\{1\}\)
    #+END_lemma

    #+BEGIN_proof
    Define \(G_i=\Gal(K_t/K_i)\). Now \(K_1=k(u)\), where \(u^{p_1}\in k\);
    since $k$ contains all the \(p_1\)th roots of unity, \(K_1/k\) is a
    splitting field of the polynomial \(f(x)=x^{p_1}-u^{p_1}\). Theorem
    ref:nthm3.17 now applies: \(G_1=\Gal(K_t/K_1)\) is a normal subgroup of
    \(G_0=\Gal(K_t/k)\) and \(G_0/G_1\cong\Gal(K_1/k)\). Now Example
    ref:nexample3.15 says that if characteristic \(k\neq p_1\), then \(f(x)\) is
    separable. By Theorem ref:nthm3.7, \(G_0/G_1\cong\I_{p_1}\) ($K_1$ is the
    splitting field of $f(x)$). If characteristic \(k=p_1\), then 
    \(G_0/G_1\cong\Gal(K_1/k)=\{1\}\)
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    A *normal series* of a group $G$ is a sequence of subgroups
    \begin{equation*}
    G=G_0\supseteq G_1\supseteq\cdots\supseteq G_t=\{1\}
    \end{equation*}
    with each \(G_{i+1}\) a normal subgroup of \(G_i\); the *factor groups* of
    this series are the quotient groups
    \begin{equation*}
    G_0/G_1,G_1/G_2,\dots,G_{t-1}/G_t
    \end{equation*}
    The *length* of this series is the number of nontrivial factor groups.

    A finite group $G$ is called *solvable* if it has a normal series each of
    whose factor groups has prime order.
    #+END_definition

    In this language, Lemma ref:nlemma3.19 says that \(\Gal(K_t/k)\) is a
    solvable group if \(K_t\) is a radical extension of $k$ and $k$ contains
    appropriate roots of unity.

    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    label:nexample3.20
    1. $S_4$ is a solvable group. Consider the chain of subgroups
       \begin{equation*}
       S_4\supseteq A_4\supseteq \bV\supseteq W\supseteq\{1\}
       \end{equation*}
       $W$ is any subgroup of \(\bV\) of order 2.
    2. A nonabelian simple group $G$, for example $G=A_5$ is not solvable, for its
       only proper normal subgroup is \(\{1\}\), and \(G/\{1\}\cong G\) is not
       cyclic of prime order
    #+END_examplle

    #+ATTR_LATEX: :options []
    #+BEGIN_lemma
    label:nlemma3.21
    Let $k$ be a field, let \(f(x)\in k[x]\) be solvable by radicals, and let 
    \(k=K_0\subseteq K_1\subseteq\cdots\subseteq K_t\) be a tower with
    \(K_i/K_{i-1}\) a pure extension of prime type $p_i$ for all $i$. If \(K_t\)
    contains a splitting field $E$ of \(f(x)\) and $k$ contains all the
    \(p_i\)th roots of unity, then the Galois group \(\Gal(E/k)\) is a quotient
    of a solvable group
    #+END_lemma

    #+BEGIN_proof
    By Lemma ref:nlemma3.18 , we may assume that \(K_t\) is a normal extension
    of $k$. By Lemma ref:nlemma3.19, \(\Gal(K_t/k)\) is a solvable group. Since
    $E$ and \(K_t\) are splitting fields over $k$, Theorem ref:nthm3.17 show
    that \(\Gal(K_t/E)\tril\Gal(K_t/k)\) and
    \(\Gal(K_t/k)/\Gal(K_t/E)\cong\Gal(E/k)\)
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    label:nprop3.22
    Every quotient of a solvable group $G$ is itself a solvable group
    #+END_proposition

    #+BEGIN_proof
    Let \(G=G_0\supseteq G_1\supseteq\cdots\supseteq G_t=\{1\}\). 
    If \(N\tril G\). There is a sequence
    \begin{equation*}
    G=G_0N\supseteq G_1N\supseteq\cdots\supseteq G_tN=N\supseteq\{1\}
    \end{equation*}
    To see that this is a normal series, we claim that
    \begin{equation*}
    (g_in)G_{i+1}N(g_in)^{-1}\subseteq
    g_iG_{i+1}Ng_i^{-1}=g_iG_{i+1}g_i^{-1}N\subseteq G_{i+1}N
    \end{equation*}
    The first inclusion holds because
    \(n(G_{i+1}N)n^{-1}\subseteq NG_{i+1}N\subseteq
    (G_{i+1}N)(G_{i+1}N)=G_{i+1}N\). 
    
    The Second Isomorphism Theorem gives
    \begin{equation*}
    \frac{G_i}{G_i\cap(G_{i+1}N)}\cong
    \frac{G_i(G_{i+1}N)}{G_{i+1}N}=
    \frac{G_iN}{G_{i+1}N}
    \end{equation*}
    Since \(G_{i+1}\tril G_i\cap G_{i+1}N\), the Third Isomorphism Theorem gives
    a surjection \(G_i/G_{i+1}\to G_i/[G_i\cap G_{i+1}N]\), and so the composite
    is a surjection \(G_i/G_{i+1}\to G_iN/G_{i+1}N\). As \(G_i/G_{i+1}\) is
    cyclic of prime order, its image is either cyclic of prime order or trivial.
    Therefore, \(G/N\) is a solvable group
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    Every subgroup $H$ of a solvable group $G$ is solvable
    #+END_proposition
    
    #+BEGIN_proof
    \begin{equation*}
    G=G_0\supseteq G_1\supseteq\dots\supseteq G_t=\{1\}
    \end{equation*}
    Consider the sequence of subgroups
    \begin{equation*}
    H=H\cap G_0\supseteq H\cap G_1\supseteq\cdots\supseteq H\cap G_t=\{1\}
    \end{equation*}
    This is a normal series. The Second Isomorphism Theorem gives
    \begin{align*}
    (H\cap G_i)/(H\cap G_{i+1})&=(H\cap G_i)/[(H\cap G_i)\cap G_{i+1}]\\
    &\cong G_{i+1}(H\cap G_i)/G_{i+1}
    \end{align*}
    But the last quotient is a subgroup of \(G_i/G_{i+1}\). Since the only
    subgroups of a cyclic group of prime order $C$ and \(\{1\}\), it follows
    that the nontrivial factor groups \((H\cap G_i)/(H\cap G_{i+1})\) are cyclic
    of prime order. Therefore, $H$ is a solvable group
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    label:nexample3.24
    If \(n\ge5\), then the symmetric group $S_n$ is not solvable. Otherwise, each
    of its subgroups would also be solvable. But \(A_5\supseteq S_5\supseteq
    S_n\), and the simple group \(A_5\) is not solvable, by Example ref:nexample3.20
    #+END_examplle

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    If \(H\tril G\) and both \(H\) and \(G/H\) are solvable groups, then $G$ is solvable
    #+END_proposition

    #+BEGIN_proof
    Since \(G/H\) is solvable, there is normal series
    \begin{equation*}
    G/H\supseteq K_1^*\supseteq\cdots\supseteq K_m^*=\{1\}
    \end{equation*}
    having factor groups of prime order. By the Correspondence Theorem for
    Groups, there are subgroups \(K_i\le G\)
    \begin{equation*}
    G\supseteq K_1\supseteq\cdots\supseteq K_m=H
    \end{equation*}
    with \(K_i/H=K_i^*\) and \(K_{i+1}\tril K_i\). By the Third Isomorphism
    Theorem
    \begin{equation*}
    K_i^*/K_{i+1}^*\cong K_i/K_{i+1}
    \end{equation*}
    for all $i$, and so \(K_i/K_{i+1}\) is cyclic of prime order for all $i$

    Since \(H\) is solvable, there is a normal series
    \begin{equation*}
    H=H_0\supseteq H_1\supseteq\cdots\supseteq H_q=\{1\}
    \end{equation*}
    
    Splice these two series together
    \begin{equation*}
    G\supseteq K_1\supseteq\cdots\supseteq K_m=H_0\supseteq H_1\supseteq\cdots
    \supseteq H_q=\{1\}
    \end{equation*}
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    If $H$ and $K$ are solvable groups, then \(H\times K\) is solvable
    #+END_corollary
    #+BEGIN_proof
    \((H\times K)/H\cong K\)
    #+END_proof

    #+ATTR_LATEX: :options [Galois]
    #+BEGIN_theorem
    Let \(f(x)\in k[x]\), where \(k\) is a field, and let \(E\) be a splitting
    field of \(f(x)\) over \(k\). If \(f(x)\) is solvable by radicals, then its
    Galois group \(\Gal(E/k)\) is a solvable group
    #+END_theorem
    #+BEGIN_proof
    Let \(p_1,\dots,p_t\) be the types of the pure extensions occurring in the
    radical extension arising from \(f(x)\) being solvable by radicals. Define
    \(m\) to be the product of all these \(p_i\), define \(E^*\) to be a
    splitting field of \(x^m-1\) over $E$, and define \(k^*=k(\Omega)\), where \Omega is
    the set of all \(m\)th roots of unity in \(E^*\). Now \(E^*/k^*\) is a
    normal extension, for it is a splitting field of \(f(x)\) over \(k^*\), and
    so \(\Gal(E^*/k^*)\) is solvable, by Lemma ref:nlemma3.21 and Proposition
    ref:nprop3.22. Consider the tower \(k\subseteq k^*\subseteq E^*\)
    \begin{center}
    \begin{tikzpicture}
    \node (1) at (0,0) {$E$};
    \node (2) at (0,-1) {$k$};
    \node (3) at (1,0.5) {$E^*$};
    \node (4) at (1,-0.5) {$k^*$};
    \draw (1) -- (2);
    \draw (1) -- (3);
    \draw (2) -- (4);
    \draw (3) -- (4);
    \end{tikzpicture}
    \end{center}
    Since \(k^*/k\) is normal, Theorem ref:nthm3.17 gives
    \(\Gal(E^*/k^*)\tril\Gal(E^*/k)\) and 
    \begin{equation*}
    \Gal(E^*/k)/\Gal(E^*/k^*)\cong\Gal(k^*/k)
    \end{equation*}
    Now\(\Gal(E^*/k^*)\) is solvable, while \(\Gal(k^*/k)\) is abelian, hence
    solvable, by Proposition ref:nprop3.12. Finally, we may use Theorem
    ref:nthm3.17 once again, for the tower \(k\subseteq E\subseteq E^*\)
    satisfies the hypothesis that both \(E\) and \(E^*\) are normal. It follows
    that \(\Gal(E^*/k)/\Gal(E^*/E)\cong\Gal(E/k)\), and so \(G/k\), being a
    quotient of a solvable group, is solvable
    #+END_proof

    #+ATTR_LATEX: :options [Abel-Ruffini]
    #+BEGIN_theorem
    If \(n\ge5\), the general polynomial
    \begin{equation*}
    f(x)=(x-y_1)(x-y_2)\cdots(x-y_n)
    \end{equation*}
    over a field $k$ is not solvable by radicals
    #+END_theorem
    #+BEGIN_proof
    If \(E=k(y_1,\dots,y_n)\) and \(F=k(a_0,\dots,a_{n-1})\) where \(a_i\) are
    the coefficients of \(f(x)\), then $E$ is the splitting field of \(f(x)\)
    over $F$.

    We claim that \(\Gal(E/F)\cong S_n\). Now if \(\sigma\in S_n\), then there
    is an automorphism \(\widetilde{\sigma}\) of \(k[y_1,\dots,y_n]\), defined by 
    \(\widetilde{\sigma}:f(y_1,\dots,y_n)\mapsto f(y_{\sigma 1},\dots,y_{\sigma n})\). Thus
    \(\widetilde{\sigma}\) extends to an automorphism \(\sigma^*\) of 
    \(E=\Frac(k[y_1,\dots,y_n])\) and \(\sigma^*\) fixes $F$; hence
    \(\sigma^*\in\Gal(E/F)\). Since \(\sigma\mapsto\sigma^*\) is an injection, 
    \(\abs{S_n}\le\abs{\Gal(E/F)}\). 
    On the other hand, Theorem ref:nthm3.3 giving the reverse inequality.
    Therefore \(\Gal(E/F)\cong S_n\). But \(S_n\) is not a solvable group if
    \(n\ge5\), by Example ref:nexample3.24
    #+END_proof




* Groups \rom{2}
** Finite Abelian Groups
   :PROPERTIES:
   :PP:       263
   :END:
*** Direct Sums
    *External direct sum*, denoted by \(S_1\times\cdots\times S_n\) is the
    \(n\)-tuples \(s_1,\dots,s_n\), where \(s_i\in S_i\) for all $i$, and its
    binary operation is 
    \begin{equation*}
    (s_1,\dots,s_n)+(s_1',\dots,s_n')=(s_1+s_1',\dots,s_n+s_n')
    \end{equation*}
    However the most useful version, isomorphic to \(S_1\times\cdots\times S_n\)
    is called their *internal direct sum*

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If $S$ and $T$ are subgroups of an abelian group $G$, then $G$ is the
    *direct sum*, denoted by
    \begin{equation*}
    G=S\oplus T
    \end{equation*}
    if \(S+T=G\) and \(S\cap T=\{0\}\)
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    The following statements are equivalent for an abelian group $G$ and
    subgroups $S$ and $T$ of $G$
    1. \(G=S\oplus G\)
    2. Every \(g\in G\) has a unique expression of the form
       \begin{equation*}
       g=s+t
       \end{equation*}
       where \(s\in S\) and \(t\in T\)
    3. There are homomorphisms \(p:G\to S\) and \(q:G\to T\), called
       *projections*, and \(i:S\to G\) and \(j:T\to G\) called *injections*, s.t.
       \begin{equation*}
       pi=1_S,\quad qj=1_T,\quad pj=0,\quad qi=0,\quad ip+jq=1_G
       \end{equation*}
    #+END_proposition

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    Let $S$ and $T$ be subgroups of an abelian group $G$. If \(G=S\oplus T\),
    then \(S\oplus T\cong S\times T\)

    Conversely, given abelian group $S$ and $T$, define subgroups \(S'\cong S\)
    and \(T'\cong T\) of \(S\times T\) by
    \begin{equation*}
    S'=\{(s,0):s\in S\}\quad\text{and}\quad T'=\{(0,t):t\in T\}
    \end{equation*}
    then \(S\times T=S'\oplus T'\)
    #+END_corollary
    
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If \(S_1,\dots,S_n,\dots\) are subgroups of an abelian group $G$, define the
    *finite direct sum* \(S_1\oplus\cdots\oplus S_n\) using induction on
    \(n\ge2\):
    \begin{equation*}
    S_1\oplus\cdots\oplus S_{n+1}=[S_1\oplus\cdots\oplus S_n]\oplus S_{n+1}
    \end{equation*}
    We also denote the direct sum by
    \begin{equation*}
    \displaystyle\sum_{i=1}^nS_i=S_1\oplus\cdots\oplus S_n
    \end{equation*}
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    Let $V$ be a two-dimensional vector space over a field $k$, which we view as
    an additive abelian group, and let \(x,y\) be a basis. It's easy to check
    that the intersection of any two of the subspaces \(\la x\ra\), 
    \(\la y\ra\) and \(\la x+y\ra\) is \(\{0\}\). On the other hand, we do not
    have
    \(V=[\la x\ra\oplus\la y\ra]\oplus\la x+y\ra\) because
    \([\la x\ra\oplus\la y\ra]\cap\la x+y\ra\neq\{0\}\)
    #+END_examplle

    In the context of abelian groups, we shall write \(S\subseteq G\) to denote
    \(S\) being a subgroup of $G$.
    

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    label:prop5.4
    Let \(G=S_1+\cdots+S_n\), where the $S_i$ are subgroups. Then the following
    conditions are equivalent
    1. \(G=S_1\oplus\cdots\oplus S_n\)
    2. Every \(a\in G\) has a unique expression of the form
       \(a=s_1+\cdots+s_n\), where \(s_i\in S_i\)
    3. For each $i$
       \begin{equation*}
       S_i\cap(S_1+\cdots+\widehat{S_i}+\cdots+S_n)=\{0\}
       \end{equation*}
       where \(\widehat{S_i}\) means that the term \(S_i\) is omitted from the
       sum
    #+END_proposition

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    Let \(G=\la y_1,\dots,y_n\ra\). If for all \(m_i\in\Z\), we have
    \(\sum_im_iy_i=0\) implies \(m_iy_i=0\); then
    \begin{equation*}
    G=\la y_1\ra\oplus\cdots\oplus\la y_n\ra
    \end{equation*}
    #+END_corollary

    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    Let $V$ be an \(n\)-dimensional vector space over a filed $k$, which we view
    as an additive abelian group. If \(v_1,\dots,v_n\) is a basis, then
    \begin{equation*}
    V=\la v_1\ra\oplus\cdots\oplus\la v_n\ra
    \end{equation*}
    where \(\la v_i\ra=\{rv_i:r\in k\}\)
    #+END_examplle

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    label:prop5.7
    label:nprop1.85
    If \(G_1,\dots,G_n\) are abelian groups and \(H_i\subseteq G_i\) are
    subgroups, then
    \begin{equation*}
    (G_1\oplus\cdots\oplus G_n)/(H_1\oplus\cdots\oplus H_n)\cong(G_1/H_1)
    \times\cdots\times(G_n/H_n)
    \end{equation*}
    #+END_proposition

    If $G$ is an abelian group and $m$ is an integer, let us write
    \begin{equation*}
    mG=\{ma:a\in G\}
    \end{equation*}

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    If $G$ is an abelian group and $p$ is a prime, then \(G/pG\) is a vector
    space over \(\F_p\)
    #+END_proposition

    #+BEGIN_proof
    If \([r]\in\F_p\) and \(a\in G\), define scalar multiplication
    \begin{equation*}
    [r](a+pG)=ra+pG
    \end{equation*}
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    Let \(F=\la x_1,\dots,x_n\ra\) be an abelian group. If 
    \begin{equation*}
    F=\la x_1\ra\oplus\cdots\oplus\la x_n\ra
    \end{equation*}
    where each \(\la x_z\ra\cong\Z\), then $F$ is called a (finitely generated) 
    *free abelian group* with *basis* \(x_1,\dots,x_n\)
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    If \(\Z^m\) denotes the direct sum of $m$ copies of \(\Z\), then
    \(\Z^m\cong\Z^n\) if and only if \(m=n\)
    #+END_proposition

    #+BEGIN_proof
    For any abelian group $G$, if $G=G_1\oplus\cdots\oplus G_n$, then
    \(2G=2G_1\oplus\cdots\oplus 2G_n\). It follows from Proposition ref:prop5.7
    that
    \begin{equation*}
    G/2G\cong(G_1/2G_1)\oplus\cdots\oplus(G_n/2G_n)
    \end{equation*}
    so that \(\abs{G/2G}=2^n\). Since \(G/2G\cong H/2H\)
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_corollary
    If $F$ is a (finitely generated) free abelian group, then any two bases of
    $F$ have the same number of elements
    #+END_corollary

    #+BEGIN_proof
    If \(x_1,\dots,x_n\) is a basis of $F$, then \(F\cong\Z^n\)
    #+END_proof
    
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If $F$ is a free abelian group with basis \(x_1,\dots,x_n\), then $n$ is
    called the *rank* of $F$, and we write \(\rank(F)=n\)
    #+END_definition

    The rank of free abelian group plays the same role as the dimension of a
    vector space.

    #+ATTR_LATEX: :options []
    #+BEGIN_theorem
    label:thm5.11
    Let $F$ be a free abelian group with basis \(X=\{x_1,\dots,x_n\}\). If $G$
    is any abelian group and if \(\gamma:X\to G\) is any function, then there exists
    a unique homomorphism \(g:F\to G\) with \(g(x_i)=\gamma(x_i)\)

    \begin{center}
    \begin{tikzcd}
    F\arrow[dr,dashed,"g"]\\
    X\arrow[u]\arrow[r,"\gamma"']&G
    \end{tikzcd}
    \end{center}
    #+END_theorem

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    Let $A$ be an abelian group containing a subset \(X=\{x_1,\dots,x_n\}\), and
    let \(A\) have the property in ref:thm5.11. Then \(A\cong\Z^n\)
    #+END_proposition

    #+BEGIN_proof
    Consider
    \begin{center}
    \begin{tikzcd}
    A\arrow[dr,dashed,"g"]\\
    X\arrow[u,"p"]\arrow[r,"q"']&\Z^n
    \end{tikzcd}
    \hspace{0.2cm}
    and
    \hspace{0.4cm}
    \begin{tikzcd}
    \Z^n\arrow[dr,dashed,"h"]\\
    X\arrow[u,"q"]\arrow[r,"\gamma"']&A
    \end{tikzcd}
    \end{center}
    #+END_proof
*** Basis Theorem
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If $p$ is a prime, then an abelian group $G$ is *\(p\)-primary* if for each
    \(a\in G\), there is \(n\ge 1\) with \(p^na=0\)

    If $G$ is any abelian group, then its *\(p\)-primary component* is 
    \begin{equation*}
    G_p=\{a\in G:p^na=0\text{ for some }n\ge1\}
    \end{equation*}
    #+END_definition


    #+ATTR_LATEX: :options [Primary Decomposition]
    #+BEGIN_theorem
    label:nthm4.15
    1. Every finite abelian group $G$ is a direct sum of its \(p\)-primary
       components
       \begin{equation*}
       G=G_{p_1}\oplus\cdots\oplus G_{p_n}
       \end{equation*}
    2. Two finite abelian groups $G$ and $G'$ are isomorphic if and only if
       \(G_p\cong G_p'\) for every prime $P$
    #+END_theorem

    #+BEGIN_proof
    Let $x\in G$ be nonzero, and let its order be $d$.
    \begin{equation*}
    d=p_1^{e_1}\dots p_n^{e_n}
    \end{equation*}
    Define \(r_i=d/p_i^{e_i}\). It follows that \(r_ix\in G_{p_i}\). But the gcd
    of \(r_1,\dots,r_n\) is 1, hence \(1=\sum_is_ir_i\). Therefore
    \begin{equation*}
    x=\displaystyle\sum_{i}s_ir_ix\in G_{p_1}+\cdots+G_{p_n}
    \end{equation*}
    Write \(H_i=G_{p_1}+\cdots+\widehat{G_{p_i}}+\cdots+G_{p_n}\). By
    Proposition ref:prop5.4, it suffices to prove that 
    \(G_{p_i}\cap H_i=\{0\}\). If \(x\in G_{p_i}\cap H_i\), \(p_i^lx=0\), 
    \(x=\sum_{j\neq i}y_j\) where \(p_j^{g_j}y_j=0\). Hence 
    \(ux=0\) where \(u=\prod_{j\neq i}p_j^{g_j}\). But \(p_i^l\) and \(u\) are
    relatively prime, so \(1=sp_i^l+tu\). Therefore
    \begin{equation*}
    x=(sp_i^l+tu)x=0
    \end{equation*}
    #+END_proof

    [[index:pure subgroup]]
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    Let $p$ be a prime and let $G$ be a \(p\)-primary abelian group. A subgroup
    \(S\subseteq G\) is a *pure subgroup* if for all \(n\ge 0\)
    \begin{equation*}
    S\cap p^nG=p^nS
    \end{equation*}
    #+END_definition

    If \(s=p^ng\), then there is a \(s'\in S\) s.t. \(s=p^ns'\)

    #+ATTR_LATEX: :options []
    #+BEGIN_examplle
    label:nexample4.16
    1. Every direct summand \(S\) of \(G\) is a pure subgroup. Let 
       \(G=S\oplus T\) and \(s\in S\), and \(s\in S\). If \(s=p^n(u+v)\) for
       \(u\in S\) and \(v\in T\), then \(p^nv=s-p^nv\in S\cap T=\{0\}\) and \(s=p^nu\)
    2. If \(G=\la a\ra\) is a cyclic group of order \(p^2\), where \(p\) is a
       prime, then \(S=\la pa\ra\) is not a pure subgroup of \(G\), for 
       \(s=pa\in S\), but there is no element \(s'\in S\) with \(s=ps'\)
    #+END_examplle

    #+ATTR_LATEX: :options []
    #+BEGIN_lemma
    If $p$ is a prime and $G$ is a finite \(p\)-primary abelian group, then $G$
    has a nonzero pure cyclic subgroup. If \(y\) is an element of largest order
    in $G$, then \(\la y\ra\) is a pure cyclic subgroup
    #+END_lemma
          
    #+BEGIN_proof
    Since \(G\) is finite, there exists \(y\in G\) of largest order, say,
    \(p^l\). We claim that \(S=\la y\ra\) is a pure subgroup of \(G\).

    If \(s\in S\), then \(s=mp^ty\), where \(t\ge 0\) and \(p\nmid m\). Suppose
    that 
    \begin{equation*}
    s=p^na
    \end{equation*}
    for some \(a\in G\); an element \(s'\in G\) with \(s=mp^ty=p^ns'\) must be
    found. We may assume that \(n<l\); otherwise \(s=p^na=0\), and we may choose
    \(s'=0\).

    We claim that \(t\ge n\). If \(t<n\), then
    \begin{equation*}
    p^l a=p^{l-n}p^na=l^{l-n}s=p^{l-n}mp^ty=mp^{l-n+t}y
    \end{equation*}
    But \(p\nmid m\) and \(l-n+t<l\) and so \(p^la\neq 0\), a contradiction.
    Thus \(t\ge n\), and we can define \(s'=mp^{t-n}y\). Now \(s'\in S\) and
    \begin{equation*}
    p^ns'=mp^ty=s
    \end{equation*}
    so that \(S\) is a pure subgroup
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If \(p\) is a prime and \(G\) is a finite \(p\)-primary abelian group, then
    \begin{equation*}
    d(G)=\dim(G/pG)
    \end{equation*}
    #+END_definition
    
    Observe that \(d\) is additive over direct sum,
    \begin{equation*}
    d(G\oplus H)=d(G)+d(H)
    \end{equation*}
    for Proposition ref:nprop1.85 gives
    \begin{equation*}
    (G\oplus H)/p(G\oplus H)=(G\oplus H)(pG\oplus pH)\cong
    (G/pG)\oplus(H/pH)
    \end{equation*}
    

    #+ATTR_LATEX: :options []
    #+BEGIN_lemma
    If \(G\) is a finite \(p\)-primary abelian group, then \(d(G)=1\) if and
    only if \(G\) is a nonzero cyclic group
    #+END_lemma

    #+BEGIN_proof
    If \(G\) is a nonzero cyclic group, then so is any nonzero quotient of
    \(G\); in particular, \(G/pG\) is _cyclic_. Now \(G/pG\neq\{0\}\), by Exercise
    ref:nex4.3, and so \(\dim(G/pG)=1\) 

    _Conversely, if \(d(G)=1\)_ , then \(G/pG\cong\I_p\); hence \(G/pG\) is
    cyclic, say, \(G/pG=\la z+pG\ra\). Assume \(\la z\ra\) is a proper subgroup
    of \(G\). The Correspondence Theorem says that \(pG\) is a maximal subgroup
    of $G$. We claim that \(pG\) is the only maximal subgroup of \(G\). If 
    \(L\subseteq G\) is any maximal subgroup, then \(G/L\cong\I_p\) for \(G/L\)
    is a simple abelian group and has order \(p\). It follows that if 
    \(a\in G\), then \(p(a+L)=0\) in \(G/L\), and so \(pa\in L\); that is, 
    \(pG\subseteq L\). But here \(pG\) is a maximal subgroup, so that \(pG=L\).
    Every proper subgroup of \(G\) is contained in \(pG\). In particular,
    \(\la z\ra\subseteq pG\), so that the generator \(z+pG\) of \(G/pG\) is
    zero, a contradiction
    #+END_proof

    #+ATTR_LATEX: :options []
    #+BEGIN_lemma
    Let \(G\) be a finite \(p\)-primary abelian group
    1. If \(S\subseteq G\), then \(d(G/S)\le d(G)\)
    2. If \(S\) is a pure subgroup of $G$, then \(d(G)=d(S)+d(G/S)\)
    #+END_lemma
    #+BEGIN_proof
    1. By the Correspondence Theorem, \(p(G/S)=(pG+S)/S\), so that
       \begin{equation*}
       (G/S)/p(G/S)=(G/S)/[(pG+S)/S]\cong G/(pG+S)
       \end{equation*}
       by the Third Isomorphism Theorem. Since \(pG\subseteq pG+S\), there is a
       surjective homomorphism (of vector spaces over \(\F_p\))
       \begin{equation*}
       G/pG\to G/(pG+S)
       \end{equation*}. Hence
       \begin{equation*}
       \dim(G)(G/pG)\ge\dim(G/(pG+S))=d(G/S)
       \end{equation*}
    2. We now analyze \((pG+S)/pG\), the kernel of \(G/pG\to G/(pG+S)\). By the
       Second Isomorphism Theorem
       \begin{equation*}
       (pG+S)/pG\cong S/(S\cap pG)
       \end{equation*}
       Since \(S\) a pure subgroup, \(S\cap pG=pS\); therefore
       \begin{equation*}
       (pG+S)/pG\cong S/pS
       \end{equation*}
       and so \(\dim[(pG+S)/pG]=d(S)\). But if \(W\) is a subspace of a
       finite-dimensional vector space \(V\), then
       \(\dim(V)=\dim(W)+\dim(V/W)\), by Exercise ref:nex2.77. Hence for
       \(V=G/pG\), \(W=(pG+S)/pG\), we have \(d(G)=d(S)=d(G/S)\)
    #+END_proof

    #+ATTR_LATEX: :options [Basis Theorem]
    #+BEGIN_theorem
    Every finite abelian group \(G\)  is a direct sum of primary cyclic groups
    #+END_theorem

    #+BEGIN_proof
    By the Primary Decomposition, Theorem ref:nthm4.15, we may assume that \(G\)
    is \(p\)-primary for some prime \(p\).
    #+END_proof

    #+BEGIN_exercise
    label:nex4.3
    Let \(G\) be a \(p\)-primary abelian group. If \(G=pG\), prove that either
    \(G=\{0\}\) or \(G\) is infinite.
    #+END_exercise

    #+BEGIN_proof
    \(G=p^kG\) for any \(k\ge 0\). If \(G\) is finite, then there exists 
    \(y\in G\) with largest order \(l\) and \(p^lG=\{0\}=G\)
    #+END_proof
* Commutative Rings \rom{2}
  :PROPERTIES:
  :PP:       333
  :END:
** Prime Ideals and Maximal Ideals
    [[index:correspondence theorem for rings]]
   #+ATTR_LATEX: :options [Correspondence Theorem for Rings]
   #+BEGIN_proposition
   If $I$ is a proper ideal in a commutative ring $R$, then there is an
   inclusion-preserving bijection \varphi from the set of all intermediate ideals $J$
   containing $I$, that is, \(I\subseteq J\subseteq R\), to the set of all the
   ideals in $R/I$, given by
   \begin{equation*}
   \varphi:J\mapsto\pi(J)=J/I
   \end{equation*}
   where \pi is the natural map
   \begin{center}
   \begin{tikzcd}
   R \arrow[d,dash] \arrow[rd]&\\
   J' \arrow[d,dash] \arrow[rd] & R/I \arrow[d,dash]\\
   J \arrow[d,dash] \arrow[rd] & J'/I \arrow[d,dash]\\
   I  \arrow[rd] & J/I \arrow[d,dash]\\
   & \{0\}
   \end{tikzcd}
   \end{center}
   #+END_proposition

   #+BEGIN_proof
   If we forget its multiplication, the commutative ring $R$ is merely an
   additive group and its ideal is a (normal) subgroup. Theorem ref:thm2.76
   gives an inclusion-preserving bijection
   \begin{equation*}
   \Phi:\{\text{all subgroups of $R$ containing $I$}\}\to 
   \{\text{all subgroups of $R/I$}\}
   \end{equation*}
   where \(\Phi(J)=\pi(J)=J/I\)

   If $J$ is an ideal, then \(\Phi(J)\) is also an ideal.
   #+END_proof

   In practice, the correspondence theorem is invoked, tacitly, by saying that
   every ideal in the quotient ring \(R/I\) has the form \(J/I\) for some ideal
   $J$ with \(I\subseteq J\subseteq R\)

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   Let \(I=(m)\) be a nonzero ideal in \(\Z\). If $J$ is an ideal in $\Z$
   containing \(I\), then \(J=(a)\) for some \(a\in\Z\) because $\Z$ is a PID
   and \((m)\subseteq(a)\) iff \(a\mid m\). The correspondence theorem shows
   that every ideal in \(\I_m\) has the form \(([a])\) for some divisor $a$ of $m$
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   An ideal $I$ in a commutative ring $R$ is called a *prime ideal* if it is a
   proper ideal and \(ab\in I\) implies \(a\in I\) or \(b\in I\)
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   We claim that the prime ideals in \(\Z\) are precisely the ideals \((p)\),
   where either \(p\) is 0 or a prime.
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   An ideal $I$ in a commutative ring $R$ is a prime ideal if and only if
   \(R/I\) is a domain
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $K$ is a field, then a nonzero polynomial \(p(x)\in k[x]\) is irreducible
   if and only if \((p(x))\) is a prime ideal
   #+END_proposition

   #+BEGIN_proof
   Suppose that \(p(x)\) is irreducible. First \((p(x))\) is a proper ideal.
   Otherwise \(1\in (p(x))\), so there is a polynomial \(f(x)\) with
   \(1=p(x)f(x)\). But \(p(x)\) has degree at least 1

   Second, if \(ab\in(p)\), then \(p\mid ab\), and Euclid's lemma in $k[x]$
   gives \(p\mid a\) or \(p\mid b\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   An ideal $I$ in a commutative ring $R$ is a *maximal ideal* if it is a proper
   ideal and there is no ideal $J$ with \(I\not\subseteq J\not\subseteq R\)
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   A proper ideal $I$ in a nonzero commutative ring $R$ is a maximal ideal if
   and only if \(R/I\) is a field
   #+END_proposition

   #+BEGIN_proof
   The correspondence theorem shows that $I$ is maximal if and only if \(R/I\)
   has no ideals other than \(\{0\}\) and \(R/I\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   label:cor6.8
   Every maximal ideal $I$ in a commutative ring $R$ is a prime ideal
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   The converse of Corollary ref:cor6.8 is false. Consider the principal ideal
   \((x)\) in \(\Z[x]\). By Exercise ref:ex3.83
   \begin{equation*}
   \Z[x]/(x)\cong\Z
   \end{equation*}
   since \(\Z\) is a domain, \(x\) is a prime ideal. Since \(\Z\) is not a field,
   \((x)\) is not a maximal ideal. Let 
   \begin{equation*}
   J=\{f(x)\in\Z[x]:f(x)\text{ has even constant term}\}
   \end{equation*}
   Since \(\Z[x]/J\cong\F_2\) is a field, it follows that $J$ is a maximal ideal
   containing \((x)\)
   #+END_examplle
   
   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   Let $k$ be a field, and let \(a=(a_1,\dots,a_n)\in k^n\). Define the
   *evaluation map*
   \begin{equation*}
   e_a:k[x_1,\dots,x_n]\to k
   \end{equation*}
   by
   \begin{equation*}
   e_a:f(x_1,\dots,x_n)\mapsto f(a)
   \end{equation*}
   \(e_a\) is a surjective ring homomorphism, and so \(\ker e_a\) is a maximal
   ideal
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   If $R$ is a PID, then every nonzero prime ideal $I$ is a maximal ideal
   #+END_theorem

   #+BEGIN_proof
   Assume that there is a proper ideal $J$ with \(I\subseteq J\). Since $R$ is a
   PID, \(I=(a)\) and \(I=(b)\) for some \(a,b\in R\). Now \(a\in J\) implies
   that \(a=rb\) for some \(r\in R\) and so \(rb\in I\). Since $I$ is prime,
   either \(r\in I\) or \(b\in I\). If \(r\in I\), then \(r=ta\) for some 
   \(t\in I\), and \(a=tab=atb\). Since $R$ is a domain, \(1=tb\). Hence \(J=R\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $k$ is a field and \(p(x)\in k[x]\) is irreducible, then the quotient ring 
   \(k[x]/(p(x))\) is a field
   #+END_corollary

   #+BEGIN_proof
   Since \(p(x)\) is irreducible, \((p(x))\) is a prime ideal. Since \(k[x]\) is
   a PID, \((p(x))\) is maximal
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let $P$ be a prime ideal in a commutative ring $R$. If $I$ and $J$ are ideals
   with \(IJ\subseteq P\), then \(I\subseteq P\) or \(J\subseteq P\)
   #+END_proposition

   #+BEGIN_proof
   Suppose, on the contrary, that \(I\not\subseteq P\) and \(J\not\subseteq P\);
   thus there are \(a\in I\) and \(b\in J\) with \(a,b\not\in P\). But 
   \(ab\in IJ\subseteq P\), contradicting $P$ being prime
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let $B$ be a subset of a commutative ring $R$ which is closed under addition
   and multiplication
   1. Let \(J_1,\dots,J_n\) be ideals in $R$, at least \(n-2\) which are prime.
      If \(B\subseteq J_1\cup\cdots\cup J_n\), then $B$ is contained in some
      $J_i$
   2. Let $I$ be an ideal in $R$ with \(I\subsetneq B\). If there are prime
      ideals $P_1,\dots,P_n$ s.t. \(B-I\subseteq P_1\cup\dots\cup P_n\), then 
      \(B\subseteq P_i\) for some $i$

   #+END_proposition

   #+BEGIN_proof
   1. Induction on \(n\ge2\). If \(B\not\subseteq J_2\), then there is \(b_1\in B\)
      with \(b_1\not\in J_2\), and hence \(b_1\in J_1\). If \(B\not\subseteq J_1\),
      there is \(b_2\in B\) with \(b_2\not\in J_1\) and \(b_2\in J_2\). However if
      \(y=b_1+b_2\), then \(y\not\in J_1\) and \(y\not\in J_2\), contradicting 
      \(B\subseteq J_1\cup J_2\)
      
      For the inductive step, assume that \(B\subseteq J_1\cup\cdots\cup J_{n+1}\),
      where at least \(n-1=(n+1)-2\) of the \(J_i\) are prime ideals. Let
      \begin{equation*}
      D_i=J_1\cup\cdots\cup\widehat{J_i}\cup\cdots\cup J_{n+1}
      \end{equation*}
      the inductive hypothesis allows us to assume that \(B\not\subseteq D_i\) for
      all \(i\). Hence for all \(i\), there exists \(b_i\in B\) with 
      \(b_i\not\in D_i\); since \(B\subseteq D_i\cup J_i\), we must have 
      \(b_i\in J_i\). Now \(n\ge3\), so that at least one of the \(J_i\) is a prime
      ideal. Assume \(J_1\) is prime. Consider the elements
      \begin{equation*}
      y=b_1+b_2b_3\dots b_{n+1}
      \end{equation*}
      \(y\not\in J_i\) for any i
   2. \(B\subseteq I\cup P_1\cup\cdots\cup P_n\)
   #+END_proof
** Unique Factorization Domains
   [[index:associate]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Elements $a$ and $b$ in a commutative ring $R$ are *associates* if there exists
   a unit \(u\in R\) with \(b=ua\)
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let $R$ be a domain and let \(a,b\in R\)
   1. \(a\mid b\) and \(b\mid a\) if and only if \(a\) and \(b\) are associates
   2. The principal ideals \((a)\)  and \((b)\) are equal if and only if \(a\)
      and $b$ are associates
   #+END_proposition

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $R$ is a PID and \(p\in R\) is irreducible, then \((p)\) is a prime ideal
   #+END_corollary

   #+BEGIN_proof
   $(p)$ is maximal
   #+END_proof

   [[index: UFD]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A domain $R$ is a *unique factorization domain* (*UFD*) if
   1. every \(r\in R\), neither 0 nor a unit, is a product of irreducibles
   2. if \(up_1\dots p_m=vq_1\dots q_n\), where $u$ and \(v\) are units and
      \(p_i,q_i\) are irreducible, then \(m=n\) and there is a permutation 
   \(\sigma\in S_n\) with \(p_i\) and \(q_{\sigma(i)}\) associates for all $i$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop6.17
   Let $R$ be a domain in which every \(r\in R\), neither 0 nor a unit, is a
   product of irreducibles. Then $R$ is a UFD if and only if \((p)\) is a prime
   ideal in $R$ for every irreducible element \(p\in R\)
   #+END_proposition
   
   #+BEGIN_proof
   Assume that $R$ is a UFD. If \(a,b\in R\) and \(ab\in(p)\), then there is
   \(r\in R\) with
   \begin{equation*}
   ab=rp
   \end{equation*}
   Factor each of \(a,b,r\) into irreducibles; by unique factorization, the left
   side of the equation must involve an associate of $p$

   Assume that 
   \begin{equation*}
   up_1\dots p_m=vq_1\dots q_n
   \end{equation*}
   where \(p_i\) and \(q_j\) are irreducibles and \(u,v\) are units. We prove,
   by induction on \(\max\{m,n\}\ge1\). If \(\max{m,n}=1\), then 
   \(up_1=v,u=vq_1\) or \(up_1=vq_1\). The first two cannot happen, and so the
   base step is true. For the inductive steop, the given equation shows that 
   \(p_1\mid q_1\dots q_n\). By hypothesis, \((p_1)\) is a prime ideal, and so
   there is some \(q_j\) with \(p_1\mid q_j\), so that \(p_1\) and \(q_j\) are
   associates. Canceling \(p_1\) from both side.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   1. If $R$ is a commutative ring and
      \begin{equation*}
      I_1\subseteq I_2\subseteq\cdots\subseteq I_n\subseteq\cdots
      \end{equation*}
      is an ascending chain of ideals of \(R\), then \(J=\bigcup_{n\ge1}I_n\) is
      an ideal in $R$
   2. If $R$ is a PID, then it has no infinite strictly ascending chain of
      ideals 
      \begin{equation*}
      I_1\subsetneq I_2\subsetneq\cdots\subsetneq I_n\subsetneq\cdots
      \end{equation*}
   3. Let $R$ be a PID. If \(r\in R\) is neither 0 nor a unit, then $r$ is a
      product of irreducibles
   #+END_lemma

   #+BEGIN_proof
   1. If \(a\in J\), then \(a\in I_n\) for some $n$; if \(r\in R\), then 
      \(ra\in I_n\); hence \(ra\in J\). If \(a,b\in J\), then \(a\in I_m\) and 
      \(a\in I_n\)...
   2. $J$ is principal ideal domain and \(J=(d)\), then
      \begin{equation*}
      J=(d)\subseteq I_n\subsetneq I_{n+1}\subseteq J
      \end{equation*}
   3. A divisor \(r\) of an element \(a\in R\) is called a *proper divisor* of $a$.
      Call a nonzero nonunit \(a\in R\) *good* if it is a product of irreducibles.
      If $a$ is bad, then \(a=rs\), where both \(r\) and $s$ are proper
      divisors. But the product of good elements is good, and so at least one of
      the factors, say $r$, is bad. It follows , by induction, that there exists
      a sequence \(a_1=a,a_2=r,a_3,\dots,a_n,\dots\) of bad elements and yields
      a strictly ascending chain
      \begin{equation*}
      (a_1)\subsetneq(a_2)\subsetneq\dots
      \end{equation*}
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   If $R$ is a PID, then $R$ is a UFD.
   #+END_theorem

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   If $R$ is a UFD, then a gcd of any finite set of elements \(a_1,\dots, a_n\)
   in $R$ exists
   #+END_proposition

   #+BEGIN_proof
   \begin{align*}
   &a=up_1^{e_1}\dots p_t^{e_t}\\
   &b=vp_1^{f_1}\dots p_t^{f_t}
   \end{align*}
   where \(e_i,f_i\ge0\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Elements \(a_1,\dots,a_n\) in a UFD $R$ is called *relatively prime* if their
   gcd is a unit
   #+END_definition

   [[index:primitive polynomial]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A polynomial \(f(x)=a_nx^n+\dots+a_1x+a_0\in R[x]\), where $R$ is a UFD, is
   called *primitive* if its coefficients are relatively prime
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   For a UFD $R$, every irreducible \(p(x)\in R[x]\) of positive degree is
   primitive. 
   #+END_examplle

   #+ATTR_LATEX: :options [Gauss's Lemma]
   #+BEGIN_lemma
   If $R$ is a UFD and \(f(x),g(x)\in R[x]\) are both primitive, then their
   product \(f(x)g(x)\) is also primitive
   #+END_lemma

   #+BEGIN_proof
   If \(\pi:R\to R/(p)\) is the natural map \(\pi:a\mapsto a+(p)\), then Proposition
   ref:prop3.48 shows that the function \(\widetilde{\pi}:R[x]\to(R/(p))[x]\) is a
   ring homomorphism. If a polynomial \(h(x)\in R[x]\) is not primitive, there is
   some irreducible $p$ s.t. all the coefficients of \(\widetilde{\pi}(h)\) are 0
   in \(R/(p)\); that is, \(\widetilde{\pi}(h)=0\) in \(R/(p)[x]\). Thus, if the
   product \(f(x)g(x)\) is not primitive, there is some irreducible $p$ with 
   \(0=\widetilde{\pi}(fg)=\widetilde{\pi}(f)\widetilde{\pi}(g)\). Since \((p)\) is a
   prime ideal, \(R/(p)\) is a domain, and hence \((R/(p))[x]\) is also a
   domain. But neither \(\widetilde{\pi}(f)\) nor \(\widetilde{\pi}(g)\) are 0 in
   \((R/(p))[x]\), a contradiction
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:lemma6.24
   Let $R$ be a UFD, let \(Q=\Frac(R)\), and let \(f(x)\in Q[x]\) be nonzero
   1. There is a factorization
      \begin{equation*}
      f(x)=c(f)f^*(x)
      \end{equation*}
      where \(c(f)\in Q\) and \(f^*(x)\in R[x]\) is primitive. This
      factorization is unique in the sense that if \(f(x)=qg^*(x)\), where
      \(q\in Q\) and \(g^*(x)\in R[x]\) is primitive, then there is a unit 
      \(w\in R\) with \(q=wc(f)\) and \(g^*(x)=w^{-1}f^*(x)\)
   2. If \(f(x),g(x)\in R[x]\), then \(c(fg)\) and \(c(f)c(g)\) are associates
      in $R$ and \((fg)^*\) and \(f^*g^*\) are associates in \(R[x]\)
   3. Let \(f(x)\in Q[x]\) have a factorization \(f(x)=qg^*(x)\), where 
      \(q\in Q\) and \(g^*(x)\in R[x]\) is primitive. Then \(f(x)\in R[x]\) if
      and only if \(q\in R\)
   4. Let \(g^*(x),f(x)\in R[x]\). If \(g^*(x)\) is primitive and 
      \(g^*(x)\mid b f(x)\), where \(b\in R\) and \(b\neq0\), then 
      \(g^*(x)\mid f(x)\)
   #+END_lemma

   #+BEGIN_proof
   1. Clearing denominators, there is \(b\in R\) with \(bf(x)\in R[x]\). If
      \(d\) is the gcd of the coefficients of \(bf(x)\), then \((b/d)f(x)\in
      R[x]\) is a primitive polynomial. If we define \(c(f)=d/b\) and 
      \(f^*(x)=(b/d)f(x)\)

      Suppose \(c(f)f^*(x)=qg^*(x)\). Exercise ref:ex6.17 allows use to write
      \(q/c(f)\) in lowest terms: \(q/c(f)=u/v\). The equation 
      \(vf^*(x)=ug^*(x)\) holds in \(R[x]\). Since \(v,u\) are relatively prime
      and \(g^*(x)\) is primitive, \(v\) is a unit

   4. [@4] Since \(bf=hg^*\), we have \(bc(f)f^*=c(h)h^*g^*=c(h)(hg)^*\). By
      uniqueness, \(f^*\) and \((hg)^*\) are associates
      
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let $R$ be a UFD with \(Q=\Frac(R)\). If \(f(x)\in Q[x]\), there is a
   factorization \(f(x)=c(f)f^*(x)\), where \(c(f)\in Q\) and 
   \(f^*(x)\in R[x]\) is primitive. We call \(c(f)\) the *content* of \(f(x)\) and
   \(f^*(x)\) the *associated primitive polynomial*
   #+END_definition

   #+ATTR_LATEX: :options [Gauss]
   #+BEGIN_theorem
   If $R$ is a UFD, then \(R[x]\) is also a UFD
   #+END_theorem

   #+BEGIN_proof
   We show first, by induction on \(\deg(f)\), that every \(f(x)\in R[x]\),
   neither 0 nor a unit, is a product of irreducibles. If \(\deg(f)>0\), then 
   \(f(x)=c(f)f^*(x)\) where \(c(f)\in R\) and \(f^*(x)\) is primitive. If
   \(f^*(x)\) is irreducible, we are done. Otherwise \(f^*(x)=g(x)h(x)\), where
   neither $g$ nor $h$ is a unit. And so each is is a product of irreducibles,
   by the inductive hypothesis

   Now Proposition ref:prop6.17 applies: \(R[x]\) is a UFD if \((p(x))\) is a
   prime ideal for every irreducible \(p(x)\in R[x]\). Let's assume 
   \(p\mid fg\) and \(p\nmid f\)

   Case (i). Suppose that \(\deg(p)=0\) . Write
   \begin{equation*}
   f(x)=c(f)f^*(x),\quad g(x)=c(g)g^*(x)
   \end{equation*}
   Now \(p\mid fg\), so that
   \begin{equation*}
   p\mid c(f)c(g)f^*(x)g^*(x)
   \end{equation*}
   Since \(f^*(x)g^*(x)\) is primitive, Lemma ref:lemma6.24 says that
   \(c(f)c(g)\) is an associate of \(c(fg)\). However, if \(p\mid fg\), then $p$
   divides each coefficient of \(fg\); that is, $p$ is a common divisor of all
   coefficients of $fg$, and hence in $R$, which is a UFD, $p$ divides the
   associates \(c(fg)\) and \(c(f)c(g)\). But Proposition ref:prop6.17 says that
   \((p)\) is a prime ideal, and so \(p\mid c(f)\) or \(p\mid c(g)\). Therefore
   \(p\mid c(g)\)

   Case (ii). Suppose that \(\deg(p)>0\). Let 
   \begin{equation*}
   (p,f)=\{s(x)p(x)+t(x)f(x):s(x),t(x)\in R[x]\}
   \end{equation*}
   Choose \(m(x)\in(p,f)\) of minimal degree. If \(Q=\Frac(R)\) is the fraction
   field of $R$, then the division algorithm in \(Q[x]\) gives polynomials
   \(q'(x),r'(x)\in Q[x]\) with
   \begin{equation*}
   f(x)=m(x)q'(x)+r'(x)
   \end{equation*}
   where either \(r'(x)=0\) or \(\deg(r')<\deg(m)\). Clearing denominators,
   there are polynomials \(q(x),r(x)\in R[x]\) and a constant \(b\in R\) with
   \begin{equation*}
   bf(x)=q(x)m(x)+r(x)
   \end{equation*}
   Since \(m\in(p,f)\), there are polynomials \(s(x),t(x)\in R[x]\) with 
   \(m=sp+tx\); hence \(r=bf-qm\in(b,f)\). Since \(m\) has minimal degree, we
   must have \(r=0\); that is, \(bf(x)=q(x)m(x)\), and so
   \(bf(x)=c(m)m*(x)q(x)\). So that \(m^*(x)\mid f(x)\) by Lemma ref:lemma6.24.
   A similar argument, replacing \(f(x)\) by \(p(x)\), gives 
   \(m^*(x)\mid p(x)\). If \(m^*(x)\) were an associate of \(p(x)\), then
   \(p(x)\mid f(x)\), contrary to hypothesis. Hence \(m^*(x)\) is a unit; that
   is, \(m(x)=c(m)\in R\), and so \(p,f\) contains the nonzero constant
   \(c(m)\). Now \(c(m)=sp+tf\), and so
   \begin{equation*}
   c(m)g(x)=s(x)p(x)g(x)+t(x)f(x)g(x)
   \end{equation*}
   Since \(p\mid fg\), we have \(p(x)\mid c(m)g(x)\). But \(p(x)\) is primitive, 
   \(p(x)\mid g(x)\)
   #+END_proof
   
   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $k$ is a field, then \(k[x_1,\dots,x_n]\) is a UFD
   #+END_corollary
   
   #+ATTR_LATEX: :options [Gauss]
   #+BEGIN_corollary
   label:ncor5.28
   Let $R$ be a UFD, let \(Q=\Frac(R)\), and let \(f(x)\in R[x]\) . If 
   \begin{equation*}
   f(x)=G(x)H(x)\in Q[x]
   \end{equation*}
   then there is a factorization
   \begin{equation*}
   f(x)=g(x)h(x)\in R[x]
   \end{equation*}
   where \(\deg(g)=\deg(G)\) and \(\deg(h)=\deg(H)\); in fact, \(G(x)\) is a
   constant multiple of \(g(x)\) and \(H(x)\) is a constant multiple of
   \(h(x)\). Therefore, if \(f(x)\) does not factor into polynomials of smaller
   degree in \(R[x]\), then \(f(x)\) is irreducible in \(Q[x]\)
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   We claim that \(f(x,y)=x^2+y^2-1\in k[x,y]\) is irreducible, where $k$ is a
   field. Write \(Q=k(y)=\Frac(k[y])\), and view \(f(x,y)\in Q[x]\). Now the
   quadratic \(g(x)=x^2+(y^2-1)\) is irreducible in \(Q[x]\) iff it has no roots
   in \(Q=k(y)\), and this is so by Exercise ref:ex3.34

   It follows from Proposition ref:prop6.17 that \((x^2+y^2-1)\) is a prime
   ideal because it is generated by an irreducible polynomial
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:nprop5.30
   Let $k$ be a field, and view \(f(x_1,\dots,x_n)\in k[x_1,\dots,x_n]\) as a
   polynomial in \(R[x_n]\), where \(R=k[x_1,\dots,x_{n-1}]\)
   \begin{equation*}
   f(x_n)=a_0(x_1,\dots,x_{n-1})+\dots+a_m(x_1,\dots,x_{n-1})x^m_n
   \end{equation*}
   If \(f(x_n)\) is primitive and cannot be factored into two polynomials of
   lower degree in \(R[x_n]\), then \(f(x_1,\dots,x_n)\) is irreducible in
   \(k[x_1,\dots,x_n]\) 
   #+END_proposition

   #+BEGIN_proof
   Suppose that \(f(x_n)=g(x_n)h(x_n)\) in \(R[x_n]\); by hypothesis, the
   degrees of \(g\) and $h$ in $x_n$ cannot both be less than \(\deg(f)\); say,
   \(\deg(g)=0\). It follows, because $f$ is primitive, that $g$ is a unit in
   \(k[x_1,\dots,x_{n-1}]\). Therefore, \(f(x_1,\dots,x_n)\) is irreducible in \(R[x_n]\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   label:ncor5.31
   If $k$ is a field and 
   \(g(x_1,\dots,x_n),h(x_1,\dots,x_n)\in k[x_1,\dots,x_n]\) are relatively
   prime, then \(f(x_1,\dots,x_n,y)=yg(x_1,\dots,x_n)+h(x_1,\dots,x_n)\) is
   irreducible in \(k[x_1,\dots,x_n,y]\)
   #+END_corollary

   #+BEGIN_proof
   Let \(R=k[x_1,\dots,x_n]\). Note that $f$ is primitive in $R[y]$, because
   \((g,h)=1\) forces any divisor of its coefficients \(g,h\) to be a unit.
   Since $f$ is linear in $y$, it is not the product of two polynomials in $R[y]$
   of smaller degree, and hence Proposition ref:nprop5.30 shows that $f$ is
   irreducible in \(R[y]=k[x_1,\dots,x_n,y]\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   The polynomials $x$ and \(y^2+z^2-1\) are relatively prime in \(\R[x,y,z]\),
   so that \(f(x,y,z)=x^2+y^2+z^2-1\) is irreducible
   #+END_examplle


   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If \alpha is an algebraic integer, then \(\irr(\alpha,\Q)\) lies in \(\Z[x]\)
   #+END_corollary


   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If \alpha is an algebraic integer, then its *minimal polynomial* is the monic
   polynomial in \(\Z[x]\) of least degree having \alpha as a root.
   #+END_definition

   #+BEGIN_remark
   We define the (algebraic) *conjugates* of \alpha to be the roots of \(\irr(\alpha,\Q)\),
   and we define the *norm* of \alpha to be the absolute value of the product of the
   conjugates \alpha
   #+END_remark

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   Let \(f(x)=a_0+a_1x+\dots+x^n\in\Z[x]\) be monic, and let $p$ be a prime. If
   \(f(x)\) is irreducible\(\mod p\), that is, if
   \begin{equation*}
   \widetilde{f}(x)=[a_0]+[a_1]x+\dots+x^n\in\F_p[x]
   \end{equation*}
   is irreducible, then \(f(x)\) is irreducible in \(\Q[x]\)
   #+END_theorem

   #+BEGIN_proof
   By Proposition ref:prop3.48, the natural map \(\varphi:\Z\to\F_p\) defines a
   homomorphism \(\widetilde{\varphi}:\Z[x]\to\F_p\). If \(g(x)\in\Z[x]\), define its
   image \(\widetilde{\varphi}(g(x))\in\F_p[x]\) by \(\widetilde{g}(x)\). Prove
   \(f(x)\) is irreducible in \(\Z[x]\). Then by Gauss's theorem, \(f(x)\) is
   irreducible in \(\Q[x]\)
   #+END_proof



   #+BEGIN_exercise
   label:ex6.17
   Let $R$ be a UFD and let \(Q=\Frac{R}\) be its fraction field. Prove that
   each nonzero \(a/b\in Q\) has an expression in lowest terms; that is, \(a\)
   and \(b\) are relatively prime.
   #+END_exercise

   #+BEGIN_proof
   there gcd exists
   #+END_proof

   #+BEGIN_exercise
   label:ex6.18
   Let $R$ be a UFD
   1. If \(a,b,c\in R\) and $a$ and $b$ are relatively prime, prove that 
      \(a\mid bc\) implies \(a\mid c\)
   2. If \(a,c_1,\dots,c_n\in R\) and \(c_i\mid a\) for all \(i\), prove that 
      \(c\mid a\), where \(c=\lcm\{c_1,\dots,c_n\}\)
   #+END_exercise

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. We show that \(f(x)=x^4-5x^3+2x+3\)  is an irreducible polynomial in
      \(\Q[x]\)

      The only candidates for rational roots are \(1,-1,3,-3\) and none of these
      is a root.
      
      Since \(\widetilde{f}(x)=x^4+x^3+1\in\F_2[x]\) is irreducible by Example
      ref:example3.35, it follows that \(f(x)\) is irreducible in \(\Q[x]\).

   2. Let \(\Phi_5(x)=x^4+x^3+x^2+x+1\in\Q[x]\)
      
      \(\widetilde{\Phi}_5(x)\) is irreducible in \(\F_2[x]\), and so 
      \(\Phi_5(x)\) is irreducible in \(\Q[x]\)
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:nlemma2.74
   Let \(g(x)\in\Z[x]\). If there is \(c\in\Z\) with \(g(x+c)\) irreducible in
   \(\Z[x]\), then \(g(x)\) is irreducible in \(\Q[x]\)
   #+END_lemma

   #+BEGIN_proof
   \(\varphi:\Z[x]\to\Z[x]\), given by \(f(x)\mapsto f(x+c)\) is an isomorphism. If
   \(g(x)=s(x)t(x)\), then \(g(x+c)=\varphi(g(x))=\varphi(s)\varphi(t)\). Therefore
   \(g(x)\) is irreducible in \(\Q\).
   #+END_proof

   #+ATTR_LATEX: :options [Eisenstein Criterion]
   #+BEGIN_theorem
   Let $R$ be a UFD with \(Q=\Frac(R)\), and let 
   \(f(x)=a_0+a_1x+\dots+a_nx^n\in R[x]\). If there is an irreducible element 
   \(p\in R\) with \(p\mid a_i\) for all \(i<n\) but with \(p\nmid a_n\) and 
   \(p^2\nmid a_0\), then \(f(x)\) is irreducible in \(Q[x]\)
   #+END_theorem

   #+BEGIN_proof
   Let \(\widetilde{\varphi}:\Z[x]\to\F_p[x]\) and let \(\bar{f}(x)\) denote 
   \(\widetilde{\varphi}(f(x))\). If \(f(x)\) is not irreducible in \(\Q[x]\), then
   Gauss's Theorem gives polynomials \(g(x),h(x)\in\Z[x]\) with
   \(f(x)=g(x)h(x)\), where \(g(x)=b_0+\dots+b_mx^m,h(x)=c_0+\dots+c_kx^k\).
   Thus \(\bar{f}=\bar{g}\bar{h}\).

   Since \(p\nmid a_n\), we have \(\bar{f}(x)\neq 0\); in fact,
   \(\bar{f}(x)=ux^n\) for some unit \(u\in\F_p\), because all its coefficients
   aside from its leading coefficient are 0. By unique factorization in
   \(\F_p[x]\), we must have \(\bar{g}(x)=vx^m\) and \(\bar{h}(x)=wx^k\) (for
   units \(v,w\in \F_p\)); equivalently, \(p\mid b_0\) and \(p\mid c_0\). But
   \(a_0=b_0c_0\), and so \(p^2\mid a_0\), a contradiction.
   #+END_proof
** Noetherian Rings
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A commutative ring $R$ satisfies the *ACC*, the *ascending chain condition*, if
   every ascending chain of ideals
   \begin{equation*}
   I_1\subseteq\cdots\subseteq I_n\subseteq\cdots
   \end{equation*}
   stops.
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $X$ is a subset of a commutative ring $R$, then the *ideal generated by $X$*
   is the set of all finite linear combinations
   \begin{equation*}
   I=(X)=\{\displaystyle\sum_{\text{finite}}r_ix_i:r_i\in R,x_i\in X\}
   \end{equation*}
   We say that $I$ is *finitely generated*, often abbreviated to f.g., if
   \(X=\{a_1,\dots,a_n\}\). We write
   \begin{equation*}
   I=(a_1,\dots,a_n)
   \end{equation*}
   and we call $I$ the *ideal generated by* \(a_1,\dots,a_n\)

   A set of generators \(a_1,\dots,a_n\) of an ideal $I$ is sometimes called a
   *basis* of $I$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   The following conditions are equivalent for a commutative ring $R$
   1. $R$ has the ACC
   2. $R$ satisfies the *maximum condition*: Every nonempty family \(\calf\) of
      ideals of $R$ has a maximal element
   3. Every ideal in $R$ is finitely generated
   #+END_proposition

   #+BEGIN_proof
   (2) \(\to\) (3). Let $I$ be an ideal in $R$, and define $\calf$ to be the
   family of all the finitely generated ideals in $I$; of course,
   \(\calf\neq\emptyset\). By hypothesis, there exists a maximal element
   \(M\in\calf\). If \(M\subsetneq I\), then there is \(a\in I\) with 
   \(a\not\in M\). The ideal
   \begin{equation*}
   J=\{m+ra:m\in M,r\in R\}\subseteq I
   \end{equation*}
   is finitely generated, and so \(J\in\calf\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A commutative ring $R$ is called *noetherian* if every ideal in $R$ is finitely generated
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $I$ is a proper ideal in a noetherian ring $R$, then there exists a
   maximal ideal $M$ in $R$ containing $I$.
   #+END_corollary

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $R$ is a noetherian ring and $I$ is an ideal in $R$, then \(R/I\) is also
   noetherian 
   #+END_corollary

   #+BEGIN_proof
   Correspondence theorem
   #+END_proof

   #+ATTR_LATEX: :options [Hilbert Basis Theorem]
   #+BEGIN_theorem
   If $R$ is a commutative noetherian ring, then \(R[x]\) is also noetherian
   #+END_theorem

   #+BEGIN_proof
   Assume that $I$ is an ideal in $R[x]$ that is not finitely generated,
   \(I\neq\{0\}\). Define \(f_0(x)\) to be a polynomial in $I$ of minimal degree
   and define, inductively, \(f_{n+1}(x)\) to be a polynomial of minimal degree
   in \(I-(f_0,\dots,f_n)\). It is clear that 
   \begin{equation*}
   \deg(f_0)\le\deg(f_1)\le\cdots
   \end{equation*}
   Let \(a_n\) denote the leading coefficient of \(f_n(x)\). Since $R$ is
   noetherian, Exercise ref:ex6.32 applies to give an integer $m$ with
   \(a_{m+1}\in(a_1,\dots,a_m)\). Define
   \begin{equation*}
   f^*(x)=f_{m+1}(x)-\displaystyle\sum_{i=0}^mx^{d_{m+1}-d_i}r_if_i(x)
   \end{equation*}
   where \(d_i=\deg(f_i)\). Now \(f^*(x)\in I-(f_0,\dots,f_m)\). It suffices to
   show that \(\deg(f^*)<\deg(f_{m+1})\), for this contradicts \(f_{m+1}(x)\)
   having minimal degree.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   1. If \(k\) is a field, then \(k[x_1,\dots,x_n]\) is noetherian
   2. The ring \(\Z[x_1,\dots,x_n]\) is noetherian
   3. For any ideal $I$ in \(k[x_1,\dots,x_n]\) where \(k=\Z\) or $k$ is a
      field, the quotient ring \(k[x_1,\dots,x_n]/I\) is noetherian
   #+END_corollary

   #+BEGIN_exercise
   label:ex6.32
   Let $R$ be a commutative ring. Prove that \(R\) is noetherian if and only if
   for every sequence \(a_1,\dots,a_n,\dots\) of elements in $R$ ,there is an
   integer \(m\ge1\) with \(a_{m+1}\) an \(R\)-linear combination of its predecessors
   #+END_exercise

   #+BEGIN_proof
   \((a_1),(a_1,a_2),(a_1,a_2,a_3),\dots\)
   #+END_proof
** Application of Zorn's Lemma
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If $A$ is a set, let \(\calp(A)^\#\) denote the family of all its nonempty
   subsets. The *axiom of choice* states that if $A$ is a nonempty set, then there
   exists a function \(\beta:\calp(A)^\#\to A\) with \(\beta(S)\in S\) for every nonempty
   subset \(S\) of $A$. Such a function \beta is called a *choice function*
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A partially ordered set $X$ is *well-ordered* if every nonempty subset $S$ of
   $X$ contains a *smallest element*;
   #+END_definition

   *Well-ordering principle*. \hspace{0.1cm} /Every set $X$ has some well-ordering/
   /of its elements/


   *Zorn's lemma*. \hspace{0.1cm} /If $X$ is a nonempty partially ordered set in/
   /which every chain has an upper bound in $X$, then $X$ has a maximal element/

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   The following statements are equivalent
   1. Zorn's lemma
   2. The well-ordering principle
   3. The axiom of choice
   #+END_theorem

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:prop6.45
   If $C$ is a chain and \(S=\{x_1,\dots,x_n\}\subseteq C\), then there exists
   some \(x_i\), for \(1\le i\le n\), with \(x_j\preceq x_i\) for all \(x_j\in S\)
   #+END_proposition
   #+BEGIN_proof
   Induction on $n\ge1$
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:nthm5.43
   If $R$ is a nonzero commutative ring, then $R$ has a maximal ideal. Indeed,
   every proper ideal $I$ in $R$ is contained in a maximal ideal
   #+END_theorem

   #+BEGIN_proof
   Let \(X\)  be the family of all the proper ideals containing $I$ and
   partially ordered by inclusion.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let $V$ be a vector space over some field $k$, and let \(Y\subseteq V\) be an
   infinite subset 
   1. $Y$ is *linearly independent* if every finite subset of $Y$ is linearly independent
   2. $Y$ *spans* $V$ if each \(v\in V\) is a linear combination of finitely many
      elements of $Y$. We write \(V=\la Y\ra\)
   3. A *basis* of a vector space $V$ is linearly independent subset that spans $V$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   Every vector space $V$ over a field $F$ has a basis. Indeed, every linearly
   independent subset $B$ of $V$ is contained in a basis of $V$; that is, there
   is a subset $B'$ so that \(B\cup B'\) is a basis of $V$
   #+END_theorem


   #+BEGIN_proof
   Let \(X\) be the family of all the linearly independent subsets of $V$ that
   contain $B$. The family $X$ is nonempty, for \(B\in X\). Let
   \(\calb=\{B_j:j\in J\}\) be a chain of $X$. It follows from Proposition 
   ref:prop6.45 that if \(B_{j_1},\dots,B_{j_n}\) is any /finite/ family of
   \(B_j\)'s, then one contains all of the others.
   
   Let \(B^*=\bigcup_{j\in J}B_j\). Clearly, \(B^*\) contains \(B\) and 
   \(B_j\subseteq B^*\). Thus \(B^*\) is an upper bound of \(\calb\) if it
   belongs to \(X\). If $B^*$ is not linearly independent, then it has a finite
   subset \(y_{i_1},\dots,y_{i_m}\) that is linearly dependent and 
   \(y_{i_k}\in B_{j_k}\) for some index \(j_k\). Since there only finitely many 
   \(y_{i_k}\), there exists \(B_{j_0}\) that 
   \(y_{i_1},\dots,y_{i_m}\in B_{j_0}\). Hence Zorn's lemma applies to say that
   there is a maximal element in $X$

   Let \(M\) be a maximal element in $X$. Since $M$ is linear independent, it
   suffices to show that \(M\) spans $V$. If \(M\) does not span $V$, then there
   is \(v_0\in V\) with \(v_0\not\in\la M\ra\). Consider the subset 
   \(M^*=M\cup\{v_0\}\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   Every subspace $W$ of a vector space $V$ is a direct summand
   #+END_corollary
   #+BEGIN_proof
   Let \(B\) be a basis of $W$. By the theorem, there is a subset \(B'\) with 
   \(B\cup B'\) is a basis of $V$. It is straightforward to check that 
   \(V=W\oplus\la B'\ra\)
   #+END_proof

   The ring of real numbers \(\R\) is a vector space over $\Q$; a basis is
   usually called a *Hamel basis*, and it is useful in constructing analytic
   counterexamples. For example, we may use a Hamel basis to prove the existence
   of a discontinuous function \(f:\R\to\R\) that satisfies the functional
   equation \(f(x+y)=f(x)+f(y)\).

   As in the finite-dimensional case, if \(B\) is a basis of a vector space $V$,
   then any function \(f:B\to V\) extends to a linear transformation \(F:V\to
   V\). A Hamel basis has cardinal \(c=\abs{\R}\), and so there are
   \(c^c=2^c>c\) functions \(f:\R\to\R\) satisfying the functional equation, for
   every linear transformation is additive. On the other hand, every continuous
   function on $\R$ is determined by its values on \(\Q\), which is countable.

   Let \(x\in\R\), then there is a sequence of rational numbers
   \((q_n)_{n=1}^\infty\) that converges to $x$. Continuity of $f$ means that
   \begin{equation*}
   \lim_{n\to\infty}f(q_n)=f(\lim_{n\to\infty}q_n)=f(x)
   \end{equation*}
   This means that the values of $f$ at rational numbers already determine $f$.
   In other words, the mapping \(\Phi:C(\R,\R)\to\R^\Q\), defined by
   \(\Phi(f)=f_{\Q}\), where \(f|_\Q:\Q\to\R\) is the restriction of $f$ to $\Q$ is
   an injection. 
   Here \(C(\R,\R)\) denotes the set of all continuous functions from \(\R\) to
   \(\R\). 

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   An *inner product* on a vector space $V$ over a field $k$ is a function 
   \begin{equation*}
   V\times V\to k
   \end{equation*}
   whose values are denoted by \((v,w)\), s.t.
   1. \((v+v',w)=(v,w)+(v',w)\) for all \(v,v',w\in V\)
   2. \((\alpha v,w)=\alpha(v,w)\) for all \(v,w\in V\) and \(\alpha\in k\)
   3. \((v,w)=(w,v)\) for all \(v,w\in V\)

      
   We say  that the inner product is *definite* if \((v,v)\neq 0\) whenever
   \(v\neq 0\)

   Regard \(\R\) is a vector space over \(\Q\), and let $Y$ be a  basis. Using 0
   coefficients if necessary, for each \(v,w\in\R\), there are \(y_i\in Y\) and
   rational \(a_i\) and \(b_i\) with \(v=\sum a_iy_i\) and \(w=\sum b_iy_i\).
   Define
   \begin{equation*}
    (v,w)=\sum a_ib_i
   \end{equation*}
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:fact1
   Let \(X\) and \(Y\) be sets, and let \(f:X\to Y\) be a function. If
   \(f^{-1}(y)\) is finite for every \(y\in Y\), then 
   \(\abs{X}\le\aleph_0\abs{Y}\); hence if \(Y\) is infinite, then
   \(\abs{X}\le\abs{Y}\) 
   #+END_lemma

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:fact2
   If $X$ is an infinite set and \(\Fin(X)\) is the family of all its finite
   subsets, then \(\abs{\Fin(X)}=\abs{X}\)
   #+END_lemma

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:fact3
   If \(X\) and \(Y\) are sets with \(\abs{X}\le\abs{Y}\) and
   \(\abs{Y}\le\abs{X}\), then \(\abs{X}=\abs{Y}\)
   #+END_lemma

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   Let $k$ be a field and let \(V\) be a vector space over $k$
   1. Any two bases of $V$ have the same number of elements; this cardinal is
      called the *dimension* of $V$ and is denoted by \(\dim(V)\)
   2. Vector spaces \(V\) and \(V'\) over $k$ are isomorphic if and only if 
      \(\dim(V)=\dim(V')\)
   #+END_theorem

   #+BEGIN_proof
   1. Let $B$ and $B'$ be bases of $V$. We may assume $B$ and $B'$ are infinite.

      Each \(v\in V\) has a unique expression of the form 
      \(v=\sum_{b\in B}\alpha_bb\), where \(\alpha_b\in k\) and almost all
      \(\alpha_b=0\)(the rule only allows finite operation)
      . Define the *support*
      of $v$ (w.r.t. $B$) by
      \begin{equation*}
      \supp(v)=\{b\in B:\alpha_b\neq0\}
      \end{equation*}
      thus \(\supp(v)\) is a finite subset of $B$ for every \(v\in V\). Define 
      \(f:B'\to\Fin(B)\) by \(f(b')=\supp(b)\). Note that if 
      \(\supp(b')=\{b_1,\dots,b_n\}\), then \(b'\in\la
      b_1,\dots,b_n\ra=\la\supp(b')\ra\). Since \(\la\supp(b')\ra\) has
      dimension $n$, it contains at most $n$ elements of $B'$, because $B'$ is
      independent. Therefore \(f^{-1}(T)\) is finite for every finite subset
      subset \(T\) of $B$. By Lemma ref:fact1, we have
      \(\abs{B'}\le\abs{\Fin(B)}\), and by Lemma ref:fact2, we have
      \(\abs{B'}\le\abs{B}\). Interchanging the roles of $B$ and $B'$ gives the
      reverse inequality, and so Lemma ref:fact3 gives \(\abs{B}=\abs{B'}\).
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   Let $R$ be a commutative ring and let \(\calf\) be the family of all those
   ideals in $R$ that are not finitely generated. If \(\calf\neq\emptyset\), then
   \(\calf\) has a maximal element
   #+END_lemma

   #+ATTR_LATEX: :options [I. S. Cohen]
   #+BEGIN_theorem
   A commutative ring $R$ is noetherian if and only if every prime ideal in $R$
   is finitely generated
   #+END_theorem

   #+BEGIN_proof
   Let \(\calf\) be the family of all ideals in $R$ that are not finitely
   generated. If \(\calf\neq\emptyset\), then the lemma provides an ideal $I$
   that is not finitely generated and that is maximal.

   Suppose \(ab\in I\) but \(a\not\in I\) and \(b\not\in I\). 
   \(I\subsetneq I+Ra\) and \(I_Ra\) is finitely generated; we may assume that 
   \begin{equation*}
   I+Ra=(i_1+r_1a,\dots,i_n+r_na)
   \end{equation*}
   where \(i_k\in I\) and \(r_k\in R\). Consider \(J=(I:a)=\{x\in R:xa\in I\}\). 
   Now \(I+Rb\subseteq J\) and hence $J$ is finitely generated. We claim that 
   \(I=(i_1,\dots,i_n,Ja)\). Clearly \((i_1,\dots,i_n,Ja)\subseteq I\). If
   \(z\in I\subseteq I+Ra\), there are \(u_k\in R\) with 
   \(z=\sum_ku_k(i_k+r_ka)\). Then \((\sum_ku_kr_k)a=z-\sum_ku_ki_k\in I\), so
   that \(\sum_ku_kr_k\in J\). Hence \(z\in(i_1,\dots,i_n,Ja)\). It follows that
   $I$ is finitely generated.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   Let $K/k$ be an extension
   1. If \(z\in K\), then \(z\) is algebraic over $k$ iff \(k(z)/k\) is finite
   2. If \(z_1,\dots,z_n\in K\) are algebraic over $k$, then 
      \(k(z_1,\dots,z_n)/k\) is a finite extension
   3. If \(y,z\in K\) are algebraic over $k$, then \(y+z,yz\) and \(y^{-1}\)
      (for \(y\neq0\)) are also algebraic
   4. Define
      \begin{equation*}
      K_{\alg}=\{z\in K:z\text{ is algebraic over }k\}
      \end{equation*}
      Then $K_{\alg}$ is a subfield of $K$
   #+END_proposition

   #+BEGIN_proof
   1. If \(k(z)/k\) is finite, then Proposition ref:prop3.117 shows that \(z\)
      is algebraic over $k$
   2. We prove this by induction on \(n\ge1\); the base step is part (1). For
      the inductive steop, there is a tower of fields
      \begin{equation*}
      k\subseteq k(z_1)\subseteq \cdots\subseteq k(z_1,\dots,z_n)
      \subseteq k(z_1,\dots,z_{n+1})
      \end{equation*}
      Now \([k(z_{n+1}):k]\) is finite by Theorem ref:nthm2.144; say,
      \([k(z_{n+1}):k]=d\), where $d$ is the degree of the monic irreducible
      polynomial in \(k[x]\) having \(z_{n+1}\) as a root. Since \(z_{n+1}\)
      satisfies a polynomial of degree $d$ over $k$, it satisfies a polynomial
      of degree \(d'\le d\) over the large field $F=k(z_1,\dots,z_n)$
   3. \(k(y+z)\subseteq k(y,z)\) and \(k(yz)\subseteq k(y,z)\)
   #+END_proof


   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Given the extension \(\C/\Q\), define the *algebraic numbers* by
   \begin{equation*}
   \A=(\C/ \Q)_{alg}
   \end{equation*}
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   We claim that \(\A/\Q\) is an algebraic extension that is not finite. Suppose 
   \([\A:\Q]=n\) for some integer $n$. There exist irreducible polynomial in
   \(\Q[x]\) of degree \(n+1\); for example, \(p(x)=x^{n+1}-2\). If \alpha is a root
   of \(p(x)\), then \(\alpha\in\A\) and so \(\Q(\alpha)\subseteq\A\). Thus
   \begin{equation*}
   n=[\A:\Q]=[\A:\Q(\alpha)][\Q(\alpha):\Q]\ge n+1
   \end{equation*}
   #+END_examplle

   #+BEGIN_lemma
   label:nlemma5.55
   1. If \(k\subseteq K\subseteq E\) is a tower of fields with \(E/K\) and \(K/k\)
      algebraic, then \(E/k\) is also algebraic
   2. Let 
      \begin{equation*}
      K_0\subseteq K_1\subseteq\cdots\subseteq K_n\subseteq K_{n+1}\subseteq\cdots
      \end{equation*}
      be an ascending tower of fields. If \(K_{n+1}/K_n\) is algebraic for all
      \(n\ge0\), then \(K^*=\bigcup_{n\ge0}K_n\) is a field algebraic over $K_0$
   3. Let \(K=k(A)\). If each element \(a\in A\) is algebraic over $k$, then
      \(K/k\) is an algebraic extension
   #+END_lemma

   [[index:algebraic closure]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A field $K$ is *algebraically closed* if every nonconstant \(f(x)\in K[x]\) has
   a root in $K$. An *algebraic closure* of a field $k$ is an algebraic extension
   \(\bar{k}\) of $k$ that is algebraically closed
   #+END_definition

   \(\bar{\Q}=\A\)

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:nlemma5.56
   Let $k$ be a field, and let \(k[T]\) be the polynomial ring in a set $T$ of
   indeterminates. If \(t_1,\dots,t_n\in T\) are distinct, where \(n\ge2\), and 
   \(f_i(t_i)\in k[t_i]\subseteq k[T]\) are nonconstant polynomials, then the
   ideal \(I=(f_1(t_1),\dots,f_n(t_n))\) in \(k[T]\) is a proper ideal
   #+END_lemma

   #+BEGIN_proof
   If \(I\) not a proper ideal, then there exists \(h_i(T)\in k[T]\) with
   \begin{equation*}
   1=h_1(T)f_1(t_1)+\cdots+h_n(T)f_n(t_n)
   \end{equation*}
   Consider the field extension \(k(\alpha_1,\dots,\alpha_n)\) whre \(\alpha_i\)
   is a root of \(f_i(t_i)\) for \(i=1,\dots,n\). Denote the variables involved
   in the \(h_i(T)\) other than \(t_1,\dots,t_n\), if any, by
   \(t_{n+1},\dots,t_m\). Evaluating when \(t_i=\alpha_i\) if \(i\le n\) and
   \(t_i=0\) if \(i\ge n+1\), then right side is 0 (evaluation is a homomorphism)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   Given a field $k$ , there exists an algebraical closure \(\bar{k}\) of $k$
   #+END_theorem

   #+BEGIN_proof
   Let $T$ be a set in bijective correspondence with the family of nonconstant
   polynomials in \(k[x]\). Let \(R=k[T]\) be the big polynomial ring, and let
   $I$ be the ideal in $R$ generated by all elements of the form \(f(t_f)\),
   where \(t_f\in T\)

   We claim that the ideal $I$ is proper; if not, \(1\in I\), and there are
   distinct \(t_1,\dots,t_n\in T\) and polynomials \(h_1(T),\dots,h_n(T)\in
   k[T]\) with \(1=h_1(T)f_1(T)+\dots+h_n(T)f_n(t_n)\), contradicting Lemma
   ref:nlemma5.56. Therefore there is a maximal ideal $M$ in $R$ containing $I$
   by Theorem ref:nthm5.43. Define \(K=R/M\). The proofs is now completed in a
   series of steps
   1. \(K/k\) is a field extension

      Because $M$ is maximal, \(R=(a)+M\) for any \(a\not\in M\). So \(1=ra+m\)
      or equivalently \(1+M=(r+M)(a+M)\). So \(K=R/M\) is a field. Let 
      \(i:k\to k[T]\) be the ring map taking \(a\in k\) to the constant
      polynomial $a$, and let \theta be the composite 
      \(k\xrightarrow{i}k[T]=R\xrightarrow{\text{nat}}R/M=K\). Now \theta is injective
      by Corollary ref:ncor2.32. We identify $k$ with \(\im\theta\subseteq K\)

   2. Every nonconstant \(f(x)\in k[x]\) splits in \(K[x]\)

      By definition, there is \(t_f\in T\) with \(f(t_f)\in I\subseteq M\), and
      the coset \(t_f+M\in R/M=K\) is a root of \(f(x)\). It now follows by
      induction on degree that \(f(x)\) splits over $K$

   3. The extension \(K/k\) is algebraic

      By Lemma ref:nlemma5.55, it suffices to show that each \(t_f+M\) is
      algebraic over $k$ (for $K=k(\text{all }t_f+M)$ follows from
      ref:prop3.117)

      
   Let \(k_1=K\) and construct \(k_{n+1}\) from $k$ in the same way $K$ is
   constructed from $k$. There is a tower of fields 
   \(k=k_0\subseteq k_1\subseteq \dots\subseteq k_n\subseteq\dots\) with each
   extension \(k_{n+1}/k_n\) algebraic and with every nonconstant polynomials in
   \(k_n[x]\) having a root in \(k_{n+1}\). By Lemma ref:nlemma5.55
   \(E=\bigcup_nk_n\)is an algebraic extension of $k$. We claim that $E$ is
   algebraiclly closed.
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If $k$ is a countable field, then it has a countable algebraic closure. In
   particular, the algebraic closures of the prime fields \(\Q\) and \(\F_p\)
   are countable
   #+END_corollary

   #+BEGIN_proof
   If $k$ is countable, then the set $T$ of all nonconstant polynomials is
   countable, say, \(T=\{t_1,t_2,\dots\}\), because \(k[x]\) is countable. Hence 
   \(k[T]=\bigcup_{l\ge1}k[t_1,\dots,t_l]\) is countable, as its quotient
   \(k_1\). It follows by induction on \(g\ge1\), that every \(k_n\) is countable
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If \(F/k\) and \(K/k\) are field extensions, then a *\(k\)-map* is a ring
   homomorphism \(\varphi:F\to K\) that fixes $k$ pointwise
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:nlemma5.59
   If $K/k$ is an algebraic extension, then every \(k\)-map
   \(\varphi:K\to K\) is an automorphism of $K$
   #+END_lemma

   #+BEGIN_proof
   By Corollary ref:ncor2.32, the \(k\)-map \varphi is injective. Let \(a\in K\) and
   there is an irreducible \(p(x)\in k[x]\) having \(a\) as a root. \varphi permutes
   all roots of \(p(x)\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:nlemma5.60
   Let $k$ be a field and let \(\bar{k}/k\) be an algebraic closure. If
   \(F/k\) is an algebraic extension, then there is an injective \(k\)-map
   \(\psi:F\to\bar{k}\)
   #+END_lemma

   #+BEGIN_proof
   If $E$ is an intermediate field, \(k\subseteq E\subseteq F\), let us call an
   ordered pair \((E,f)\) an *approximation* if \(f:E\to\bar{k}\) is a \(k\)-map.
   In the following diagram, all arrows other than $f$ are inclusions

   
   \begin{center}
   \begin{tikzcd}
   \bar{k}\\
   k \arrow[u,"i"]\arrow[r]&E\arrow[ul,"f"']\arrow[r]&F
   \end{tikzcd}
   \end{center}

   Define \(X=\{\text{approximations }(E,f):k\subseteq E\subseteq F\}\). Note that
   \(X\neq\emptyset\) because \((k,i)\in X\). Partially order $X$ by
   \begin{equation*}
   (E,f)\preceq(E',f')\text{ if }E\subseteq E'\text{ and }f|E'=f
   \end{equation*}

   Chain
   \begin{equation*}
   \cals=\{(E_j,f_j):j\in J\}
   \end{equation*}
   has an upper bound \((\bigcup E_j,\bigcup f_j=\Phi)\). \Phi is a \(k\)-map

   By Zorn's Lemma, there exists a maximal element \((E_0,f_0)\) in $X$. We
   claim that \(E_0=F\) and this will complete the proof (take \(\psi=f_0\)).
   If \(E_0\subsetneq F\), then there is \(a\in F\) with \(a\not\in E_0\). Since
   \(F/k\) is algebraic, we have \(F/E_0\) algebraic, and there is an
   irreducible \(p(x)\in E_0[x]\) having $a$ as a root. Since \(\bar{k}/k\) is
   algebraic and \(\bar{k}\) is algebraically closed, we have a factorization
   in \(\bar{k}[x]\):
   \begin{equation*}
   f_0^*(p(x))=\displaystyle\prod_{i=1}^n(x-b_i)
   \end{equation*}
   where \(f_0^*:E_0[x]\to\bar{k}[x]\) is the map
   \(f_0^*:e_0+\dots+e_nx^n\mapsto f_0(e_0)+\dots+f_0(e_n)x^n\). If all the
   \(b_i\) lie in \(f_0(E_0)\subseteq\bar{k}\), then
   \(f_0^{-1}(b_i)\in E_0\subseteq F\) for all $i$, and there is a factorization
   of \(p(x)\) in \(F[x]\), namely, \(p(x)=\prod_{i=1}^n[x-f_0^{-1}(b_i)]\). But
   \(a\not\in E_0\) implies \(a\neq f_0^{-1}(b_i)\) for all $i$. Thus \(x-a\) is
   another factor of \(p(x)\) in \(F[x]\), _contrary to unique factorization._
   . We conclude that there is some \(b_i\not\in\im f_0\). By Theorem
   ref:nthm2.144, we may define \(f_1:E_0(a)\to\bar{k}\) by
   \begin{equation*}
    c_0+c_1a+c_2a^2+\cdots\mapsto f_0(c_0)+f_0(c_1)b_i+\dots
   \end{equation*}
   A straight forward check shows that \(f_1\) is a \(k\)-map extending \(f_0\).
   Hence \((E_0,f_0)\prec(E_0(a),f_1)\), contradicting the maximality of \((E_0,f_0)\).
   #+END_proof
   

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   Any two algebraic closure of a field $k$ are isomorphic via a \(k\)-map
   #+END_theorem

   #+BEGIN_proof
   Let $K$ and $L$ be two algebraic closure of a field $k$. By Lemma
   ref:nlemma5.60, there are \(k\)-maps \(\psi:K\to L\) and \(\theta:L\to K\). By Lemma
   ref:nlemma5.59, both composites \(\theta\psi\) and \(\psi\theta\) are
   automorphisms. It follows that \psi (and \theta) is a \(k\)-isomorphisms
   #+END_proof

   [[index:degree]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If \(\varphi\in k(x)\), then there are polynomials \(g(x),h(x)\in k[x]\) with
   \((g,h)=1\) and \(\varphi=g(x)/h(x)\). Define the *degree* of \varphi by
   \begin{equation*}
   \degree(\varphi)=\max\{\deg(g),\deg(h)\}
   \end{equation*}
   A rational function \(\varphi\in k(x)\) is called a *linear fractional
   transformation* if
   \begin{equation*}
   \varphi=\frac{ax+b}{cx+d}
   \end{equation*}
   where \(a,b,c,d\in k\) and \(ad-bc\neq 0\). Let
   \begin{equation*}
   \LF(k)
   \end{equation*}
   denote the group of all linear fractional transformations in \(k(x)\) with
   binary operation composition: if \(\varphi:x\mapsto(ax+b)/(cx+d)\) and 
   \(\psi:x\mapsto(rx+s)/(tx+u)\), then 
   \begin{equation*}
   \psi\varphi:x\mapsto\frac{r\varphi(x)+s}{t\varphi(x)+u}=
   \frac{(ra+sc)x+(rb+sd)}{(ta+ud)x+(tb+ud)}
   \end{equation*}
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:nprop5.62
   If \(\varphi\in k(x)\) is nonconstant, then \varphi is transcendental over $k$ and
   \(k(x)\) is a finite extension of \(k(\varphi)\) with
   \begin{equation*}
   [k(x):k(\varphi)]=\degree(\varphi)
   \end{equation*}
   Moreover, if \(\varphi=g(x)/h(x)\) and \((g,h)=1\), then 
   \begin{equation*}
   \irr(x,k(\varphi))=g(y)-\varphi h(y)
   \end{equation*}
   where \(\varphi h(y)\) denotes the product of \varphi and \(h(y)\) in \(k(\varphi)[y]\)
   #+END_proposition

   #+BEGIN_proof
   Let \(g(x)=\sum a_ix^i\) and \(h(x)=\sum b_ix^i\in k[x]\). Define
   \begin{equation*}
   \theta(y)=g(y)-\varphi h(y)
   \end{equation*}
   Now \(\theta(y)\) is a polynomial in \(k(\varphi)[y]\): 
   \(\theta(y)=\sum a_iy^i-\varphi\sum b_iy^i=\sum(a_i-\varphi b_i)y^i\). If \(\theta(y)\)
   were the zero polynomial, then all its coefficients would be 0. But if $b_i$
   is a nonzero coefficient of \(h(y)\), then \(a_i-\varphi b_i=0\) gives
   \(\varphi=a_i/b_i\), contradicting \varphi not being a constant; that is,
   \(\varphi\not\in k\).
   \begin{equation*}
   \deg(\theta(y))=\deg(g(y)-\varphi h(y))=\max\{\deg(g),\deg(h)\}=\degree(\varphi)
   \end{equation*}
   Now \(x\) is a root of \(\theta(y)\), so that \(x\) is algebraic over \(k(\varphi)\).
   Were \varphi algebraic over $k$, then \(k(\varphi)/k\) would be finite, giving 
   \([k(x):k]=[k(x):k(\varphi)][k(\varphi):k]\) finite, a contradiction. Therefore \varphi
   is transcendental over $k$.

   We claim that \(\theta(y)\) is an irreducible polynomial in \(k(\varphi)[y]\). If not,
   then \(\theta(y)\) factors in \(k[\varphi][y]\) by Gauss's Corollary ref:ncor5.28. 
   But \(\theta(y)=g(y)-\varphi h(y)\) is linear in \varphi, and so Corollary ref:ncor5.31
   shows that \(\theta(y)\) is irreducible. Finally, since \(\deg(\theta)=\degree(\varphi)\), we
   have \([k(x):k(\varphi)]=\degree(\varphi)\)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   label:ncor5.63
   Let \(\varphi\in k(x)\), where \(k(x)\) is the field of rational functions
   over a field $k$. Then \(k(\varphi)=k(x)\) if and only if \varphi is a linear fractional
   transformations 
   #+END_corollary

   #+BEGIN_proof
   By Proposition ref:nprop5.62, \(k(\varphi)=k(x)\) if and only if \(\degree(\varphi)=1\);
   that is, \varphi is a linear fractional transformation
   #+END_proof

   Define a map \(\zeta:\GL(2,k)\to\LF(k)\) by 
   \(\begin{psmallmatrix}a&b\\c&d\end{psmallmatrix}\mapsto(ax+b)/(cx+d)\). In
   Exercise ref:nex5.45, \(\ker\zeta=Z(2,k)\), the center of \(\GL(2,k)\)
   consisting of all nonzero \(2\times 2\) scalar matrices. Hence if
   \begin{equation*}
   \PGL(2,k)=\GL(2,k)/Z(2,k)
   \end{equation*}
   then \(\LF(k)\cong\PGL(2,k)\)
   #+ATTR_LATEX: :options []
   #+BEGIN_corollary
   If \(k(x)\) is the field of rational functions over a field $k$, then 
   \begin{equation*}
   \Gal(k(x)/k)\cong\LF(k)\cong\PGL(2,k)
   \end{equation*}
   #+END_corollary
   
   #+BEGIN_proof
   Let \(\sigma:k(x)\to k(x)\) be an automorphism of \(k(x)\) fixing $k$. Since
   \(k(\sigma(x))=k(x)\), Corollary ref:ncor5.63 says that \(\sigma(x)\) is a linear
   fractional transformation. Define \(\gamma:\Gal(k(x)/k)\to\LF(k)\) by 
   \(\gamma:\sigma\mapsto\sigma(x)\). Now \gamma is a homomorphism:
   \(\gamma(\sigma\tau)=\gamma(\sigma)\gamma(\tau)\). Finally \gamma is an isomorphisms
   #+END_proof

   #+ATTR_LATEX: :options [Lüroth's theorem]
   #+BEGIN_theorem
   If \(k(x)\) is a simple transcendental extension, then every intermediate
   field $B$ with \(k\subsetneq B\subseteq k(x)\) is also a simple
   transcendental extension of $k$: there is \(\varphi\in B\) with \(B=k(\varphi)\)
   #+END_theorem

   [[index:purely transcendental]]
   [[index:algebraically dependent]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let \(E/k\) be a field extension. A subset \(U\) of $E$ is 
   *algebraically dependent* over $k$ if there exists a finite subset 
   \(u_1,\dots,u_n\subseteq U\) and a nonzero polynomial
   \(f(x_1,\dots,x_n)\in k[x_1,\dots,x_n]\) with \(f(u_1,\dots,u_n)=0\). A field
   extension \(E/k\) is *purely transcendental* if either \(E=k\) or $E$ contains
   an algebraically independent $B$ and \(E=k(B)\)
   #+END_definition

   Let \(E/k\) be a field extension, let \(u_1,\dots,u_n\in E\) and let 
   \(\varphi:k[x_1,\dots,x_n]\to E\) be the evaluation map
   \(f(x_1,\dots,x_n)\mapsto f(u_1,\dots,u_n)\). Now
   \(\{u_1,\dots,u_n\}\)is algebraically dependent if and only if \(\ker\varphi\neq\{0\}\).
   If \(\{u_1,\dots,u_n\}\) is algebraically independent, then \varphi extends to an
   isomorphism \(\widetilde{\varphi}:k(x_1,\dots,x_n)\to k(u_1,\dots,u_n)\subseteq E\), where
   \(k(x_1,\dots,x_n)\) is the field of rational functions
   \begin{center}
   \begin{tikzcd}
   k(x_1,\dots,x_n)\arrow[r,"\widetilde{\varphi}"]&
   \Frac(E)=E\\
   k[x_1,\dots,x_n]\arrow[u]\arrow[r,"\varphi"]&
   E\arrow[u]
   \end{tikzcd}
   \end{center}

   In particular, if \(\{u_1,\dots,u_n\}\) is algebraically independent and
   \(E=k(u_1,\dots,u_n)\), then \(\widetilde{\varphi}\) is an isomorphism 
   \(k(x_1,\dots,x_n)\to k(u_1,\dots,u_n)\) with \(x_i\mapsto u_i\). Therefore,
   a purely transcendental extension \(k(u_1,\dots,u_n)/k\) is isomorphic to the 
   *function field in $n$ variables*


   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   label:nprop5.66
   Let \(E/k\) be a field extension. Then \(U\subseteq E\) is algebraically
   dependent over $k$ if and only if there is \(v\in U\) with $v$ algebraic
   over \(k(U-\{v\})\)
   #+END_proposition

   #+BEGIN_proof
   If $U$ is algebraically dependent over $k$, then there is a finite
   algebraically dependent subset \(\{u_1,\dots,u_n\}\subseteq U\); thus, we may
   assume that \(U\) is finite. We prove, by induction on \(n\ge1\), that some
   \(u_i\) is algebraic over \(k(U-\{u_i\})\). If \(n=1\), it's obvious.

   Let \(U=\{u_1,\dots,u_{n+1}\}\) be algebraically dependent. We may assume
   that \(\{u_1,\dots,u_n\}\) is algebraically independent. Since $U$ is
   algebraically dependent, there is a nonzero 
   \(f(X,y)\in k[x_1,\dots,x_n,y]\) with \(f(u_1,\dots,u_n,u_{n+1}=0)\), where
   \(X=\{x_1,\dots,x_n\}\) and $y$ is a new variable. We may write 
   \(f(X,y)=\sum_ig_i(X)y^i\), where \(g_i(X)\in k[X]\).Since \(f(X,y)\neq0\),
   some \(g_i(X)\neq0\), and it follows from algebraic independence of
   \(\{u_1,\dots,u_n\}\) that \(g_i(u_1,\dots,u_n)\neq0\). Therefore,
   \(h(y)=\sum_ig(u_1,\dots,u_n)y^i\in k(U)[y]\). But \(0=h(u_{n+1})\), so that
   \(u_{n+1}\) is algebraic over \(k(u_1,\dots,u_n)\)

   For the converse, assume that $v$ is algebraic over \(k(U-\{v\})\). We may
   assume that \(U-\{v\}\) is finite, say, \(U-\{v\}=\{u_1,\dots,u_n\}\), where
   \(n\ge0\). We prove by induction on \(n\ge0\). For the inductive step, let 
   \(U-\{u_{n+1}\}=\{u_1,\dots,u_n\}\) and assume it is algebraically
   independent. By hypothesis, there is a nonzero polynomial 
   \(f(y)=\sum_ic_iy^i\in k(u_1,\dots,u_n)[y]\) with \(f(u_{n+1})=0\). As
   \(f(y)\neq0\), we may assume that at least one of its coefficients is
   nonzero. For all $i$, the coefficient \(c_i\in k(u_1,\dots,u_n)\), so there
   are rational functions \(c_i(x_1,\dots,x_n)\) with
   \(c_i(u_1,\dots,u_n)=c_i\) [because 
   \(k(u_1,\dots,u_n)\cong k(x_1,\dots,x_n)\)]
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   A *dependency relation on a set* \Omega is a relation \(\preceq\) from \Omega
   to \(2^\Omega\), pronounced "is dependent on", satisfying the following 
   *Dependency Axioms*:
   1. if \(S\subseteq\Omega\) and \(u\in S\), then \(u\preceq S\)
   2. if \(u\preceq S\), then there exists a finite subset \(S'\subseteq S\)
      with \(u\preceq S'\)
   3. (*Transitivity*) if \(u\preceq S\) and, for some \(T\subseteq\Omega\), we
      have \(s\preceq T\) for every \(s\in S\), then \(u\preceq T\)
   4. (*Exchange Axiom*) if \(u\preceq S\) and \(u\not\preceq S-\{v\}\), then
      \(v\preceq (S-\{v\})\cup\{u\}\)
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   If \Omega is a vector space, define \(u\preceq S\) to mean \(u\in\la S\ra\), the
   subspace spanned by \(S\). We claim that \(\preceq\) is a dependency relation.
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   If \(E/k\) is a field extension, then \(u\preceq S\), defined by $u$ being
   algebraic over \(k(S)\) is a dependency relation on $E$
   #+END_lemma

   #+BEGIN_proof
   If \(u\preceq S\), then $u$ is algebraic over \(k(S)\); that is, 
   \(u\in(E/k(S))_{\alg}=\{e\in E:e\text{ is algebraic over }k(S)\}\). Suppose
   there is some \(T\subseteq E\) with \(s\preceq T\) for every \(s\in S\); that
   is, \(S\subseteq(E/k(T))_{\alg}\). It follows from Lemma ref:nlemma5.55 that 
   \((E/k(S))_{\alg}\subseteq(E/k(T))_{\alg}\); and so \(u\preceq T\)

   The Exchange Axiom assumes that \(u\preceq S\) and $u$ is transcendental over
   \(k(S-\{v\})\). Note that \(v\in S\) and \(u\not\in S\). Let us apply
   Proposition ref:nprop5.66 to the subsets \(U'=\{u,v\}\) and \(S'=S-\{v\}\) of
   $E$ and the subfield \(k'=k(S')\). With this notation,
   \(k'(U'-\{u\})=k'(v)=k(S',v)=k(S)\), so that $u$ is algebraic over \(k(S)\)
   can be restated as \(u\)  algebraic over \(k'(U'-\{u\})\). Thus Proposition
   ref:nprop5.66 says that \(U'=\{u,v\}\) is algebraically dependent over
   \(k'=k(S)\): there is a nonzero polynomial \(f(x,y)\in k(S')[x,y]\) with 
   \(f(u,v)=0\).
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let \(\preceq\) be a dependency relation on a set \Omega. Call a subset
   \(S\subseteq\Omega\) *dependent* if there exists \(s\in S\) with 
   \(s\preceq S-\{s\}\); call $S$ *independent* if it is not dependent
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   Let \(\preceq\) be a dependency relation on a set \Omega. We say that a subset $S$
   *generates* \Omega if \(x\preceq S\) for all \(x\in\Omega\). A *basis* of \Omega
   is an independent subset that generates \Omega
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_lemma
   label:nlemma5.69
   1. Let \(\preceq\) be a dependency relation on a set $\Omega$. If
      \(T\subseteq\Omega\) is independent and \(z\not\preceq T\) for some
      \(z\in\Omega\), then \(T\cup\{z\}\supsetneq T\)is a strictly larger
      independent set.
   2. Let \(E/k\) be a field extension. If \(T\supseteq E\) is algebraically
      independent over $k$ and \(z\in E\) is transcendental over \(k(T)\), then 
      \(T\cup\{z\}\) is algebraically independent
   #+END_lemma

   #+BEGIN_proof
   1. not hard
   2. By Proposition ref:nprop5.66, this is a special case of (1)
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   If \(E/k\) is a field extension, then a *transcendence basis* is a maximal
   algebraically independent subset of $E$ over $k$
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   label:nthm5.70
   1. If \(\preceq\) is a dependency relation on a set \Omega, then \Omega has a basis. In
      fact, every independent subset $B$ of \Omega is part of basis
   2. If \(E/k\) is a field extension, then $E$ has a transcendence basis. In
      fact, every algebraically independent subset is part of a transcendence basis
   #+END_theorem

   #+BEGIN_proof
   1. Since the empty set \(\emptyset\) is independent, the second statement
      implies the first

      We use Zorn's Lemma to prove the existence of maximal independent subsets
      of \Omega containing $B$. Let $X$ be the family of all independent subsets of
      \Omega containing $B$. $X$ is nonempty. Suppose that \(\calb=(B_j)_{j\in J}\)
      is a chain in $X$. It is clear that \(B^*=\bigcup_{j\in J}B_j\) is an
      upper bound of $\calb$ if it lies in $X$. If \(B^*\) is dependent, then
      there is \(y\in B^*\) with \(y\preceq B^*-\{y\}\). By _Dependency Axiom
      (2)_, there is a finite subset \(\{x_1,\dots,x_n\}\subseteq B^*-\{y\}\)
      with \(y\preceq\{x_1,\dots,x_n\}-\{y\}\). There is \(B'\) s.t. 
      \(\{x_1,\dots,x_n,y\}\subseteq B'\) dependent
   #+END_proof

   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   If $B$ is a transcendence basis, then \(k(B)/k\) is purely transcendental and
   \(E/k(B)\) is algebraic
   #+END_theorem

   #+BEGIN_proof
   By Theorem ref:nthm5.70, it suffices to show that if $B$ is a transcendence
   basis, then \(E/k(B)\) is algebraic
   #+END_proof
   
   #+ATTR_LATEX: :options []
   #+BEGIN_theorem
   1. If \Omega is a set with a dependency relation \(\preceq\), then any two bases
      $B$ and $C$ have the same cardinality
   2. If $B$ and $C$ are transcendence basis of a field extension \(E/k\) , then 
      \(\abs{B}=\abs{C}\)
   #+END_theorem

   #+BEGIN_proof
   If \(B=\emptyset\), we claim that \(C=\emptyset\). Otherwise there exists
   \(y\in C\), and since \(C\) is independent, \(y\not\preceq C-\{y\}\). But
   \(y\preceq B=\emptyset\) and \(\emptyset\subseteq C-\{y\}\), so that 
   _Dependency Axiom (3)_ \(y\preceq C-\{y\}\), a contradiction.

   Now assume \(B\) is finite; say, \(B=\{x_1,\dots,x_n\}\). We prove, by
   induction on \(k\ge0\), that there exists 
   \(\{y_1,\dots,y_{k-1}\subseteq C\}\) with
   \begin{equation*}
   B_k=\{y_1,\dots,y_{k-1},x_k,\dots,x_n\}
   \end{equation*}
   a basis; that is , the elements \(x_1,\dots,x_{k-1}\) can be exchanged with
   elements \(y_1,\dots,y_{k-1}\in C\) so that \(B_k\) is a basis.

   Assume \(B_k=\{y_1,\dots,y_{k-1},x_k,\dots,x_n\}\) is a basis, we want to
   show that \(B_{k+1}\) is basis. We claim that there is \(y_k\in C\) with 
   \(y_k\not\preceq B_k-\{x_k\}\). Otherwise, \(y\preceq B_k-\{x_k\}\) for all 
   \(y\preceq C\). But \(x_k\preceq C\), and so _Dependency Axiom (3)_ gives 
   \(x_k\preceq B_k-\{x_k\}\).

   By Lemma ref:nlemma5.69, \(B_{k+1}\) is independent. To see that \(B_{k+1}\)
   is a basis, it suffices to show that it generates \Omega. Now 
   \(y_k\preceq B_k\) and \(y_k\not\preceq B_k-\{x_k\}\); the Exchange Axiom
   gives
   \(x_k\preceq (B_k-\{x_k\}\cup\{y_k\})=B_{k+1}\)

   If \(\abs{C}>n=\abs{B}\). Then \(B_n\subseteq C\). Thus a proper subset of
   $C$ generates \Omega, contradicting the independence of $C$
   #+END_proof

   [[index:transcedence degree]]
   #+ATTR_LATEX: :options []
   #+BEGIN_definition
   The *transcendence degree* of \(E/k\)is defined by 
   \begin{equation*}
   \trdeg(E/k)=\abs{B}
   \end{equation*}
   where $B$ is a transcendence basis of \(E/k\)
   #+END_definition

   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. If \(E/k\) is a field extension, then \(\trdeg(E/k)=0\) if and only if
      \(E/k\) is algebraic
   2. If \(E=k(x_1,\dots,x_n)\) is the function field in $n$ variables over a
      field $k$, then \(\trdeg(E/k)=n\), because \(\{x_1,\dots,x_n\}\) is a
      transcendence basis of $E$
   #+END_examplle

   #+ATTR_LATEX: :options []
   #+BEGIN_proposition
   There are nonisomorphic fields each of which is isomorphic to a subfield of
   the other
   #+END_proposition

   #+BEGIN_proof
   Clearly \(\C\) is isomorphic to a subfield of \(\C(x)\). However we claim
   that \(\C(x)\) is isomorphic to a subfield of $\C$. Let $B$ be a
   transcendence basis of \(\C\) over \(\Q\), and discard one of its element,
   say, $b$. The algebraic closure \(F\) of \(\Q(B-\{b\})\) is a proper subfield
   of \(\C\)
   #+END_proof
*** Exercise
    #+BEGIN_exercise
    label:nex5.44
    Prove that \(\varphi\in k(x)\) has degree 1 if and only if \varphi is a linear
    fractional transformation
    #+END_exercise

    #+BEGIN_exercise
    label:nex5.45
    For any field $k$, define a map \(\zeta:\GL(2,k)\to LF(k)\) by
    \begin{equation*}
    \zeta:\begin{psmallmatrix}a&b\\c&d\end{psmallmatrix}\mapsto
    (ax+b)/(cx+d)
    \end{equation*}
    1. Prove that \zeta is a surjective group homomorphism
    2. Prove that \(\ker\zeta=Z(2,k)\), the subgroup of \(\GL(2,k)\) consisting
       of all nonzero scalar matrices
    #+END_exercise

    #+BEGIN_exercise
    label:nex5.46
    Prove that the set $\A$ of all algebraic numbers is an algebraic closure of \(\Q\)
    #+END_exercise
** Varieties
*** Varieties and Ideals

    Let \(k\) be a field and let \(k^n\) denoted the set of all \(n\)-tuples:
    \begin{equation*}
    k^n=\{a=(a_1,\dots,a_n):a_i\in k\text{ for all }i\}
    \end{equation*}
    We use the abbreviation
    \begin{equation*}
    X=(x_1,\dots,x_n)
    \end{equation*}
    so that the polynomial ring \(k[x_1,\dots,x_n]\) in several variables may be
    denoted by \(k[X]\) and a polynomial \(f(x_1,\dots,x_n)\) in \(k[X]\) may be
    abbreviated by \(f(X)\)
    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If \(f(X)\in k[X]\), its *polynomial function* \(f^{\flat}:k^n\to k\) is
    defined by evaluation
    \begin{equation*}
    f^\flat:(a_1,\dots,a_n)\mapsto f(a_1,\dots,a_n)
    \end{equation*}
    #+END_definition

    *For the remainder of this section, we assume that all fields are infinite*

    #+ATTR_LATEX: :options []
    #+BEGIN_definition
    If \(f(X)\in k[X]=k[x_1,\dots,x_n]\) and \(f(a)=0\), where \(a\in k^n\),
    then \(a\) is called a *zero* of \(f(X)\).
    #+END_definition

    #+ATTR_LATEX: :options []
    #+BEGIN_proposition
    If \(k\) is an algebraically closed field and \(f(X)\in k[X]\) is not a
    constant, then \(f(X)\) has a zero
    #+END_proposition

    #+BEGIN_proof
    Induction on \(n\ge1\), where \(X=(x_1,\dots,x_n)\). Write
    \begin{equation*}
    f(X,y)=\displaystyle\sum_{i}g_i(X)y^i
    \end{equation*}
    For each \(a\in k^n\), define \(f_a(y)=\sum_ig_i(a)y^i\). If \(f(X,y)\) has
    no zeros, then each \(f_a(y)\in k[y]\) has no zeros, hence \(f_a(y)\) is a
    nonzero constant.
    #+END_proof
    

* Rings
** Modules
   #+ATTR_LATEX: :options []
   #+BEGIN_examplle
   1. If \(k\) is any nonzero commutative ring, then \(\Mat_n(k)\), all
      \(n\times n\)  matrices with entries in \(k\), is a ring under matrix
      multiplication and matrix addition; it is commutative if and only if
      \(n=1\).

      Let \(k\) be any ring. If \(A=[a_{ip}]\) is an \(n\times l\) matrix and
      \(B=[b_{pj}]\) is an \(l\times m\) matrix, then their product \(AB\) is
      defined to be the \(n\times m\) matrix whose \(ij\) entry has the usual
      formula: \((AB)_{ij}=\sum_pa_{ip}b_{pj}\). Thus \(\Mat_n(k)\) is a ring,
      even if \(k\) is not commutative

   2. If $G$ is a group, we define the *group ring* \(\Z G\) as follows. Its
      additive group is the free abelian group having a basis
   #+END_examplle


* Index  
  \renewcommand{\indexname}{}
  printindex:nil
* TODO Some statements need to be verified
  ref:nlemma5.60