\documentclass[11pt]{article}

\usepackage{amsmath,amssymb,amsthm}

\usepackage{xcolor}

%\usepackage{systeme}

%\usepackage[textheight=8.5in, margin=1.6in]{geometry}
\usepackage[total={6in,8in}]{geometry}


\theoremstyle{definition}
\newtheorem{thm}{Theorem}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{eg}[thm]{Example}
\newtheorem{ex}[thm]{Exercise}
\newtheorem{defn}[thm]{Definition}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{remark}[thm]{Remark}
\newtheorem{question}[thm]{Question}

\usepackage{pgfplots}
\pgfplotsset{compat=newest}
\usepackage{tikz}
\usepackage{tikz-cd}
\tikzstyle{ball} = [circle,shading=ball, ball color=black,
    minimum size=1mm,inner sep=1.3pt]
\tikzstyle{miniball} = [circle,shading=ball, ball color=black,
    minimum size=1mm,inner sep=0.5pt]
\tikzstyle{mminiball} = [circle,shading=ball, ball color=black,
    minimum size=0.6mm,inner sep=0.1pt]
\usetikzlibrary{arrows.meta}
\tikzset{>={Latex[length=2mm,width=1.5mm]}}
\tikzset{->-/.style={decoration={markings, mark=at position #1 with
  {\arrow{>}}},postaction={decorate}}}
\usetikzlibrary{decorations.pathmorphing}    
\usetikzlibrary{decorations.pathreplacing}
\usetikzlibrary{arrows.meta,calc}
\usetikzlibrary{bending}
\usetikzlibrary{decorations.markings,shapes.geometric}
\tikzset{->-/.style={decoration={markings, mark=at position #1 with
  {\arrow{>}}},postaction={decorate}}}
\tikzset{-|-/.style={decoration={markings, mark=at position #1 with
  {\arrow{stealth}}},postaction={decorate}}}
\tikzset{movearrow/.style 2 args ={
        decoration={markings,
	  mark= at position {#1} with {\arrow{#2}} ,
        },
        postaction={decorate}
    }
}

%\usepackage{graphicx}

\usepackage{needspace}  % <-- to prevent bad pagebreak over environment titles
% Use:  \Needspace{3\baselineskip} before opening an environment

\usepackage{enumitem}
%\setlist[enumerate]{leftmargin=*}
\setlist[enumerate]{itemsep=1pt}

%\allowdisplaybreaks

\usepackage[colorlinks=true,citecolor=black,linkcolor=black,urlcolor=blue]{hyperref}

%\usepackage{algorithm}
%\usepackage[noend]{algpseudocode}

\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\W}{\mathbb{W}}
\newcommand{\sn}{\mathfrak{S}}
\newcommand{\ve}{\varepsilon}
\newcommand{\ips}[2]{\langle{#1}, {#2}\rangle}
\newcommand{\incid}{\mathcal{I}(P)}
\newcommand{\tL}{\widetilde{L}}
\newcommand{\tV}{\widetilde{V}}
\newcommand{\F}{\mathcal{F}}
\DeclareMathOperator{\tr}{tr}
\DeclareMathOperator{\Span}{\mathrm{Span}}
\DeclareMathOperator{\rk}{\mathrm{rank}}
\DeclareMathOperator{\wt}{\mathrm{wt}}
\DeclareMathOperator{\trace}{\mathrm{tr}}
\DeclareMathOperator{\im}{\mathrm{im}}
\DeclareMathOperator{\Sym}{\mathrm{Sym}}
\DeclareMathOperator{\Hom}{\mathrm{Hom}}
\DeclareMathOperator{\diag}{\mathrm{diag}}
\DeclareMathOperator{\glb}{\mathrm{glb}}
\DeclareMathOperator{\lub}{\mathrm{lub}}
\DeclareMathOperator{\Div}{\mathrm{Div}}
\DeclareMathOperator{\Pic}{\mathrm{Pic}}
\DeclareMathOperator{\Jac}{\mathrm{Jac}}
\DeclareMathOperator{\cok}{\mathrm{cok}}
\DeclareMathOperator{\sgn}{\mathrm{sgn}}
\DeclareMathOperator{\outdeg}{\mathrm{outdeg}}
\DeclareMathOperator{\supp}{\mathrm{supp}}

\parskip = 0.05in
\parindent = 0.0in

%\pagestyle{empty}

\begin{document}
\centerline{Math 361 lecture for Monday, Week 3}
\bigskip

\centerline{Discriminants}
\bigskip

\begin{thm}
  (Fundamental theorem of algebra.)  Let $h\in\C[x]$ be a nonconstant
  polynomial.  Then there exists $\alpha\in\C$ such that $h(\alpha)=0$.
\end{thm}
Using polynomial division, we get the following (equivalent) formulation of the
fundamental theorem of algebra:
\begin{cor}
  A polynomial $h\in \C[x]$ of degree $n$ has $n$ complex roots
  $\theta_{1},\ldots,\theta_{n}$ counting multiplicities (i.e., the $\theta_{i}$
  are not necessarily distinct), and
  \[
    h=\beta\prod_{n=1}^{n} (x-\theta_{i}) 
  \]
  for some~$\beta\in\C$.
\end{cor}

Let $K$ be a subfield of $\C$.  

\begin{ex}
 Show that $\Q\subseteq K$.  (Hint: since~$K$ is a field,~$1\in K$.)
\end{ex}

Let $\Q\colon K\to\C$ be a homomorphism of fields, i.e., for all $a,b\in K$,
\[
  \sigma(a+b)=\sigma(a)+\sigma(b)\quad\text{and}\quad
  \sigma(ab)=\sigma(a)\sigma(b).
\]

\begin{prop}
   With notation as above,
   \begin{enumerate}
     \item The homomorphism $\sigma$ is either injective or identically~$0$.
     \item If $\sigma$ is injective, then $\sigma$ is the identity mapping when
       restricted to $\Q\in K$.
     \item Suppose that $\alpha\in K$ and $h\in \Q[x]$ with $h(\alpha)=0$.
       If~$\sigma\neq0$, then
       $h(\sigma(\alpha))=0$.  Thus, $\sigma$  permutes the roots of~$h$ in~$\C$.
   \end{enumerate}
\end{prop}
\begin{proof}\ 
 \begin{enumerate}[leftmargin=*] 
   \item First, as an exercise, check that $\ker\sigma$ is an ideal in $K$. (Hint:
     $\ker\sigma$ is nonempty since~$\sigma(0)=0$, then check that if
     $\alpha,\beta\in\ker\sigma$ and~$\gamma\in K$, then $\alpha+\beta,\gamma
     \alpha\in\ker\sigma$.)  Next,
     suppose~$\ker\sigma\neq(0)=\{0\}$.  Let~$\alpha$ be a nonzero element of
     $\ker\sigma$.  Since~$K$ is a field, $\frac{1}{\alpha}\in K$, and since
     $\ker\sigma$ is an ideal, $\frac{1}{\alpha}\cdot\alpha=1\in\ker\sigma$.
     So~$\ker\sigma=(1)=K$, i.e., $\sigma=0$.
   \item Suppose $\sigma$ is injective.  Then the standard argument shows
     that~$\sigma(1)=1$:
     \[
       \sigma(1)=\sigma(1\cdot 1)=\sigma(1)\sigma(1).
     \]
     Since $\sigma$ is injective, $\sigma(1)\neq0$.  Multiplying the above
     equation through by $1/\sigma(1)$, gives~$\sigma(1)=1$.

     Then, for each $n\in\N$,
     \[
       \sigma(n)=\sigma(\underbrace{1+\cdots+1}_{n \text{ times}})
       =\underbrace{\sigma(1)+\cdots+\sigma(1)}_{n \text{ times}}
       =\underbrace{1+\cdots+1}_{n \text{ times}}
       =n.
      \]
     The rest is left as an exercise: if $n\in\N$ with $n\neq0$, show
     $\sigma(1/n)=1/\sigma(n)$ by applying $\sigma$ to the identity $(1/n)n=1$;
     next show $\sigma(m/n)=\sigma(m)/\sigma(n)$ for all $m,n\in\N$	with~$n\neq0$; finally, show
     that for any $\alpha\in K$, we have $\sigma(-\alpha)=-\sigma(\alpha)$.
   \item Suppose $\sigma\neq0$, in which case~$\sigma$ is injective.
     Say $h=\sum_{i=1}^{n} a_i x^{i}$ and that~$h(\alpha)=0$.  Then, using
     the fact that $\sigma$ preserves sums and products, $\sigma$ is the identity on
     $\Q$, and the $a_{i}\in\Q$,
     \[
       0=\sigma(0)=\sigma (\textstyle\sum_{i=1}^{n} a_i \alpha^{i})
      = \sum_{i=1}^{n} \sigma(a_{i}) (\sigma(\alpha))^{i}
      =\sum_{i=1}^{n} a_i (\sigma(\alpha))^{i}=h(\sigma(\alpha)).
     \]
 \end{enumerate}
\end{proof}
\bigskip

\textbf{Embeddings of number fields.} Let $K$ be a number field (finite
extension of~$\Q$ inside $\C$).  By an \emph{embedding} of $K$ into~$\C$, we
mean an injective homomorphism $\sigma\colon K\to\C$.  It turns out the problem
of describing all of the embedding of $K$ into~$\C$ has a beautiful solution,
which we now describe.

By the primitive element theorem there exists an algebraic number~$\theta\in K$
such that $K=\Q(\theta)=\Q[\theta]$ (we could even take~$\theta$ to be an
algebraic \emph{integer}, but that is not important here).  Let $p\in\Q[x]$ be the
minimal polynomial for $\theta$ over~$\Q$, and say~$\deg(p)=n$.  We have seen
that $[K\mathbin{:}\Q]=n$.  Then we have the following characterization of all
of the embeddings of $K$ into $\C$ (whose proofs appear in a course in algebra):
\begin{enumerate}
  \item The number of embeddings of~$K$ into $\C$ is
    $n=[K\mathbin{:}\Q]=\deg(p)$.
  \item Let $\sigma_{1},\ldots\sigma_{n}$ be the embeddings of $K$ into~$\C$ and
    define $\theta_{i}:=\sigma_{i}(\theta)$ for $i=1,\ldots,n$.  Then
    the~$\theta_{i}$ are distinct, and they are precisely the roots of~$p$.  So
    \[
      p=\prod_{i=1}^{n} (x-\theta_{i})=\prod_{i=1}^{n}(x-\sigma_{i}(\theta)).
    \]
  \item If $\theta_{i}$ is any root of~$p$, then $\theta\mapsto\theta_{i}$
    determines the embedding~$\sigma_{i}$.  (To see this, recall that
    $\{1,\theta,\ldots,\theta^{n-1}\}$ is a basis for~$K$ over~$\Q$.  Then, any homomorphism sending
    $\theta\mapsto\theta_{i}$ will send $\sum_{j=1}^{n} \alpha_{j}\theta^{j}$ to
    $\sum_{j=1}^{n} \alpha_{j}\theta_{i}^{j}$.  So the value of the homomorphism
    is determined for all elements of $K$.)
\end{enumerate}

\begin{eg}
\begin{enumerate}[leftmargin=*]
  \item What are the embeddings of $\Q(\sqrt{5})$ into~$\C$?  The minimal
    polynomial for~$\sqrt{5}$ is
    \[
      p = x^2-5=(x-\sqrt{5})(x+\sqrt{5}).
    \]
    The roots of~$p$ are $\sqrt{5}$ and $-\sqrt{5}$.  So we get two embeddings:
    \begin{align*}
      \sigma_{1}(r+s\sqrt{d})&:=\mathrm{id}(r+s\sqrt{5})=r+s\sqrt{5}\\
      \sigma_{2}(r+s\sqrt{d})&:=r-s\sqrt{5}.
    \end{align*}
    Note that in this case, the images of both embeddings~$\sigma_{i}\colon
    \Q(\sqrt{5})\to\C$ actually lie in $\Q(\sqrt{5})$.  So each embedding is an
    isomorphism of~$\Q(\sqrt{5})$ with itself.
  \item What are the embeddings of $\Q(\sqrt[3]{5})$?  Let 
    \[
      \omega=e^{2\pi i/3}=\cos(2\pi/3)+i\sin(2\pi/3)=\frac{-1+i\sqrt{3}}{2},
    \]
    a cube root of unity.  Then the minimal polynomial for $\sqrt[3]{5}$ is
    \[
      p=x^3-5=(x-\sqrt[3]{5})(x-\omega\sqrt[3]{5})(x-\omega^{2}\sqrt[3]{5}).
    \]
    The three embeddings of $\Q(\sqrt[3]{5})$ are given by
    \begin{align*}
      \sigma_{1}(1+a\sqrt[3]{5}+b(\sqrt[3]{5})^{2})&:=\mathrm{id}(1+a\sqrt[3]{5}+b(\sqrt[3]{5})^{2})=1+a\sqrt[3]{5}+b(\sqrt[3]{5})^{2}\\
      \sigma_{2}(1+a\sqrt[3]{5}+b(\sqrt[3]{5})^{2})&:=1+a \omega\sqrt[3]{5}+b(\omega\sqrt[3]{5})^{2}
      =1+a \omega\sqrt[3]{5}+b\omega^{2}(\sqrt[3]{5})^{2} \\
      \sigma_{3}(1+a\sqrt[3]{5}+b(\sqrt[3]{5})^{2})
      &:=1+a \omega^{2}\sqrt[3]{5}+b(\omega^{2}\sqrt[3]{5})^{2}
      =1+a \omega^{2}\sqrt[3]{5}+b\omega(\sqrt[3]{5})^{2}.
    \end{align*}
    Unlike the previous example, note that neither $\im(\sigma_{2})$ nor
    $\im(\sigma_{3})$ are contained in $\Q(\sqrt[3]{5})$.
\end{enumerate}  
\end{eg}
\bigskip

\textbf{The discriminant.}  Let $K=\Q(\theta)$ be a number field with
$[K\mathbin{:}\Q]=n$.  Let $p\in\Q[x]$ be the minimal polynomial of~$\theta$.
Then $p$ has~$n$ distinct complex roots~$\theta_{1},\ldots,\theta_{n}$, and $p$
factors as follows:
\[
  p =\prod_{i=1}^{n}(x-\theta_{i}).  
\]
Let $\sigma_{i}$ be the embedding of~$K$ defined by letting
$\theta\mapsto\theta_{i}$.
\begin{defn}
  The \emph{discriminant} for a basis $\alpha_{1},\ldots,\alpha_{n}$ for~$K$
  over~$\Q$ is the square of the determinant of the~$n\times n$ matrix with
  $i,j$-th entry $\sigma_{i}(\alpha_{j})$:
  \[
    \Delta[\alpha_{1},\ldots,\alpha_{n}] :=
    \left(\det(\sigma_{i}(\alpha_{j}))\right)^{2}.
  \]
\end{defn}

\begin{eg}
  Let $K=\Q(\sqrt{d})$ where $d$ is a square-free integer~$\neq 0,1$.  Then
  \[
    \Delta[1,\sqrt{d}]=\left(\det
    \left(\begin{array}{rr}
    1 & \sqrt{d} \\
    1 & -\sqrt{d}
    \end{array}\right)\right)^2
    =(-2\sqrt{d})^{2}=4d.
  \]
\end{eg}

\begin{prop}
  Let $\alpha_{1},\ldots,\alpha_{n}$ and $\beta_{1},\ldots,\beta_{n}$ be bases
  for the number field $K$ over~$\Q$.  Let $C$ be the change of basis matrix
  from the $\alpha_{i}$ to the $\beta_{i}$.  Then
  \[
    \Delta[\beta_{1},\ldots,\beta_{n}]=(\det
    C)^{2}\Delta[\alpha_{1},\ldots,\alpha_{n}].
  \]
\end{prop}
\begin{proof}
  We have~$\beta_{j}=\sum_{k=1}^{n} c_{kj}\alpha_{k}$ with~$c_{kj}\in\Q$.
  Let $A = (\sigma_{i}(\alpha_{j}))$ and $B=(\sigma_{i}(\beta_{j}))$.  The
  $i,j$-th element of $AC$ is
  \[
    \sum_{k=1}^{n} a_{ik}c_{kj}=\sum_{k=1}^{n} \sigma_{i}(\alpha_{k})c_{kj}
    =\sum_{k=1}^{n} \sigma_{i}(c_{kj}\alpha_{k})
    =\sigma_{i}(\textstyle\sum_{k=1}^{n} c_{kj}\alpha_{k})
    =\sigma_{i}(\beta_{j}).
  \]
  (We have used the fact that $\sigma_{i}$ is the identity when restricted
  to~$\Q$ in order to bring $c_{kj}$ inside $\sigma_{i}$, above.)
  Therefore
  \[
    \det(\sigma_{i}(\beta_{j}))=\det(AC)=\det
    A\det(C)=\det(\sigma_{i}(\alpha_{j}))\det C.
  \]
  Squaring both sides yields the result.
\end{proof}

\begin{thm} Let $K=\Q(\theta)$ be a number field, with embeddings $\sigma_{i}$
  and with $\theta_{i}=\sigma_{i}(\theta)$ for $i=1,\ldots,n$. Let
  $\alpha_{1},\ldots,\alpha_{n}$ be a basis for~$K$ over~$\Q$.  Then the
  discriminant $\Delta[\alpha_{1},\ldots,\alpha_{n}]$ is a nonzero rational
  number.  It is positive if all of the $\theta_{i}$ are real. It is a rational
  integer if the $\alpha_{i}$ are algebraic integers.
\end{thm}
\begin{proof}[Sketch of proof]
   We have seen that $1,\theta,\ldots,\theta^{n-1}$ is a basis for~$K$
   over~$\Q$, and we have $\sigma_{i}(\theta^{j})=\theta_{i}^{j}$.  Therefore,
   \[
     \Delta[1,\theta,\ldots,\theta^{n-1}]=\det(\theta_{i}^{j})
     =\prod_{1\leq i<j\leq n}(\theta_{i}-\theta_{j})^{2}.
   \]
   (For the final step, see, for example, \href{https://en.wikipedia.org/wiki/Vandermonde_matrix}{Vandermonde
   matrix} in Wikipedia  This may appear as homework, too.) 
   Letting~$C$ be the change of basis matrix from $1,\theta,\ldots,\theta^{n-1}$
   to $\alpha_{1},\ldots,\alpha_{n}$, we have
   \[
     \Delta[\alpha_{1},\ldots,\alpha_{n}]=(\det
     C)^{2}\Delta[1,\theta,\ldots,\theta^{n-1}]
     =(\det C)^{2}\prod_{1\leq i<j\leq n}(\theta_{i}-\theta_{j})^{2}.
   \]
   Since the $\theta_{i}$ are distinct, the discriminant is nonzero, and
   if the~$\theta_{i}$ are real, the discriminant is positive.
\end{proof}
\end{document}
