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

\centerline{Discriminants}
\bigskip

Our goal is to prove the theorem on discriminants stated last time:

\begin{thm}~\label{thm:03wed}
  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}

We first introduce the necessary tools.
\bigskip

\textbf{Vandermonde matrix.} Let $x_{1},\ldots,x_{n}$ be indeterminates, and
consider the~$n\times n$ matrix
\[
  V=
  \left(\begin{array}{ccccc}
  1 & x_{1} & x_{1}^{2} & \ldots & x_{1}^{n-1} \\
  1 & x_{2} & x_{2}^{2} & \ldots & x_{2}^{n-1} \\
  1 & x_{3} & x_{3}^{2} & \ldots & x_{3}^{n-1} \\
  1 & \vdots & \vdots & \ddots & \vdots \\
  1 & x_{n} & x_{n}^{2} & \ldots & x_{n}^{n-1}
  \end{array}\right).
\]
Then 
\[
  \det V = \prod_{1\leq i<j\leq n}(x_{j}-x_{i}).
\]
\begin{proof}[Sketch of proof]
  Think of $\det V$ as an element of the polynomial ring
  $\Q[x_{1},\ldots,x_{n}]$.  Note that if we set $x_{i}=x_{j}$, then $V$ has two
  equal rows, and hence, the determinant becomes~$0$.  It turns out the this
  means that $x_{j}-x_{i}$ divides $\det V$ for all $1\leq i<j\leq n$.  Next,
  compare degrees.  We find $\deg V=\binom{n}{2}=\deg\prod_{1\leq i<j\leq
  n}(x_{j}-x_{i})$.  Hence, 
  \[
    \det V=r\prod_{1\leq i<j\leq n}(x_{j}-x_{i})
  \]
  for some $r\in\Q$. Next, the coefficient of $x_{2}x_{3}^{2}\cdots x_{n}^{n-1}$
  is $1$ on the left-hand side of the above equation and $r$ on the other.  So
  it follows that $r=1$.
\end{proof}
\bigskip

\textbf{Symmetric polynomials.}  Let $R$ be a ring, and consider
$R[x_{1},\ldots,x_{n}]$, the ring of polynomials in $n$ variables with
coefficients in $R$.  If $\pi$ is a permutation of the numbers $1,\ldots,n$,
and $f\in R[x_{1},\ldots,x_{n}]$, define a new polynomial~$f^{\pi}\in
R[x_{1},\ldots,x_{n}]$ by
\[
  f^{\pi}(x_{1},\ldots,x_{n})= f(x_{\pi(1)},\ldots,x_{\pi(n)}).
\]
In this way, we get an action of the symmetric group~$S_{n}$ of permutations
of~$1,\ldots,n$ on the polynomial ring $R[x_{1},\ldots,x_{n}]$.  
A polynomial~$f$ is \emph{symmetric} if $f^{\pi}=f$ for all $\pi\in S_{n}$.

\begin{eg}
  Suppose that $n=4$ and $\pi$ is the permutation that $\pi(1)=3$, $\pi(2)=1$, $\pi(3)=4$,
  and $\pi(4)=2$,  Let $f=3x_{1}^{2}-5x_{2}x_{3}+x_{1}x_{4}^{3}$.  Then 
  \[
    f^{\pi}=3x_{3}^{2}-5x_{1}x_{4}+x_{3}x_{2}^{3}.
  \]
  Since $f\neq f^{\pi}$, the polynomial~$f$ is not symmetric.   On the other
  hand, the polynomial $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}$ is symmetric,
  as is $x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}$.
\end{eg}

\begin{defn}
  For $1\leq r\leq n$, the \emph{elementary symmetric polynomials}
  in~$x_{1},\ldots,x_{n}$ are $s_{r}(x_{1},\ldots,x_{n})$ formed by summing all
  products of exactly $r$ of the indeterminates $x_{1},\ldots,x_{n}$:
  \begin{align*}
    s_{1}
    &=x_{1}+x_{2}+\cdots+x_{n}\\
    s_{2}
    &=x_{1}x_{2}+x_{1}x_{3}+\cdots+x_{n-1}x_{n}\\
    &\vdots\\
    s_{n}
    &=x_{1}x_{2}\ldots x_{n}.
  \end{align*}
\end{defn}
Let $h\in R[x_{1},\ldots,x_{n}]$ be any polynomial. Note that
$h(s_{1},\ldots,s_{n})$ is symmetric.  For
instance,~$2s_{1}^{2}-5s_{2}^{2}x_{5}^{4}$ is symmetric.  A first theorem in
the theory of symmetric functions is that the converse holds:
\begin{thm}
  (I.~Newton) Let~$R$ be a ring, and let $f\in R[x_{1},\ldots,x_{n}]$.  Then~$f$
  is symmetric if and only if there exists~$h\in R[x_{1},\ldots,x_{n}]$ such
  that
  \[
    f=h(s_{1},\ldots,s_{n})
  \]
  where the~$s_{i}$ are the elementary symmetric polynomials.
\end{thm}
\begin{proof}
 See our textbook, Theorem~1.12.
\end{proof}

We now explain how the theory of symmetric functions is crucially connected to
our subject.  Suppose, for example that $f\in \Q[x]$ is a monic polynomial of
degree~$4$.  By the fundamental theorem of arithmetic, there exists
$\theta_{1},\ldots,\theta_{4}\in\C$ such that
\[
  f=(x-\theta_{1})(x-\theta_{2})(x-\theta_{3})(x-\theta_{4}).
\]
Expand the product to find that
\begin{align*}
  f &= x^{4}
  -(\theta_{1}+\cdots+\theta_{4})x^{3} 
  + (\theta_{1}\theta_{2}+\cdots+\theta_{3}\theta_{4})x^{2} 
  - (\theta_{1}\theta_{2}\theta_{3}+\cdots+\theta_{2}\theta_{3}\theta_{4})x 
  + (\theta_{1}\theta_{2}\theta_{3}\theta_{4})\\[5pt]
  &= x^4
  -s_{1}(\theta_{1},\theta_{2},\theta_{3},\theta_{4})x^{3}
  +s_{2}(\theta_{1},\theta_{2},\theta_{3},\theta_{4})x^{2}
  -s_{3}(\theta_{1},\theta_{2},\theta_{3},\theta_{4})x
  +s_{4}(\theta_{1},\theta_{2},\theta_{3},\theta_{4}).
\end{align*}
Thus, the coefficients of $f$ are, up to sign, the elementary functions of the
roots of $f$.  Further, since $f$ has rational coefficients, we see that the
elementary functions of the roots of~$f$ are rational numbers.  If $f$ has
integer coefficients, then these elementary functions would be integers.

\begin{eg}
  Let $\omega = e^{2\pi i/3}=\frac{-1+i\sqrt{3}}{2}$ be a cube root of~$1$.  We
  have
  \begin{align*}
    x^{3}-1 &= (x-1)(x-\omega)(x-\omega^{2})\\[5pt]
    &=x^{3}-(1+\omega+\omega^{2})x^{2}
    +(1\cdot \omega+1\cdot\omega^{2}+\omega\cdot\omega^{2})x
    -(1\cdot\omega\cdot\omega^{2})\\[5pt]
    &=x^3-s_{1}(1,\omega,\omega^{2})x^{2}
    +s_{2}(1,\omega,\omega^{2})x
    -s_{3}(1,\omega,\omega^{2}).
  \end{align*}
  Comparing coefficients, we see that
  \begin{align*}
    s_{1}(1,\omega,\omega^{2})&=1+\omega\cdot+\omega^{2}=0 \\[5pt]
    s_{2}(1,\omega,\omega^{2})&=
       1\cdot \omega+1\cdot\omega^{2}+\omega\cdot\omega^{2}
     =\omega+\omega^{2}+1=0 \\[5pt]
    s_{3}(1,\omega,\omega^{2})&=1\cdot\omega\cdot\omega^{2}=1.
  \end{align*}
  Since $x^3-1$ has integer coefficients, the elementary functions of its roots
  are all integers, too.
\end{eg}
\bigskip

\begin{proof}[Proof of Theorem~\ref{thm:03wed}]
  We have seen that $1,\theta,\ldots,\theta^{n-1}$ is a basis for~$K$
  over~$\Q$. We have
  $\sigma_{i}(\theta^{j})=\sigma_{i}(\theta)^{j}=\theta_{i}^{j}$.  Therefore,
  calculating the discriminant involve taking the determinant of a Vandermonde
  matrix:
  \[
    \Delta[1,\theta,\ldots,\theta^{n-1}]=
    \left(\begin{array}{ccccc}
	1 & \theta_{1} & \theta_{1}^{2} & \ldots & \theta_{1}^{n-1} \\
	1 & \theta_{2} & \theta_{2}^{2} & \ldots & \theta_{2}^{n-1} \\
	1 & \theta_{3} & \theta_{3}^{2} & \ldots & \theta_{3}^{n-1} \\
	\vdots & \vdots & \vdots & \ddots & \vdots \\
	1 & \theta_{n} & \theta_{n}^{2} & \ldots & \theta_{n}^{n-1}
    \end{array}\right)^{2}
    =\prod_{1\leq i<j\leq n}(\theta_{j}-\theta_{i})^{2}.
  \]
  Letting~$C$ be the change of basis matrix from $1,\theta,\ldots,\theta^{n-1}$
  to $\alpha_{1},\ldots,\alpha_{n}$, we have
  \[
    \Delta[\alpha_{1},\ldots,\alpha_{n}]=(\det
    C)^{2}\Delta[1,\theta,\ldots,\theta^{n-1}]
    =(\det C)^{2}\prod_{1\leq i<j\leq n}(\theta_{j}-\theta_{i})^{2}.
  \]
  Since the $\theta_{i}$ are distinct, the discriminant is nonzero, and
  if the~$\theta_{i}$ are real, the discriminant is positive.
  To see that the discriminant is rational in general, note that the above
  expression is a symmetric polynomial in the $\theta_{i}$.  Therefore, the
  discriminant can be written as a rational polynomial combination of the
  elementary symmetric functions in the $\theta_{i}$.   Recall that
  the~$\theta_{i}$ are the roots to minimal polynomial for $\theta$, which is a
  monic polynomial with rational coefficients.  Therefore, as we have seen, the elementary
  symmetric functions in~$\theta_{i}$ are rational numbers. 

  Finally, suppose that $\alpha_{1},\ldots,\alpha_{n}$ are algebraic integers.
  This means that for each~$j$ there exists a monic polynomial $p_{j}=p_{j}(x)$ with
  \emph{integer} coefficients such that $p_{j}(\alpha_{j})=0$.  For each $i$ and
  $j$, we
  claim that $\sigma_{i}(\alpha_{j})$ is an algebraic integers.  To see this,
  say that $p_{j}=x^{k}+c_{1j}x^{k-1}+\cdots+c_{1k}$.  Then
  \begin{align*}
    p_{j}(\sigma_{i}(\alpha_{j}))
    &=(\sigma_{i}(\alpha_{j}))^{k}+c_{1j}(\sigma_{i}(\alpha_{j}))^{k-1}+\cdots+c_{1k}
    \\[5pt]
    &=\sigma_{i}(\alpha_{j}^{k})+\sigma_{i}(c_{1j}(\alpha_{j}))^{k-1}+\cdots+c_{1k}
    \\[5pt]
    &=\sigma_{i}(\alpha_{j}^{k}+c_{1j}\alpha_{j}^{k-1}+\cdots+c_{1k}) \\[5pt]
    &= \sigma_{i}(p_{j}(\alpha_{j}))=0.
  \end{align*}
  We are using the fact that~$\sigma_{i}$ is a homomorphism of fields, hence
  preserves algebraic operations, and that $\sigma_{i}$ is the identity when
  restricted to $\Q$, and hence~$\sigma_{i}(a_{ik})=a_{ik}$ for all~$k$.  We
  have just demonstrated that $\sigma_{i}(\alpha_{j})$ satisfies a monic
  polynomial with integer coefficients.  Hence, $\sigma_{i}(\alpha_{j})$ is an
  algebraic integer.

  Next, we know that the algebraic integers form a ring.  Therefore, the
  discriminant
  \[
    \Delta[\alpha_{1},\ldots,\alpha_{n}]=(\det(\sigma_{i}(\alpha_{j})))^{2}
  \]
  is an algebraic integer.  However, we also know that the discriminant is a
  rational number.  Therefore, it must be a (rational) integer.
\end{proof}
\end{document}
