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\begin{document}

\chapter*{September 28:}

\begin{center}
{\Huge \bf Math 431 Class Lecture Notes}
\end{center}

\bigskip

\begin{itemize}
\item
Discriminants, continued
\item
Examples
\end{itemize}

\section{Discriminants redux}

Let $F$ be a number field of degree~$n$ over~$\Q$.  If 
$\alpha_1, \cdots , \alpha_n$ is a basis of $F$ over~$\Q$
then its discriminant is defined by

$$\disc(\alpha_1, \alpha_2, \cdots , \alpha_n) = \det[\alpha_i^{(j)}]^2
= \det[\trr_{F/\Q}(\alpha_i \alpha_j)].$$
If $\beta_1, \cdots , \beta_n$ is a new basis, it can be described in
terms of the basis of $\alpha_i$ by a ``change of basis matrix'' 
$A = [a_{ij}]$ whose entries are determined by expressing each of
the new basis vectors in terms of the old:
$$\beta_i = \sum_{j=1}^n a_{ij} \alpha_j, \quad 1 \le i \le n.$$
The entries $a_{ij}$ are in the ground field~$\Q$, and the matrix~$A$ is
nonsingular since the change of basis matrix $B$ from the new basis to
the old satisfies $AB = BA = I_n$, i.e., $B$ is the inverse of the
matrix~$A$.  Indeed, from the linear independence of the $\alpha$'s and the
formula
$$\alpha_ i = \sum_{j=1}^n b_{ij} \sum _{k=1}^n a_{jk} \alpha_k$$
we see that the $ij$ entry of $BA$ is the same as the $ij$ entry of
the $n$ by~$n$ identity matrix~$I_n$.

If $M = [\alpha_i^{(j)}]$ is the matrix in the definition of the
discriminant of the original basis, then it is easy to work out that
the ``discriminant matrix'' for the new basis of $\beta$'s is $AM$.
It follows that
$$\disc(\beta_1, \cdots , \beta_n) = \det(AM)^n = \det(A)^2 \det(M)^2 =
\det(A)^2 \disc(\alpha_1, \cdots , \alpha_n).$$
Thus when bases are changed the discriminant changes by the square of the 
determinant of the change of basis matrix.

\begin{theorem}
The discriminant of any basis is nonzero.  Its sign depends only
on the field, and not on the choice of basis.
\end{theorem}

\begin{proof}
Let $F = \Q(\alpha)$.  Then the powers of $\alpha$, from $\alpha^0 = 1$
up to $\alpha^{n-1}$ form a basis, and the discriminant is the
discriminant of the minimal polynomial of $\alpha$.  The discriminant
of any polynomial with distinct roots is nonzero, and the first
statement in the theorem follows immediately.  The second follows
from the fact that the sign of the discriminant is unchanged by
multiplying by a square.
\end{proof}

\begin{remark}
Later we will see that the sign of the discriminant is 
$(-1)^{r_2}$ where $r_2$ is the number of pairs of complex conjugate
roots of the minimal polynomial of some (or any) generator~$\alpha$.
\end{remark}

\section{Examples}

\begin{example}
Let $F = \Q(\alpha)$ be a quadratic extension, where
$$m_\alpha(x) = x^2 + ax + b = (x-\alpha)(x-\alpha').$$
Then
$$\disc(1,\alpha) = \det \left [
\begin{array}{cc}
\trr(1) & \trr(\alpha) \\
\trr(\alpha) & \trr(\alpha^2)\end{array} \right]
= \det \left[\begin{array}{cc}
2 & -a \\
-a & a^2-2b\end{array}\right] = a^2-4b.$$
In calculating the traces we can find $\trr(\alpha^2)$ by
doing symmetric function calculations (the trace of a power is
a power sum, and we know the elementary symmetric functions), 
or by using linearity of the
trace, starting with $\alpha^2 = -a\alpha-b$ so that
$$\trr(\alpha^2) = \trr(-a\alpha-b) = -a \trr(\alpha) - b \trr(1)
= a^2-2b.$$

Note that the final form of the discriminant is what is classically
called the discriminant of the quadratic polynomial $x^2+ax+b$.
\end{example}

\begin{example}
We calculate the discriminant of a special form of a basis of  a cubic 
field.  Next time we will how to do these calculations with less
effort.

Suppose that $F = \Q(\alpha)$ is a cubic extension, with the minimal
polynomial of $\alpha$ having the form
$$\alpha^3 + a \alpha + b = 0.$$
If the three roots of the cubic are $\alpha_1 = \alpha, \alpha_2, \alpha_3$
then the discriminant of the basis $1, \alpha, \alpha^2$ is
$$\prod_{i<j}(\alpha_i-\alpha_j)^2$$
as we saw earlier by considering a Vandermonde matrix.

To find this explicitly in terms of $a$ and~$b$, it is convenient to
compute some traces, since
$$\disc(1,\alpha,\alpha^2) = \det \left[
\begin{array}{ccc}
\trr(1) & \trr(\alpha) & \trr(\alpha^2)\\
\trr(\alpha) & \trr(\alpha^2) & \trr(\alpha^3)\\
\trr(\alpha^2) & \trr(\alpha^3) & \trr(\alpha^4)\end{array}\right]. 
$$
We find the first three traces by simple symmetric function exercises:
$$\trr(1) = 3, \quad \trr(\alpha) = e_1 = 0, \quad
\trr(\alpha^2) = p_2 = e_1^2 - 2 e_2 = 0 - 2a = -2a.$$
The remaining traces can be computed by the convenient device of using
linearity (which, in effect, gives a recursion on the traces of powers
of~$\alpha$).
Specifically, from $\alpha^3 = - a \alpha - b$ we get
$$\trr(\alpha^3) = -a \trr(\alpha) - b\trr(1) = -3b$$
and from $\alpha^4 = - a \alpha^2 - b \alpha$ we get
$$\trr(\alpha^4) = -a \trr(\alpha^2) - b\trr(\alpha) = 2a^2.$$
Finally, taking the indicated determinant gives the desired
discriminant
$$\disc(1,\alpha,\alpha^2) = -4a^3-27b^2.$$
This is called the discriminant of the cubic polynomial.  With more
work one can calculate the discriminant of the general cubic
polynomial (i.e., one not missing the quadratic term); this discriminant
has 5 terms that are monomials in the three variables.
\end{example}

\begin{remark}
For a non-monic polynomial $f(x) = a \prod_{i=1}^n (x-\alpha_i)$ it turns
out to be best (for non-obvious reasons that might be touched on next time)
to define the discriminant to be
$$\disc(f) = a^{2n-2} \prod_{i<<j} (\alpha_i - \alpha_j)^2.$$
\end{remark}

\begin{remark}
The discriminant plays a famous role in galois theory in the following way.
Let $F$ be the splitting field of a polynomial~$f$ of degree~$n$ with
rational coefficients.  Then elements of the galois group 
$G := \mbox{Aut}(f)$ permute
the roots of~$f$, and distinct elements induce distint permutations of
the roots.  It follows that $G$ can be identified with a subgroup of
the symmetric group~$S_n$.

\begin{theorem}
The galois group $G$ is contained in the alternating group $A_n$ if and
only if the discriminant of~$f$ is a perfect square in~$\Q$.
\end{theorem}
\end{remark}

\begin{example}
Why does the determinant of the Vandermonde matrix have the specific
value that it does?  Well, as is often a bright idea in algebraic
environments, we replace numbers by variables, i.e., let $a_i$, $1\le i \le n$
be indeterminates and consider
$$\det \left[ \begin{array}{ccccc}
1 & a_1 & a_1^2 & \cdots & a_1^{n-1} \\
1 & a_2 & a_2^2 & \cdots & a_2^{n-1} \\
\vdots & \vdots & \vdots & \ddots & \\
1 & a_n & a_n^2 & \cdots & a_n^{n-1} \end{array}\right].$$
Clearly this is zero if equal values are substituted for any of the
variables.  This means that the polynomial in question is divisible
by $a_i - a_j$ for all distinct $i$ and~$j$.  However, the degree of
the product
$$\prod_{i<j} (a_j - a_i)$$
is $n(n-1)/2$, and this is also the degree of the indicated determinant.
Thus the Vandermonde determinant is equal to this product times a constant.
The diagonal term in the determinant expansion is
$$a_2 a_3^2 \cdots a_n^{n-1}$$
and one checks that this occurs in the product with coefficient~$1$.
Thus the Vandermonde determinant is equal to the product.
\end{example}


\end{document}

