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\begin{document}

\chapter*{September 26:}

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

\bigskip

\begin{itemize}
\item
Traces and Norms
\item 
Discriminants
\end{itemize}

\section{Traces and Norms}

If $E/F$ is a finite extension of fields, then for every element $x \in E$
we get a ``multiplication by~$x$'' mapping from $E$ to $E$ that is an
endomorphism of an $F$-vector space of dimension $n := [E:F]$.
The trace and norm of this element were defined to be the trace
and determinant of this linear transformation, respectively.

From these definitions it follow immediately that the trace and
norm are elements of~$F$. It also follows that the trace is
a homomorphism for addition, and the norm is a homomorphism for
multiplication.

\begin{theorem}
For all $x,y$ in $E$,
$$\trr_{E/F}(x+y) = \trr_{E/F}(x) + \trr_{E/F}(y), \qquad
N_{E/F}(xy) = N{E/F}(x) N{E/F}(y).$$
\end{theorem}

Now we express the trace and norm of an element of~$E$ in terms of the 
minimal polynomial of the element.

\begin{theorem} Let $\beta$ be any element of $E$, and assume that
its minimal polynomial over~$F$ is
$$\m_\beta(x) = x^k- a_1 x^{k-1} + \cdots + (-1)^k a_k.$$
Then $n$ is divisible by~$k$ and
$$\trr_{E/F}(\beta) = \frac{n}{k} a_1, \qquad N_{E/F}(\beta) = a_k^{n/k}.$$
\end{theorem}

\begin{proof}
It is fairly easy to verify this directly, but more amusing to do it by
using linear algebra.

First note that $F(\beta)$ has degree $k$ over~$F$, so that $n = [E:F]$
is divisible by~$k$.

It is known (under the rubric ``Rational Canonical Form'') that every
square matrix $A$ over a field is similar to a matrix that is a direct sum
of companion matrices
$$ A = C_1 + C_2 + \cdots + C_m$$
where $C_i$ is the companion matrix of a polynomial $f_i$ and
$f_i$ divides $f_{i+1}$ for $ 1 \le i < m$.

\begin{remark}
%The direct sum of matrices $A$ and $B$ is the block matrix
%
%\vspace*{2in}
%
%The companion matrix of a polynomial $f(x) = x^4+ax^3+bx+cx+d$ is
$$
\left[ \begin{array}{cccc}
0&0& 0&-d\\
1&0&0&-c\\
0&1&0&-b\\
0&0&1&-a
\end{array}
\right]
$$

This generalizes in an obvious way to arbitrary monic polynomials, and
the key property is that the characteristic polynomial of the companion
matrix of~$f$ is $f$ itself.  
\end{remark}

\begin{remark}
The $f_i$ in the rational canonical form are called invariant factors
of $A$.  Moreover, they are unique: two matrices 
have the same rational canonical form if and only if they 
are similar.
\end{remark}

\begin{remark}
In the rational canonical form, the characteristic polynomial of~$A$ is
the product of the $f_i$.  The minimal polynomial, i.e., the monic
polynomial of smallest degree that vanishes when $A$ is substituted,
is the last invariant factor $f_m$.
\end{remark}

Let $mul_\beta$ denote the ``multiplication by~$\beta$'' linear transformation
on $E$, as an $F$ vector space.
Since the minimal polynomial of $mul_\beta$ is irreducible, the 
invariant factors are of the form $1,1, \cdots, \m_\beta,\m_\beta, \cdots , \m_\beta$.
Thus the characteristic polynomial of $mul_\beta$ is a power of $\m_\beta$, say
$$f_\beta = \m_\beta^r.$$
By comparing degrees, we see that $n = kr$.

The trace of $mul_\beta$ is the negative of
the coefficient of $x^{n-1}$ in the characteristic polynomial, so
$$\trr_{E/F}(\beta) = r a_1.$$
The determinant of $mul_\beta$ is $(-1)^n$ times the constant coefficient,
so
$$(-1)^n N_{E/F}(\beta) =  \left ( (-1)^k a_k \right )^r.$$
The relationships in the theorem now follow from the fact that $n = kr$.
\end{proof}

Now we express the trace and norm in terms of the embeddings of~$E$ into
a splitting field.  Using the Primitive Element Theorem, there is an
$\alpha$ such that $E = F(\alpha)$.  If $K$ is the splitting field
of $\m_\alpha$ then by the Embedding Theorem there are~$n$ embeddings
of $E$ into~$K$; label them
$\sigma_1, \sigma_2, \cdots , \sigma_n$ where we let $\sigma_1$ be
the identity.
To simplify notation slightly we write
$$\beta^{(i)} := \sigma_i(\beta)$$
where $\beta$ is any element of~$F$.

\begin{theorem}
$$\trr_{E/F}(\beta) = \sum_i \beta^{(i)}, \qquad
N_{E/F}(\beta) = \prod_i \beta^{(i)}.$$
\end{theorem}

\begin{corollary}
If $\beta \in F$ then
$$\trr_{E/F} = n \beta, \qquad N_{E/F}(\beta) = \beta^n.$$
\end{corollary}

\begin{proof} (of the theorem)
Let $[F(\beta):F] = k$, and $[E:F(\beta)] = r$, so that $kr = n$.
There are $k$ embeddings of $F(\beta)$ into $K$ and, by the
generalized form of the Emedding Theorem, each of those embeddings
has $r$ extensions to an embedding of~$E$ into~$K$.
We find that
$$\trr_{E/F}(\beta) = r a_1 = r \sum_{\tau} \tau(\beta)$$
since coefficient $a_1$ of $\m_\beta$ is the sum of the distinct
roots of~$\m_\beta$.  It follows that
$$\trr_{E/F}(\beta) =  \sum_{\tau} \tau(\beta) = 
\sum_{\sigma_i \mbox{ extends } \tau } \sum_\tau \tau(\beta) =
\sum_{i} \sigma_i(\beta) = \sum_i \beta^{(i)}$$
as desired.  The formula for the norm follows from a very
similar calculation.
\end{proof}

This result reduces the calculation of norms and traces to exercises
in elementary symmetric functions.  

\begin{corollary}
If $\beta = g(\alpha)$ for some polynomial~$g(x) \in F[x]$ then
$$\trr_{E/F}(\beta) = \sum_i g(\alpha^{(i)}), \qquad
N_{E/F}(\beta) = \prod_i g(\alpha^{(i)}).$$
\end{corollary}

The sum and product of $g$ evaluated at the conjugates of $\alpha$
are symmetric functions in the $\alpha^{(i)}$ and, by the basic
theorem on elementary symmetric functions, are polynomials in the
coefficients of~$\m_\alpha$.

\begin{example}
If $m_\alpha(x) = x^n - a_1 x^{n-1} + a_2 x^{n-2} - \cdots$ then
$$\trr(\alpha^2) = \sum_i \left ( \alpha^{(i)} \right) ^ 2
= \left ( \sum_i \alpha^{(i)} \right ) ^2 - 2 \sum_{i<j} \alpha^{(i)} \alpha^{(j)}
= a_1^2 - 2 a_2.$$
\end{example}

\begin{example}
If $u$ is an element of~$F$ then
$$N(\alpha-u) = \prod_i ( \alpha^{(i)}-u) = (-1)^n \prod_i (u - \alpha^{(i)}
= (-1)^n \m_\alpha(u).$$
\end{example}

\section{The discriminants of a basis}

Let $F$ be a number field of degree~$n$ over~$\Q$.

\begin{defin}
If $\alpha_1, \alpha_2, \cdots , \alpha_n$ is a basis of~$F$ over~$\Q$ then
the \defn{discriminant} of the basis is defined to be the square of
the determinant 
$$\disc(\alpha_1, \alpha_2, \cdots , \alpha_n) := \det[\alpha_i^{(j)}]^2$$
of the matrix whose entry in the $i$-th row and $j$-th column is the image of
$\alpha_i$ under the $j$-th embedding.
\end{defin}

\begin{remark}
Although the numbering of the basis elements is fixed, the numbering of
the embeddings is arbitrary.  However, if the embeddings are permuted the
determinant only changes at most by a sign, so the square of the determinant
is independent of the choice of numbering.
\end{remark}

\begin{theorem}
\begin{description}
\item[(a)]
The discriminant of a basis $\{\alpha_i\}$ is an element of~$\Q$.
\item[(b)] The discriminant is given by
$$\disc(\alpha_1, \alpha_2, \cdots , \alpha_n) = 
\det[\trr_{F/\Q}(\alpha_i \alpha_j)].
$$
\end{description}
\end{theorem}

\begin{proof}
The product of matrices $M = [a_{ij}]$ and $N = [b_{ij}]$ is
$$MN = \left[\sum_k a_{ik} b_{kj}\right]$$
where in each case the index~$i$ refers to the row and the index~$j$ refers
to the column.  Applying this to the matrix $M = [\alpha_i^{(j)}]$ in
the definition of the discriminant, and its transpose, we find that
\begin{eqnarray}
\disc(\alpha_i) & = &\det(M)^2 = \det(M) \det(M^t) = \det(MM^t)\\
& =  &\det([\sum_k \alpha_i^{(k)} \alpha_j^{(k)}])
= \det[\trr_{F/\Q}(\alpha_i \alpha_j].
\end{eqnarray}
This proves the second claim in the theorem, and the first claim
follows immediately since each trace, and therefore the determinant
of the matrix of traces, is a rational number.
\end{proof}

There is a special case where the discriminant is especially easy to
compute, namely when the basis is a ``power basis'' consisting of
the powers of an element $\alpha$:
$$1, \alpha, \alpha^2, \cdots , \alpha^{n-1}.$$

In this case the matrix $M$ in the definition of the discriminant is
$$M = \left[\left(\alpha^{(j)} \right)^{i}\right]_{0 \le i < n, 1 \le j \le n}.$$
This is the famous Vanderonde matrix, whose determinant is
$$\det(M) = \prod_{i<j} (\alpha^{(j)} - \alpha^{(i)} ).$$
Therefore
$$\disc(1,\alpha, \alpha^2, \cdots , \alpha^{n-1}) =
\prod_{i<j} \left ( \alpha^{(j)} - \alpha^{(i)} \right ) ^2$$
which is called the \defn{discriminant} of the (monic) polynomial~$\m_\alpha$.
In the near future we will learn techniques for calculating this
discriminant efficiently.

\begin{remark}
The word ``discriminant'' will be used in several different ways, though
there isn't much real chance of ambiguity, since we will always be taking
the discriminant of something specific.  So far, we know the discriminant
of a basis, and of a polynomial.  Later, we will learn that the ring
of integers has a basis, and will define the discriminant of a number
field to the the discriminant of such a basis.
\end{remark}
\end{document}
