\documentclass[12pt]{article}
\usepackage{amsfonts, amsthm, amsmath}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{8.5in}
\setlength{\topmargin}{0in}
\setlength{\headheight}{0in}
\setlength{\headsep}{0in}
\setlength{\parskip}{0pt}
\setlength{\parindent}{20pt}

\def\CC{\mathbb{C}}
\def\FF{\mathbb{F}}
\def\PP{\mathbb{P}}
\def\QQ{\mathbb{Q}}
\def\RR{\mathbb{R}}
\def\ZZ{\mathbb{Z}}
\def\gotha{\mathfrak{a}}
\def\gothb{\mathfrak{b}}
\def\gothm{\mathfrak{m}}
\def\gotho{\mathfrak{o}}
\def\gothp{\mathfrak{p}}
\def\gothq{\mathfrak{q}}
\DeclareMathOperator{\Disc}{Disc}
\DeclareMathOperator{\Gal}{Gal}
\DeclareMathOperator{\Norm}{Norm}
\DeclareMathOperator{\Trace}{Trace}
\DeclareMathOperator{\Cl}{Cl}

\def\head#1{\medskip \noindent \textbf{#1}.}
\def\fixme#1{\textbf{FIXME! #1}}

\begin{document}

\begin{center}
\bf
Math 254A, UC Berkeley, Fall 2001 (Kedlaya) \\
Addendum on the Relative Discriminant \\
October 31, 2001
\end{center}

A lot of you had trouble with the homework problems on the relative
discriminant last week. One big problem was typos on the problem set.
The correct statement of problem 2 was:
If $L/K$ is a number field, prove that the ideal $(\Disc(L)/\Disc(K)^{[L:K]})$
is the ideal generated by the absolute norm of the relative discriminant
$\Disc(L/K)$.
(I forgot the factor of $[L:K]$ in the exponent. Mea maxima culpa.)

One approach to computing this is the one taken by Neukirch in Section 3.2,
using the different. I had in mind a more direct argument, something like the
following.

It is true (but I didn't say this explicitly in class, and should have)
that trace, norm and the discriminant can be computed two different ways for
local fields just like for number fields. Namely, if $L/K$ is an extension
of finite extensions of $\QQ_p$ of degree $n$, there will be exactly
$n$ embeddings $\tau_1, \dots, \tau_n$ of $L$ into $\overline{K}$
(that much is true for any separable extension), and we can write
\[
\Trace_{L/K}(\alpha) = \sum_i \tau_i(\alpha), \quad
\Norm_{L/K}(\alpha) = \prod_i \tau_i(\alpha), \quad
\Disc(\alpha_1, \dots, \alpha_n)
= (\det (\tau_i(\alpha_j))_{i,j=1}^n)^2;
\]
the equivalence of these definitions with the original definitions
works the same way as in the number field case.

Using problem 1 and the relationship between the absolute discriminant and
Neukirch's form of the relative discriminant,
problem 2 reduces to the following:
if $M/L$ and $L/K$ are extensions of finite extensions of $\QQ_p$, then
\begin{equation}
\Disc(M/K) = \Disc(L/K)^{[M:K]} \Norm_{L/K} \Disc(M/L),
\end{equation}
where $\Norm_{L/K}$ sends an ideal of $\gotho_L$ generated by $\alpha$
to the ideal $(\Norm_{L/K}(\alpha))$. Here's one way to prove this.

Put $m = [M:L]$
and $n = [L:K]$, choose a basis $\alpha_1, \dots, \alpha_m$ of
$\gotho_M$ over $\gotho_L$ 
and a basis $\beta_1, \dots, \beta_m$ of $\gotho_L$ over $\gotho_K$,
and choose an isomorphism $f: \overline{L} \to
\overline{K}$ of fields
once and for all. (Note: $\overline{L}$ is a field together with a fixed
map of $L$ into it, so I'll push elements of $L$ into it without comment.
Likewise, $\overline{K}$ is a field together with a fixed
map of $K$ into it.)

Let $\tau_1, \dots, \tau_m$ be the embeddings of
$M$ into $\overline{L}$ over $L$, and let $\sigma_1, \dots, \sigma_n$ be the
embeddings of $L$ into $\overline{K}$ over $K$. Choose automorphisms
$\rho_1, \dots, \rho_n$ of $\overline{K}$ such that
$\rho_i \circ f = \sigma_i$ for each $i$.
Now $\rho_i \circ \tau_j$ gives all of the embeddings of $M$ into
$\overline{K}$ over $K$, so using the basis
$\alpha_k \beta_l$ of $\gotho_M$ over $\gotho_K$ (for
$k=1, \dots, m$ and $l=1, \dots, n$), I can write
$\Disc(M/K) = (\det C)^2$, where the matrix $C$ has rows and columns
indexed by ordered pairs and is given by
\begin{equation}
C_{(i,j)(k,l)} = \rho_i(\tau_j((\alpha_k \beta_l)))
= \rho_i(\tau_j(\alpha_k)) \cdot \rho_i(\beta_l)).
\end{equation}
(The $\tau_j$ dropped out in the second term because
we have a fixed identification of
$L$ as a subfield of $\overline{L}$, and $\tau_j$ is $L$-linear.)

I'd like to factor this determinant using the identity
$\det (A \otimes B) = (\det A)^n (\det B)^m$ for the determinant of
the tensor product of an $m \times m$ matrix $A$ and an
$n \times n$ matrix $B$. (The \emph{tensor product} of $A$ and $B$ is
the matrix with rows and columns indexed by pairs $(i,j)$ and
$(A \otimes B)_{(i,j)(k,l)} = A_ik B_{jl}$. Why it's called that is
left to the reader.) Unfortunately, I haven't quite got that yet, but
I can modify the determinant in question to bring this about.

For ease of notation, let $T$ be the matrix with $T_{ij} = \tau_i(\alpha_j)$,
so that $C_{(i,j)(k,l)} = \rho_i(M_{jk}) \rho_i(\beta_l)$.
Now define the matrix $D$ by
\[
D_{(i,j)(k,l)} = \begin{cases} (T \rho_i(T)^{-1})_{jl} & i=k \\
  0 & i \neq k. \end{cases}
\]
We have
\begin{align*}
(DC)_{(i,j)(k,l)} &= \sum_{(r,s)}
D_{(i,j)(r,s)} C_{(r,s)(k,l)} \\
&= \sum_s D_{(i,j)(i,s)} C_{(i,s)(k,l)} \\
&= \sum_s (T\rho_i(T)^{-1})_{js} \rho_i(T)_{jk} \rho_i(\beta_l) \\
&= T_{jk} \rho_i(\beta_l).
\end{align*}
That is, $DC$ is the tensor product of the matrices $M$ and
$N$, where $N_{il} = \rho_i(\beta_l)$. Since $D$ is a block matrix, we
can compute its determinant as
\[
\det(D) = \prod_i \frac{\det T}{\det \rho_i(T)}.
\]
We conclude
\begin{align*}
\det C &= (\det T^{-m} \prod_i \det \rho_i(T)) 
\det T^m \det N^m) \\
&= (\det N)^m \prod_i \rho_i(\det T) \\
&= (\det N)^m \Norm_{L/K} (\det T).
\end{align*}
To conclude, observe that $(f^{-1}(\det T))^2 = (\Disc(M/L))$ and
$(\det N)^2 = (\Disc(L/K))$.

\end{document}