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\begin{document}

\begin{center}
\bf
Math 254A, UC Berkeley, Fall 2001 (Kedlaya) \\
Lectures 28-29: Gauss Sums and Class Numbers \\
November 7-9, 2001
\end{center}

\head{Reference} Lang, \textit{Algebraic Number Theory},
Chapter IV, or Washington, \textit{Introduction to Cyclotomic Fields},
Chapter 6. (Washington gives slightly more general results than we will.)

\head{Outline of lectures}

Throughout, let $\zeta_n$ denote a primitive $n$-th root of unity,
let $p$ be a prime, let $m$ be an integer not divisible by $p$,
let $l$ be the order of $p$ mod $m$, let $q = p^l$, and let
$\FF_q$ be the finite field of $q$ elements.

\begin{enumerate}
\item
By a \emph{character} we will mean a homomorphism
$\chi: \FF_q^* \to \QQ(\zeta_{p(q-1)})^*$,
which we extend to $\FF_q$ by setting $\chi(0)=0$.
For any character $\chi$, define the Gauss sum
\[
G(\chi) = \sum_{a \in \FF_q^*} \chi(a) \zeta_p^{\Trace(a)}
\]
where the trace is from $\FF_q$ to $\FF_a$. For example, if $q=p$, we
could take $\chi$ to be the quadratic character $\chi(a) = \left( \frac{a}{p} 
\right)$ and get back the usual Gauss sum. Note that if $\chi$ has order
dividing $m$, then $G(\chi)$ takes values in $\QQ(\zeta_{pm})$.
\item
Work out some basic properties of the Gauss sums:
\begin{enumerate}
\item[(a)] $G(\overline{\chi}) = \chi(-1) \overline{G(\chi)}$;
\item[(b)] for $\chi \neq 1$,  $g(\chi) g(\overline{\chi}) = \chi(-1) q$;
\item[(c)] for $\chi \neq 1$, $g(\chi) \overline{g(\chi)} = q$;
\item[(d)] $G(\chi^p) = G(\chi)$.
\end{enumerate}
\item
Given characters $\chi_1$ and $\chi_2$, define the Jacobi sum
\[
J(\chi_1, \chi_2) = - \sum_{a \in \FF_q} \chi_1(a) \chi_2(1-a).
\]
Work out some basic properties of the Jacobi sums:
\begin{enumerate}
\item[(a)] $J(1,1) = 2-q$;
\item[(b)] $J(1,\chi) = J(\chi,1) = 1$ if $\chi \neq 1$;
\item[(c)] $J(\chi, \overline{\chi}) = \chi(-1)$ if $\chi \neq 1$;
\item[(d)] $G(\chi_1) G(\chi_2) = J(\chi_1, \chi_2) G(\chi_1\chi_2)$
if $\chi_1 \chi_2 \neq 1$.
\end{enumerate}
\item
Note that if $\chi_1$ and $\chi_2$ have order dividing $m$
that $G(\chi_1) G(\chi_2) / G(\chi_1 \chi_2)$
is an algebraic integer in $\QQ(\zeta_m)$ (not just in $\QQ(\zeta_{pm})$).
\item
Choose a prime $\gothp$ of $\QQ(\zeta_{pm})$ above $(p)$
and a prime $\gothq$ of $\QQ(\zeta_{p(q-1)})$ above $\gothp$,
choose an isomorphism of the residue field of $\gothq$ with $\FF_q$,
and let $\chi_\gothq$ be the character such that $\chi_\gothq(a) \equiv a^{-1}
\pmod{\gothq}$.
\item
For $1 \leq \nu < q-1$, let $s(\nu)$ be the sum of the base $p$ digits of
$\nu$. For future reference, observe that
\[
s(\nu) = (p-1) \sum_{i=0}^{l-1} \left\{ \frac{p^i \nu}{q-1} \right\},
\]
where the braces denote fractional part.
\item
Show that $v_{\gothq}(G(\chi_{\gothq}^{\nu})) = s(\nu)$, by induction
on $\nu$, by checking that $v_{\gothq}(\chi_\gothq) = 1$
and that $v_{\gothq}(J(\chi_\gothq^{\nu-1},
\chi_\gothq)) = 0$ if $\nu$ is not divisible by $p$.
This gives us an
explicit formula for the prime factorization of the ideal $(G(\chi))$ for
any character $\chi$.
\item
Let $G$ be the Galois group of $\QQ(\zeta_m)$ and let $\ZZ[G]$ and $\QQ[G]$
be the group algebras of $G$ over $\ZZ$ and $\QQ$, respectively. (That is,
these algebras are free with generators $[g]$ for each $g \in G$, multiplied
by the rule $[g_1][g_2] = [g_1g_2]$.) Define $\gotha^{\beta}$
for $\gotha$ a fractional ideal of $\QQ(\zeta_m)$ and $\beta \in \ZZ[G]$
as follows: if $\beta = \sum z_i [g_i]$, then
$\gotha^\beta = \prod g_i(\gotha)^{z_i}$.
Define the \emph{Stickelberger element}
of $\QQ[G]$ by
\[
\theta = \sum_{a \in (\ZZ/m\ZZ)^*} \left\{ \frac{a}{m} \right\}
\sigma_a^{-1},
\]
where $\sigma_a$ denotes the
automorphism $\zeta_m \mapsto \zeta_m^a$.
\item
Check that
\[
(G(\chi_{\gothq}^{(q-1)/m})) = \prod_{a \in (\ZZ/m\ZZ)^*} \sigma_a^{-1}(\gothp)^{s(a(q-1)/m)},
\]
where $a$ runs over $(\ZZ/m\ZZ)^*$.
Deduce that $(G(\chi_{\gothq}^{(q-1)/m})^m) = \gothp^{m\theta}$.
Then note that the generator of the ideal on the left
actually lies in $\QQ(\zeta_m)$.
(It generates an extension of $\QQ(\zeta_{m})$ which is only ramified
above primes dividing $m$, whereas $\QQ(\zeta_{pm})/\QQ(\zeta_m)$ is totally
ramified above primes dividing $p$.)
\item
Prove Stickelberger's theorem: if $\beta \in \ZZ[G]$ is such that
$\beta \theta \in \ZZ[G]$ as well, then $\gotha^{\beta \theta}$ is
principal for any fractional ideal $\gotha$ of $\QQ(\zeta_m)$.
It suffices to check this for $\gotha$ prime and coprime to $m$ (since
such ideals generate the full class group; see a prior homework).
In fact, if $\gotha = \gothp \cap \QQ(\zeta_m)$, we have now shown that
$\gotha^{\beta \theta}$ is generated by $G(\chi_{\gothq}^{(q-1)/m})^{\beta
\theta}$.
(The more general statement of the theorem is that the same thing happens
in any subfield of $\QQ(\zeta_m)$.)
\end{enumerate}

\end{document}