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\begin{center}
\bf
Math 254A, UC Berkeley, Fall 2001 (Kedlaya) \\
Lecture 25: Cyclotomic Fields (Part 1) \\
October 31, 2001
\end{center}

\head{Reference} Neukirch, Section 1.10. But we will supplant some of the
proofs with the
local arguments in Section 2.7, which we partially covered in class already.

\head{Outline of lecture}

Throughout, let $\zeta_n$ denote a primitive $n$-th root of unity.

\begin{enumerate}
\item
Prove that $\ZZ[\zeta_n]$ is the ring of integers of $\QQ(\zeta_n)$
by showing that $\ZZ_p[\zeta_n]$ is the valuation ring of $\QQ_p(\zeta_n)$
for each $p$, using Proposition~II.7.13 in the case $n = p^l$ is a prime
power. One can also give a global proof (which is pretty similar);
see Section 1.10.
\item
Calculate the discriminant of $\QQ(\zeta_n)$, up to sign, in case $n = p^l$,
by explicitly calculating the discriminant of the $n$-th cyclotomic
polynomial. 
See the homework for the general case.
\item
Prove
the explicit formula for the splitting of a rational prime $p$
in $\QQ(\zeta_n)$: if $v_p(n) = m$ and $f$ is the smallest positive integer
such that $p^f \equiv 1 \pmod{n/p^m}$, then $p$ splits into prime ideals
with ramification index $\phi(p^m)$ and inertia degree $f$.
\item
For $n = p$ an odd prime, Galois theory tells us that $\QQ(\zeta_p)$
contains a unique quadratic subfield. Identify this subfield explicitly:
the Gauss sum $\tau = \sum_{a \in \FF_p^*} \left( \frac{a}{p} \right) 
\zeta_p^a$ satisfies $\tau^2 = (-1)^{(p-1)/2}p$.
\item
Now suppose $q$ is another odd prime. Show that
$q$ splits in $\QQ(\sqrt{(-1)^{(p-1)/2}p})$ if and only if
$q$ splits into an even number of prime ideals in $\QQ(\zeta_p)$.
Reobtain from this the law of quadratic reciprocity.
\end{enumerate}

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