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\begin{center}
\bf
Math 254A, UC Berkeley, Fall 2001 (Kedlaya) \\
Problem Set 10: Cyclotomic Fields \\
Due in class \emph{Wednesday}, November 14
\end{center}

Please submit nine of the following problems.

\head{Leftover Problems about Ramification/Splitting}

\begin{enumerate}
\item
Prove that $\QQ(\sqrt{-5})$ has exactly one quadratic extension which
is everywhere unramified, and find it. (This has something to do with
class numbers; we'll get back to this next term.)
\item
Let $f$ be a monic polynomial over $\ZZ$ of degree greater than 1. Suppose
that $f$ has a root modulo $p$ for all primes $p$. Prove that $f$
cannot be irreducible. (Hint: if $G$ is a finite group and $H$ is a proper
subgroup, the union of the conjugates of $H$ cannot be all of $G$.)
\end{enumerate}

\head{Cyclotomic Fields}

\begin{enumerate}
\item
Prove that the sign of the discriminant of the number field $K$ with
signature $(r,s)$ equals $(-1)^s$. (You'll need this for the next problem.)
\item
Calculate the discriminant of $\QQ(\zeta_n)$
for $n$ arbitrary. (Hint: choose an integral basis by multiplying
together integral
bases of $\QQ(\zeta_{p^m})$ for each prime power $p^m$ dividing $n$.
Then reduce to the computation we did in class.)
\item
Neukirch exercise I.10.1: Prove that for every natural number $n$, there
are infinitely many primes congruent to 1 modulo $n$, without invoking
Dirichlet's theorem on primes in arithmetic progressions. (Hint: let
$\phi_n$ be the minimal polynomial of $\zeta_n$. Prove that every prime
divisor of $\phi_n(m)$, for $m \in \ZZ$, is congruent to 1 modulo $n$.)
\item
Neukirch exercise I.10.2: Prove that for every finite abelian group $A$,
there exists a Galois extension $L/\QQ$ with Galois group $A$.
\end{enumerate}

\head{Interlude on Fermat's Last Theorem}

Throughout this section, let $p \geq 5$ be a regular prime, that is,
a prime that does not divide the class number of $\QQ(\zeta_p)$.
Our goal is to prove part of Fermat's Last Theorem for the exponent
$p$. Namely, if $x,y,z$ are rational integers, none divisible by $p$,
and $p$ is regular as assumed above, we will prove that $x^p + y^p + z^p 
\neq 0$. As usual, we may assume $x,y,z$ are pairwise coprime (if 
$x^p+y^p+z^p=0$ and two of $x,y,z$ have a common prime factor,
so does the third).

\begin{enumerate}
\item
Let $x$ and $y$ be relatively prime rational integers such that
$x, y$ and $x^p+y^p$ are not divisible by $p$. Prove that
the ideals $(x + \zeta^i y)$ are pairwise coprime for $i=0, \dots, p-1$.
\item
Suppose $x^p + y^p + z^p = 0$ and that $p$ is regular.
Prove that for $i=0, \dots, p-1$, there
exists a unit $\epsilon_i$ and an integer $\alpha_i \in \ZZ[\zeta_p]$ such that
$x+ \zeta^i y = \epsilon_i \alpha_i^p$. (Hint:
the assumption that $p$ is regular means
that the $p$-th power map is a bijection
on the class group.)
\item
Let $\lambda = 1 - \zeta$. Show that $\Norm(\lambda) = p$, so that
every element of $\ZZ[\zeta_p]$ is congruent to a rational integer
modulo $\lambda$. Then show that if $a \equiv b \pmod{\lambda}$
for $a,b \in \ZZ[\zeta_p]$, then $a^p \equiv b^p \pmod{p}$. (Hint:
use the fact that as ideals, $(\lambda)^{p-1} = (p)$.)
\item
Suppose $x+\zeta y = \epsilon \alpha^p$ for some unit $\epsilon$ and
some integer $\alpha$ in $\ZZ[\zeta_p]$. Prove that there exists a
rational integer $g$ such that $x\zeta^g + y \zeta^{g+1} \equiv x \zeta^{-g}
+ y \zeta^{-g-1} \pmod{p}$. (Hint: write $\epsilon$ as a power of $\zeta$
times an element of $\ZZ[\zeta_p + \zeta_p^{-1}]$, i.e., a real number; we
saw this on a previous homework.
Also note that $\alpha^p$ is congruent to a
rational integer modulo $p$.)
\item
Assume the conditions of the previous problem, and also assume
$x$ and $y$ are not divisible by $p$. Prove that if
$x \zeta^g + y \zeta^{g+1} \equiv x \zeta^{-g} + y \zeta^{-g-1}
\pmod{p}$ for some rational integer $g$, then $2g \equiv -1 \pmod{p}$.
(Hint: first rule out the possibilities $g \equiv 0 \pmod{p}$ and
$g \equiv -1 \pmod{p}$. Then
use the fact that $\zeta, \zeta^2, \dots, \zeta^{p-1}$ form an
integral basis to conclude that $g, g+1, -g, -g-1$ cannot be
distinct modulo $p$.)
\item
Suppose that $p \geq 5$ is regular and that $x^p + y^p + z^p = 0$ for 
pairwise coprime rational integers $x,y,z$, none divisible by $p$.
Conclude from the above that $x \equiv y, y \equiv z, z \equiv x \pmod{p}$
and thus obtain a contradiction.
\end{enumerate}




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