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\begin{center}
\bf
Math 254A, UC Berkeley, Fall 2001 (Kedlaya) \\
Problem Set 9: Ramification in Galois Extensions \\
Due in class Monday, November 5
\end{center}

Please do six of the following problems, including the first problem.

\head{Leftover Problem about Ramification}

\begin{enumerate}
\item
Let $K$ be a finite extension of $\QQ_p$ and
$L/K$ be a tamely ramified (but not necessarily totally ramified) extension.
Compute the discriminant $\Disc(L/K)$ in terms of $e(L/K)$ and $f(L/K)$.
(The point: the discriminant only gets really nasty in the wildly 
ramified case.)
\end{enumerate}

\head{Ramification in Galois Extensions}

\begin{enumerate}
\item
Neukirch exercise I.9.1: if $L/K$ is a Galois extension of number fields
with noncyclic Galois group, then there are at most finitely many totally
nonsplit prime ideals of $K$ (i.e., primes of $\gotho_K$ with only one prime
of $\gotho_L$ over them). Also, use this argument to explicitly construct
an irreducible polynomial over $\ZZ$ which is not irreducible modulo any prime.
\item
Neukirch exercise I.9.2 (corrected):
if $L/K$ is a Galois extension of number fields
and $\gothb$ is a prime ideal which is unramified over $L$ (i.e.,
for $\gothp = \gothb \cap K$, $e(\gothb/\gothp) = 1$), there exists
a unique automorphism $\phi_{\gothb}$ of $L$ over $K$ such that
$\phi_{\gothb}(a) \equiv a^q \pmod{\gothb}$,
where $q = \Norm(\gothp)$. It is called the \emph{Frobenius automorphism}
of $\gothb$; observe also that $\phi_{\gothb}$ generates the decomposition
group of $\gothb$ in this case.
\item
Neukirch I.9.3:
let $L/K$ be an extension of prime degree $p$ with solvable normal closure.
If the unramified prime ideal $\gothp$ in $L$ has two prime factors
$\gothb$ and $\gothb'$ of inertia degree 1, then it is totally split.
(Hint from Neukirch: use a theorem of Galois, that if $G$ is a transitive
solvable permutation group of prime degree $p$, then there is no nontrivial
permutation of $G$ that fixes two distinct letters.)
\item
Neukirch I.9.4: Let $L/K$ be a finite, but not necessarily Galois, extension
of number fields and $N/K$ the normal closure of $L/K$. Show that
a prime ideal $\gothp$ of $K$ is totally split in $L$ if and only if it is
totally split in $N$. (Hint from Neukirch: use the double coset decomposition
$H \backslash G/G_{\gothb}$, where $G=G(N/K)$, $H = G(N/L)$, and $G_{\gothb}$ is the
decomposition groups of a prime ideal $\gothb$ over $\gothp$.)
\item
Prove that the previous exercise remains true if ``totally split'' is
replaced by ``unramified'' or by ``tamely ramified''. Optional:
describe more generally how to compute the
ramification and inertia degrees of the primes of $L$ above $\gothp$ 
in terms of the double coset decomposition.
%\item
%Let $P(x)$ be an irreducible polynomial of degree $n$ over $\ZZ$ with
%exactly two complex roots. Prove that the Galois group of the normal
%closure of $\QQ[x]/(P(x))$ is the full symmetric group $S_n$.
\item
Let $P(x)$ be a polynomial of degree $n$
over $\gotho_K$ for some number field $K$ such that modulo some prime $\gothp$,
$P(x)$ is irreducible, while modulo some other prime $\gothq$, $P(x)$
factors as an irreducible quadratic times pairwise coprime linear factors.
Prove that the Galois group of
the normal closure of $K[x]/(P(x))$ is the full symmetric group $S_n$.
(Hint: use Frobenius automorphisms.)
\item
Formulate a sensible definition of ``the decomposition/inertia group of an
archimedean embedding of a number field''. As a test, your definition should
allow the proof of the preceding problem to pass through unchanged to
prove the following: if $P(x)$ is a polynomial over $\ZZ$
which is irreducible modulo some prime $p$ and which has exactly two
nonreal roots, then the Galois group of the normal closure of
$\QQ[x]/(P(x))$ is the full symmetric group $S_n$.
\end{enumerate}



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