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\begin{document}

\begin{center}
\bf
Math 254B, UC Berkeley, Spring 2002 (Kedlaya) \\
The Hilbert Class Field
\end{center}

\head{Reference} Milne, Introduction; Neukirch, VI.6.

Recall that the field $\QQ$ has no extensions which are everywhere
unramified (Minkowski's theorem from last semester, which we used in proving
the Kronecker-Weber theorem). This is quite definitely not true of other
number fields; we begin with an example illustrating this.

In the number field $K = \QQ(\sqrt{-5})$, the ring of integers is
$\ZZ[\sqrt{-5}]$ and the ideal $(2)$ factors as $\gothp^2$,
where the ideal $\gothp = (2, 1 + \sqrt{-5})$ is not principal.

Now let's see what happens when we adjoin a square root of $-1$,
obtaining $L = \QQ(\sqrt{-5}, \sqrt{-1})$. The
extension $\QQ(\sqrt{-1})/\QQ$ only ramifies over 2, so
$\QQ(\sqrt{-5}, \sqrt{-1})/K$ can only be ramified over $\gothp$.
Moreover, in $K_{\gothp}$, $-1 \equiv (2 + \sqrt{-5})^2 \pmod{8}$,
so $-1$ is a square in $K_{\gothp}$. Thus $L/K$ is also unramified
over $\gothp$!

We've now seen that $\QQ(\sqrt{-5})$ admits both a nonprincipal ideal and
an unramified abelian extension. It turns out these are not unrelated events.
Caution: until further notice, the phrase ``$L/K$ is unramified'' will mean
that $L/K$ is unramified over all finite places in the usual sense, \emph{and}
that every real embedding of $K$ extends to a real embedding of $L$. 
(Get used to this. The real and complex embeddings of a number field will
be treated like primes throughout the course.)
\begin{theorem}
Let $L$ be the maximal unramified
abelian extension of a number field $K$.
Then $L/K$ is finite, and its Galois group is isomorphic to
the ideal class group of $K$.
\end{theorem}
In fact, there is a canonical isomorphism, given by the Artin reciprocity
law. We'll see this a bit later. The field $L$ is called the \emph{Hilbert
class field} of $K$.

Warning: there can be infinite unramified \emph{nonabelian} extensions.
In fact, Golod and Shafarevich used unramified abelian extensions to 
construct these! Namely, starting from a number field $K = K_0$,
let $K_1$ be the Hilbert class field of $K_0$, let $K_2$ be the Hilbert
class field of $K_1$, and so on. Then $K_i$ is a unramified but not
necessarily abelian extension of $K_0$, and for a suitable choice of 
$K_0$, $[K_i:K_0]$ can be unbounded.

\head{Exercises}

\begin{enumerate}
\item
Let $K$ be an imaginary
quadratic extension of $\QQ$ in which $t$ finite primes ramify.
Prove that $\#(\Cl(K)/2\Cl(K)) = 2^{t-1}$, assuming Theorem~1.
(Hint: if an odd prime $p$
ramifies in $K$,
show that $K(\sqrt{p})/K$ is unramified; if 2 ramifies in $K$, show that
$K(\sqrt{-1})/K$ is unramified.)
This was first proved by Gauss using
binary quadratic forms, as alluded to last semester.
\item
Give an example, using a real quadratic field, to illustrate that
\begin{enumerate}
\item[(a)] Theorem~1 fails if we don't require the extensions to be
unramified above the real place;
\item[(b)] The previous exercise fails for real quadratic fields.
\end{enumerate}
(Hint: see the hint for the previous exercise.)
Optional: prove that the previous exercise works if
you replace the class group by the \emph{narrow class group}, in which you
only mod out by principal ideals having a totally positive generator.
This gives an example of a \emph{ray class group}; more on those next time.
\item
The field $\QQ(\sqrt{-23})$ admits an ideal of order 3 in the class group and
an unramified
abelian extension of degree 3. Find both; feel free to use a computer if
that helps. (Hint: the extension contains a cubic field of discriminant -23.)
\end{enumerate}

\end{document}