existence of Hilbert class fieldExistenceOfHilbertClassField2002-04-23 22:02:022006-06-15 21:49:07TheoremHilbert class field% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\rai}[1]{\mathcal{O}_{#1}}Let $K$ be a number field. There exists a finite extension $E$ of $K$ with the following properties:
\begin{enumerate}
\item $[E:K]=h_K$, where $h_K$ is the class number of $K$.
\item $E$ is Galois over $K$.
\item The ideal class group of $K$ is isomorphic to the Galois group of
$E$ over $K$.
\item Every ideal of $\rai{K}$ is a principal ideal of the ring extension $\rai{E}$.
\item Every prime ideal ${\cal P}$ of $\rai{K}$ decomposes into the product of
$\frac{h_K}{f}$ prime ideals in $\rai{E}$, where $f$ is the \PMlinkname{order}{Order}
of $[{\cal P}]$ in the ideal class group of $\rai{E}$.
\end{enumerate}
There is a unique field $E$ satisfying the above five properties, and it is known as the {\em Hilbert class field} of $K$.
The field $E$ may also be characterized as the \PMlinkname{maximal abelian unramified}{AbelianExtension} extension of $K$. Note that in this context, the term `unramified' is meant not only for the finite places (the classical ideal theoretic \PMlinkescapetext{ interpretation}) but also for the infinite places. That is, every real embedding of $K$ extends to a real embedding of $E$. As an example of why this is necessary, consider some real quadratic field.