Herbrand's theorem
HerbrandsTheorem
2004-02-27 17:25:44
2004-03-05 11:35:54
Theorem
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Let $\mb{Q}(\zeta_p)$ be a cyclotomic extension of $\Q$, with $p$ an odd prime, let $A$ be the Sylow $p$-subgroup of the ideal class group of $\mb{Q}(\zeta_p)$, and let $G$ be the Galois group of this extension. Note that the character group of $G$, denoted $\hat{G}$, is given by
\begin{align*}
\hat{G}=\{\chi^i\mid0\leq i\leq p-2\}
\end{align*}
For each $\chi\in\hat{G}$, let $\varepsilon_\chi$ denote the corresponding orthogonal idempotent of the group ring, and note that the $p$-Sylow subgroup of the ideal class group is a $\mathbb{Z}[G]$-module under the typical multiplication. Thus, using the orthogonal idempotents, we can decompose the module $A$ via $A=\sum_{i=0}^{p-2}A_{\omega^i}\equiv\sum_{i=0}^{p-2}A_i$.
Last, let $B_k$ denote the $k$th Bernoulli number.
\begin{Theo}[Herbrand]
Let $i$ be odd with $3\leq i\leq p-2$. Then $A_i\neq 0 \iff p\mid B_{p-i}$.
\end{Theo}
Only the first direction of this theorem ($\implies$) was proved by Herbrand himself. The converse is much more intricate, and was proved by Ken Ribet.