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Herbrand's theorem (Theorem)

Let $ \mathbb{Q}(\zeta_p)$ be a cyclotomic extension of $ \mathbb{Q}$, with $ p$ an odd prime, let $ A$ be the Sylow $ p$-subgroup of the ideal class group of $ \mathbb{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

$\displaystyle \hat{G}=\{\chi^i\mid0\leq i\leq p-2\}$    

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.

Theorem 1 (Herbrand)   Let $ i$ be odd with $ 3\leq i\leq p-2$. Then $ A_i\neq 0 \iff p\mid B_{p-i}$.

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.



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Cross-references: converse, Bernoulli number, module, orthogonal idempotents, multiplication, subgroup, orthogonal idempotent of the group ring, group, character, extension, Galois group, ideal class group, prime, odd, cyclotomic extension
There are 3 references to this entry.

This is version 2 of Herbrand's theorem, born on 2004-02-27, modified 2004-03-05.
Object id is 5647, canonical name is HerbrandsTheorem.
Accessed 2230 times total.

Classification:
AMS MSC11R29 (Number theory :: Algebraic number theory: global fields :: Class numbers, class groups, discriminants)

Pending Errata and Addenda
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The character group $\hat{G}$ by micah on 2007-06-15 15:02:12
Forgive me if this suggestion is too pedantic, but I think that it might be useful to specify some $\chi$ that generates the group $\hat{G}$. While I can certainly deduce that such a $\chi$ needs to be some non-trivial complex character, and can then deduce that any such non-trivial character will do, it might be useful to state something to this effect.
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