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Herbrand-Ribet theorem

The Herbrand-Ribet theorem is a strengthening of Kummer's theorem to the effect that the prime p divides the class number of the cyclotomic field of p-th roots of unity if and only if p divides the denominator of the nth Bernoulli number Bn for some n, 0 < n < p − 1. The Herbrand-Ribet theorem specifies what, in particular, it means when p divides such an Bn.

The Galois group Σ of the cyclotomic field of pth roots of unity for an odd prime p, \Bbb{Q}(\zeta) with ζp = 1, consists of the p − 1 group elements σa, where σa is defined by the fact that σa(ζ) = ζa. As a consequence of the little Fermat theorem, in the ring of p-adic integers \Bbb{Z}_p we have p − 1 roots of unity, each of which is congruent mod p to some number in the range 1 to p − 1; we can therefore define a Dirichlet character ω (the Teichmüller character) with values in \Bbb{Z}_p by requiring that for n relatively prime to p, ω(n) be congruent to n modulo p. The p part of the class group is a \Bbb{Z}_p-module, and we can apply elements in the group ring \Bbb{Z}_p[\Sigma] to it and obtain elements of the class group. We now may define an idempotent element of the group ring for each n from 1 to p − 1, as

\epsilon_n = \frac{1}{p-1}\sum_{a=1}^{p-1} \omega(a) \sigma_a^{-1}.

We now can break up the p part of the ideal class group G of \Bbb{Q}(\zeta) by means of the idempotents; if G is the ideal class group, then Gn = εn(G).

Then we have the theorem of Herbrand-Ribet: Gn contains no elements if and only if p divides the Bernoulli number Bpn. The part saying p divides Bpn if Gn is not trivial is due to Herbrand. The converse, that if p divides Bpn then Gn is not trivial is due to Ribet, and is considerably more difficult. By class field theory, this can only be true if there is an unramified extension of the field of pth roots of unity by a cyclic extension of degree p which behaves in the specified way under the action of Σ; Ribet proves this by actually constructing such an extension.

01-04-2007 01:18:14
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