# Easy cases of Herbrand's theorem

$$\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$$ I recall Herbrand's theorem about class groups of cyclotomic fields: Let $$p$$ be an odd prime, let $$\zeta$$ be a primitive $$p$$-th root of $$1$$ and let $$K = \QQ(\zeta)$$, so $$\mathrm{Gal}(K/\QQ)$$ is canonically isomorphic to $$(\ZZ/p \ZZ)^{\times}$$. Let $$A$$ be the class group of $$K$$ and let $$V = A/p A$$, an $$\mathbb{F}_p$$ vector space. Then $$\mathrm{Gal}(K/\QQ)$$ acts on $$V$$ and $$V$$ splits accordingly into characters of $$(\ZZ/p \ZZ)^{\times}$$; let $$V = \bigoplus V_r$$ where $$a \in (\ZZ/p \ZZ)^{\times}$$ acts by $$a^r$$ on $$V_r$$.

For $$1 \leq r \leq p-2$$ odd, Herbrand's theorem tells us that, if $$V_r \neq 0$$, then $$p$$ divides the numerator of the Bernoulli number $$B_{p-r}$$. For example, since $$B_2 = \frac{1}{6}$$, we always have $$V_{p-2}=0$$. First question:

Is there a way to see that $$V_{p-2}=0$$ without understanding Herbrand's proof?

As an example of what I'm hoping for, it is straightforward to see that $$V_{p-1}=V_0=0$$. If $$V_{p-1}$$ were nonzero, class field theory would give an unramified extension $$K/\mathbb{Q}(\zeta_p)$$ so that $$K/\QQ$$ is Galois with Galois group $$(\ZZ/p) \times (\ZZ/(p-1))$$. But then the fixed field of $$\ZZ/(p-1)$$ is an unramified degree $$p$$ extension of $$\QQ$$, violating Minkowski's theorem.

I ask because I am still thinking about this very challenging question. If $$V_{p-2}=V_{-1}$$ were nonzero, I believe I could show that the ring of integers in the corresponding $$(\ZZ/p) \rtimes (\ZZ/(p-1))$$ extension of $$\QQ$$ would give a counter-example to this question. With similar motivation, I ask:

Is there a straightforward way to see that the eigenspace $$V_1$$ is zero?

This last occurs as Proposition 6.16 in Washington's Introduction to Cyclotomic Fields, but I can't figure out whether it is straightforward or whether it needs the 6 chapters that precede it.

• It should be possible to calculate one of $V_1$ or $V_{-1}$ (and therefore show it vanishes I guess) by Kummer theory. – Will Sawin Nov 16 '18 at 1:59
• Short answer: No, there is no "easy" way to prove that $V_{p-2} = 0$. And yes, there is an easy way to prove $V_1 = 0$ --- as Sawin says, Kummer theory says any such Galois extension (without any ramification conditions) is of the form $\mathbf{Q}(\zeta_p, N^{1/p})$ where $N \in \mathbf{Q}$, at which point it is easy to see (using unique factorization) that all such extensions are ramified. – user131093 Nov 18 '18 at 3:28