$\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$ 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.
