# Higher ramification groups of Galois extension of order $p^2$

Let $p\in \mathbb{Z}$ be a prime number and $K/\mathbb{Q}$ be a Galois extension of degree $p^2$ over $\mathbb{Q}$. Suppose that $P\subset \mathcal{O}_K$ is the only prime ramified over $p$. Let $G:=\text{Gal}(K/\mathbb{Q})$ and recall the definition of the higher ramification groups, for $m\ge 0$ $$V_m(P\mid p):=\{\sigma\in G: \sigma(x)\equiv x\pmod {P^{m+1}} \;\forall x\in \mathcal{O}_K\}$$ (in particular $V_0$ is the inertia group).

It's easy to show that $|V_0|=p^2$ (because $P$ is totally ramified over $p$) and that $|V_1|=p^2$ (since $V_1$ is the $p$-Sylow of $V_0$). Now I have to show that the first ramification group having order $<p^2$ has order $p$. I know that $V_{m}/V_{m+1}$ is sum of cyclic groups of order $p$, but this is not sufficient to conclude. Any ideas?

• It’s because each $V_n/V_{n+1}$ is injected into the additive group of the residue field. In your situation, the residue field is $\Bbb F_p$. – Lubin Dec 31 '15 at 21:38
• @Lubin you should consider posting this as an answer – Stella Biderman Jan 2 '16 at 19:48

Remember that $V_m/V_{m+1}$ is not merely a sum of cyclic groups of order $p$, but is injected into $\kappa^+$, where $\kappa$ is the common residue field of the rings of integers. Here $\kappa=\Bbb F_p$, so that $V_m/V_{m+1}$ is either trivial or cyclic of order $p$.