I don't know if the next thing is true, but I'm not able to find a counterexample: suppose you have a surjective group homomorphism of finite groups $f:G \rightarrow G'$ and normal subgroups $H \lhd G, H' \lhd G'$, such that the induced homomorphism $f':G/H \rightarrow G'/H'$ is also surjective. Is it true that $f\mid_H: H \rightarrow H'$ is also surjective ? In particular, is this true for decomposition groups and inertia groups in the case of Galois extensions $L/K/\mathbb{Q}_p$ ?

  • $\begingroup$ I'm not entirely clear on what your second question is. You want $\mathbf{Q}_p\subseteq K\subseteq L$ to be a tower of Galois extensions, and then you're asking if the inertia group of $L/\mathbf{Q}_p$ surjects onto the inertia group of $K/\mathbf{Q}_p$? Also, since you're working over $\mathbf{Q}_p$, "decomposition group" just means "Galois group." $\endgroup$ – Keenan Kidwell Dec 9 '11 at 15:12

This is not true; in fact, for any surjective group homomorphism $f:G\to G'$, we can choose $H=\{e\}\triangleleft G$ and $H'=G'\triangleleft G'$, so that $G/H\cong G$ and $G'/H'$ is trivial, hence the induced homomorphism $f':G/H\to G'/H'$ is certainly surjective, but $f|_H:H\to H'$ is not surjective unless the group $G'$ is itself trivial.

I think that if

  • $L/\mathbb{Q}$ and $K/\mathbb{Q}$ are Galois extensions with $L\supseteq K$
  • $p$ is a prime of $\mathbb{Q}$, $\mathcal{P}$ is a prime of $L$ lying over $p$, and $\mathfrak{p}=\mathcal{P}\cap K$ is the prime of $K$ beneath $\mathcal{P}$
  • $\begin{align*}G&=D(\mathcal{P}/p)=\{\sigma\in\operatorname{Gal}(L/\mathbb{Q})\mid \sigma(\mathcal{P})=\mathcal{P}\}\\ H&= I(\mathcal{P}/p)=\{\sigma\in D(\mathcal{P}/p)\mid \overline{\sigma}:\mathcal{O}_L/\mathcal{P}\to\mathcal{O}_L/\mathcal{P}\text{ is the identity} \}\end{align*}$

  • $\begin{align*}G'&=D(\mathfrak{p}/p)=\{\tau\in\operatorname{Gal}(K/\mathbb{Q})\mid \tau(\mathfrak{p})=\mathfrak{p}\}\\ H'&= I(\mathfrak{p}/p)=\{\tau\in D(\mathfrak{p}/p)\mid \overline{\tau}:\mathcal{O}_K/\mathfrak{p}\to\mathcal{O}_K/\mathfrak{p}\text{ is the identity} \}\end{align*}$

  • $f:G\to G'$ is the restriction of the restriction map $r:\operatorname{Gal}(L/\mathbb{Q})\to\operatorname{Gal}(K/\mathbb{Q})$, which is well-defined because if $\tau(\mathfrak{p})=\mathfrak{q}\neq\mathfrak{p}$ then, for any lift of $\tau$ to a $\sigma\in \operatorname{Gal}(L/\mathbb{Q})$, we would have that $\sigma(\mathcal{P})=\mathcal{Q}$ for some prime $\mathcal{Q}$ lying over $\mathfrak{q}$, so $\sigma(\mathcal{P})\neq\mathcal{P}$, and hence $f$ is surjective because $r$ is

then in fact $f':G/H\to G'/H'$ is always surjective, as it is equivalent to the restriction map of Galois groups $\operatorname{Gal}(\mathcal{O}_L/\mathcal{P}\;\;/\;\;\mathbb{Z}/(p) )\to\operatorname{Gal}(\mathcal{O}_K/\mathfrak{p}\;\;/\;\;\mathbb{Z}/(p))$.

So I think your question just becomes, is $f|_H: H\to H'$ always surjective? I think the answer is again yes, intuitively because of the multiplicativity of ramification indices, but I am having trouble proving it. Hopefully someone will be able to sort this out.

  • $\begingroup$ Alright, maybe a stupid question in the general case. But what about decomposition and inertia groups ? $\endgroup$ – KevinDL Dec 9 '11 at 9:26
  • $\begingroup$ @KevinDL: I've posted my thoughts about that, but unfortunately I don't have an answer. My number theory is a bit rusty, and it is 5AM where I am, so I don't know if what I wrote even makes sense. I am sure that someone will come along soon who will be able to answer your question. $\endgroup$ – Zev Chonoles Dec 9 '11 at 10:18
  • $\begingroup$ Thanks, but my question was just about p-adic extensions. Anyway, I think I proved it using Galois correspondence and looking at fixed fields. Thanks for your help. $\endgroup$ – KevinDL Dec 9 '11 at 10:54
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    $\begingroup$ Haha, I didn't even see the little ${}_p$ down there :) Oh well. By the way, you don't have to accept my answer; you can write up your solution and accept it, so that the whole web can get the benefit of the answer. This is explicitly encouraged by the SE network of sites; see here and here. $\endgroup$ – Zev Chonoles Dec 9 '11 at 11:00

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