Surjective homomorphisms on decomposition and inertia groups

Assume that $L/\mathbb{Q}$ and $K/\mathbb{Q}$ are Galois extensions with $K \subset L$. Let $p$ be a prime of $\mathbb{Q}$, $\mathfrak p$ be a prime of $K$ lying over $p$, and $P$ be a prime of $L$ lying over $\mathfrak p$. $\renewcommand{\Gal}{\mathrm{Gal}}$

Let $$I(P/p) \trianglelefteq D(P/p) ≤ \Gal(L/\Bbb Q)$$ be respectively the decomposition and inertia groups of the prime $P$ over $p$. Similarly for $I(\mathfrak p/p) \trianglelefteq D(\mathfrak p/p) ≤ \Gal(K/\Bbb Q)$.

My questions are:

1. Why is the morphism $$f : D(P/p) \to D(\mathfrak p/p) \qquad f(\sigma) = \sigma\vert_K$$ surjective?
2. Why is the morphism $$g : I(P/p) \to I(\mathfrak p/p) \qquad g(\sigma) = \sigma\vert_K$$ surjective?

I know that the restriction map $r:\Gal(L/\mathbb{Q}) \to \Gal(K/\mathbb{Q})$ is surjective, and that $f,g$ are well-defined. I also know that $$r' : \dfrac{D(P/p)}{I(P/p)} \to \dfrac{D(\mathfrak p/p)}{I(\mathfrak p/p)}$$ is surjective.

The order of $D(P/p)$ is $$e(P/p)f(P/p) = e(P/\mathfrak p) e(\mathfrak p/p) f(P/\mathfrak p) f(\mathfrak p/p) \geq e(\mathfrak p/p)f(\mathfrak p/p) = |D(\mathfrak p/p)|$$ so this is a necessary condition for surjectivity.

In other words (for 1.), I would like to show that if $\sigma(\mathfrak p) = \mathfrak p$ then $\sigma(P)=P$, where $\mathfrak p = P \cap K$ and $\sigma \in \Gal(L/\Bbb Q)$.

This is a related question.

• $e(P/p),f(P/p)$ are the ramification index and intertia degree, not the same as $f(\sigma) = \sigma_K$ – reuns Dec 14 '16 at 22:34
• And since $L/\mathbb{Q}$ is Galois and $\mathbb{Q} \subset K \subset L$ then $L/K$ is a Galois extension – reuns Dec 14 '16 at 22:56
• @user1952009 : what do you mean by "not the same as $f(σ)=σ_K$"? And yes, $L/K$ is Galois, so that $D(P/\mathfrak p)$ makes sense, for instance. – Watson Dec 15 '16 at 10:05
• it is not the same $f$ – reuns Dec 15 '16 at 10:52
• Ah ok… but I think the context is clear. Even if my homomorphism is called $f$, the notation $f(P/p)$ is only for inertia index. – Watson Dec 15 '16 at 10:56

It is not true that $\sigma(\mathfrak p) = \mathfrak p$ implies $\sigma(P) = P$, for obvious reasons. (Try this on a concrete example.)
Note that the kernel of the map $D(P/p) \to D(\mathfrak p/p)$ given by restriction to $K$ is exactly $\textrm{Gal}(L/K) \cap D(P/p) = D(P/\mathfrak p)$. Thus, there is an embedding $D(P/p)/D(P/\mathfrak p) \to D(\mathfrak p/p)$. However, by multiplicativity of ramification index and inertia degree, these groups have equal order, thus the injection must actually be an isomorphism, i.e the initial map must be surjective. I leave (2) as an exercise, the proof idea is similar.
• Thank you for your answer! So actually, we don't even need to use directly the surjectivity of the restriction morphism $\Gal(L/\Bbb Q) \to \Gal(K/\Bbb Q)$ ? – Watson Dec 15 '16 at 14:38
• Notice that you can use \Gal here (I defined a renewcommand). – Watson Dec 15 '16 at 14:39
• The surjectivity of the restriction morphism can be derived in the same way in the finite case if one does not want to use isomorphism extension - indeed, restriction gives an embedding $\Gal(L/\mathbf Q) / \Gal(L/K) \to \textrm{Gal}(K/\mathbf Q)$, which must be surjective by order considerations. Thus, while we are not using the result directly, we are still using the "spirit" of it, in a sense. – Starfall Dec 15 '16 at 15:25