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.

Thank you for your help!

  • $\begingroup$ $e(P/p),f(P/p)$ are the ramification index and intertia degree, not the same as $f(\sigma) = \sigma_K$ $\endgroup$ – reuns Dec 14 '16 at 22:34
  • $\begingroup$ And since $L/\mathbb{Q}$ is Galois and $\mathbb{Q} \subset K \subset L$ then $L/K$ is a Galois extension $\endgroup$ – reuns Dec 14 '16 at 22:56
  • $\begingroup$ @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. $\endgroup$ – Watson Dec 15 '16 at 10:05
  • $\begingroup$ it is not the same $f$ $\endgroup$ – reuns Dec 15 '16 at 10:52
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ Thank you for your answer! So actually, we don't even need to use directly the surjectivity of the restriction morphism $\Gal(L/\Bbb Q) \to \Gal(K/\Bbb Q)$ ? $\endgroup$ – Watson Dec 15 '16 at 14:38
  • $\begingroup$ Notice that you can use \Gal here (I defined a renewcommand). $\endgroup$ – Watson Dec 15 '16 at 14:39
  • 1
    $\begingroup$ 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. $\endgroup$ – Starfall Dec 15 '16 at 15:25

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