Lemma 15.110.11. Let $A$ be a normal domain with fraction field $K$. Let $M/L/K$ be a tower of (possibly infinite) Galois extensions of $K$. Let $H = \text{Gal}(M/K)$ and $G = \text{Gal}(L/K)$ and let $C$ and $B$ be the integral closure of $A$ in $M$ and $L$. Let $\mathfrak r \subset C$ and $\mathfrak q = B \cap \mathfrak r$. Set $D_\mathfrak r = \{ \tau \in H \mid \tau (\mathfrak r) = \mathfrak r\} $ and $I_\mathfrak r = \{ \tau \in D_\mathfrak r \mid \tau \bmod \mathfrak r = \text{id}_{\kappa (\mathfrak r)}\} $ and similarly for $D_\mathfrak q$ and $I_\mathfrak q$. Under the map $H \to G$ the induced maps $D_\mathfrak r \to D_\mathfrak q$ and $I_\mathfrak r \to I_\mathfrak q$ are surjective.
Proof. Let $\sigma \in D_\mathfrak q$. Pick $\tau \in H$ mapping to $\sigma $. This is possible by Fields, Lemma 9.22.2. Then $\tau (\mathfrak r)$ and $\mathfrak r$ both lie over $\mathfrak q$. Hence by Lemma 15.110.10 there exists a $\sigma ' \in \text{Gal}(M/L)$ with $\sigma '(\tau (\mathfrak r)) = \mathfrak r$. Hence $\sigma '\tau \in D_\mathfrak r$ maps to $\sigma $. The case of inertia groups is proved in exactly the same way using surjectivity onto automorphism groups. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #2536 by Mathias on
Comment #2570 by Johan on
There are also: