Lemma 15.98.8. 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 15.98.7 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. $\square$

## Comments (2)

Comment #2536 by Mathias on

I see a few typos in the proof of this lemma. First and foremost we should pick an element $\tau\in H$ that maps to $\sigma$ ($\sigma\in G$ already)

Secondly, it should be $\tau(\mathfrak{r})$ that lies above $\mathfrak{q}$ and not $\sigma(\mathfrak{r})$ ($\sigma$ is an element of $\operatorname{Gal}(L/K)$ not $\operatorname{Gal}(M/K)$).

Thirdly, it should be $\sigma'(\tau(\mathfrak{r}))=\mathfrak{r}$ and not as the current version in the lemma.

Finally, the line "Hence $\sigma'\sigma \in D_\mathfrak r$ maps to $\tau$." should be "Hence $\sigma'\tau\in D_{\mathfrak{r}}$ maps to $\sigma$."

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