**Proof.**
Proof of (1). Consider pairs $(M, \sigma )$ where $K \subset M \subset L$ is a subfield such that $M/K$ is Galois, $\sigma \in \text{Gal}(M/K)$ with $\sigma (\mathfrak q \cap M) = \mathfrak q' \cap M$. We say $(M', \sigma ') \geq (M, \sigma )$ if and only if $M \subset M'$ and $\sigma '|_ M = \sigma $. Observe that $(K, \text{id}_ K)$ is such a pair as $A = K \cap B$ since $A$ is a normal domain. The collection of these pairs satisfies the hypotheses of Zorn's lemma, hence there exists a maximal pair $(M, \sigma )$. If $M \not= L$, then we can find $M \subset M' \subset L$ with $M'/M$ nontrivial and finite and $M'/K$ Galois (Fields, Lemma 9.16.5). Choose $\sigma ' \in \text{Gal}(M'/K)$ whose restriction to $M$ is $\sigma $ (Fields, Lemma 9.22.2). Then the primes $\sigma '(\mathfrak q \cap M')$ and $\mathfrak q' \cap M'$ restrict to the same prime of $B \cap M$. Since $B \cap M = (B \cap M')^{\text{Gal}(M'/M)}$ we can use Lemma 15.110.5 to find $\tau \in \text{Gal}(M'/M)$ with $\tau (\sigma '(\mathfrak q \cap M')) = \mathfrak q' \cap M'$. Hence $(M', \tau \circ \sigma ') > (M, \sigma )$ contradicting the maximality of $(M, \sigma )$.

Part (2) is proved in exactly the same manner as part (1). We write out the details. Pick $\overline{\sigma } \in \text{Aut}(\kappa (\mathfrak q)/\kappa (\mathfrak p))$. Consider pairs $(M, \sigma )$ where $K \subset M \subset L$ is a subfield such that $M/K$ is Galois, $\sigma \in \text{Gal}(M/K)$ with $\sigma (\mathfrak q \cap M) = \mathfrak q \cap M$ and

\[ \xymatrix{ \kappa (\mathfrak q \cap M) \ar[r] \ar[d]_\sigma & \kappa (\mathfrak q) \ar[d]_{\overline{\sigma }} \\ \kappa (\mathfrak q \cap M) \ar[r] & \kappa (\mathfrak q) } \]

commutes. We say $(M', \sigma ') \geq (M, \sigma )$ if and only if $M \subset M'$ and $\sigma '|_ M = \sigma $. As above $(K, \text{id}_ K)$ is such a pair. The collection of these pairs satisfies the hypotheses of Zorn's lemma, hence there exists a maximal pair $(M, \sigma )$. If $M \not= L$, then we can find $M \subset M' \subset L$ with $M'/M$ finite and $M'/K$ Galois (Fields, Lemma 9.16.5). Choose $\sigma ' \in \text{Gal}(M'/K)$ whose restriction to $M$ is $\sigma $ (Fields, Lemma 9.22.2). Then the primes $\sigma '(\mathfrak q \cap M')$ and $\mathfrak q \cap M'$ restrict to the same prime of $B \cap M$. Adjusting the choice of $\sigma '$ as in the first paragraph, we may assume that $\sigma '(\mathfrak q \cap M') = \mathfrak q \cap M'$. Then $\sigma '$ and $\overline{\sigma }$ define maps $\kappa (\mathfrak q \cap M') \to \kappa (\mathfrak q)$ which agree on $\kappa (\mathfrak q \cap M)$. Since $B \cap M = (B \cap M')^{\text{Gal}(M'/M)}$ we can use Lemma 15.110.6 to find $\tau \in \text{Gal}(M'/M)$ with $\tau (\mathfrak q \cap M') = \mathfrak q \cap M'$ such that $\tau \circ \sigma $ and $\overline{\sigma }$ induce the same map on $\kappa (\mathfrak q \cap M')$. There is a small detail here in that the lemma first guarantees that $\kappa (\mathfrak q \cap M')/\kappa (\mathfrak q \cap M)$ is normal, which then tells us that the difference between the maps is an automorphism of this extension (Fields, Lemma 9.15.10), to which we can apply the lemma to get $\tau $. Hence $(M', \tau \circ \sigma ') > (M, \sigma )$ contradicting the maximality of $(M, \sigma )$.
$\square$

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