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 \ref{fields-lemma-normal-closure-inside-normal}). Choose $\sigma' \in \text{Gal}(M'/K)$ whose restriction to $M$ is $\sigma$ (Fields, Lemma \ref{fields-lemma-galois-infinite}). 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 \ref{lemma-one-orbit} 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 $(M, \sigma)$. \medskip\noindent 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 \ref{fields-lemma-normal-closure-inside-normal}). Choose $\sigma' \in \text{Gal}(M'/K)$ whose restriction to $M$ is $\sigma$ (Fields, Lemma \ref{fields-lemma-galois-infinite}). 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 \ref{lemma-one-orbit-geometric} 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 \ref{fields-lemma-normal-embeddings-differ-by-aut}), to which we can apply the lemma to get $\tau$. Hence $(M', \tau \circ \sigma') > (M, \sigma)$ contradicting the maximality of $(M, \sigma)$.