**Proof.**
We will use the results of the discussion in Remark 15.111.6 without further mention.

Proof of (1). Let $C/B/A$ be the integral closures of $A$ in $M/L/K$. Since $C$ is a finite ring extension of $B$, we see that $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(B)$ is surjective. Hence for ever maximal ideal $\mathfrak m \subset B$ there is a maximal ideal $\mathfrak m' \subset C$ lying over $\mathfrak m$. By the multiplicativity of ramification indices (Lemma 15.111.3) and the assumption, we conclude that the ramification index of $B_\mathfrak m$ over $A$ is $1$. Since $\kappa (\mathfrak m')/\kappa _ A$ is finite separable, the same is true for $\kappa (\mathfrak m)/\kappa _ A$.

Proof of (2). Let $M$ be the normal closure of $L$ over $K$, see Fields, Definition 9.16.4. Then $M/K$ is Galois by Fields, Lemma 9.21.5. On the other hand, there is a surjection

\[ L \otimes _ K \ldots \otimes _ K L \longrightarrow M \]

of $K$-algebras, see Fields, Lemma 9.16.6. Let $B$ be the integral closure of $A$ in $L$ as in Remark 15.111.6. The condition that $L$ is unramified with respect to $A$ exactly means that $A \to B$ is an étale ring map, see Algebra, Lemma 10.143.7. By permanence properties of étale ring maps we see that

\[ B \otimes _ A \ldots \otimes _ A B \]

is étale over $A$, see Algebra, Lemma 10.143.3. Hence the displayed ring is a product of Dedekind domains, see Lemma 15.44.4. We conclude that $M$ is the fraction field of a Dedekind domain finite étale over $A$. This means that $M$ is unramified with respect to $A$ as desired.

Proof of (3). Let $B_ i \subset L_ i$ be the integral closure of $A$. Argue in the same manner as above to show that $B_1 \otimes _ A B_2$ is finite étale over $A$. Details omitted.
$\square$

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