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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