Proof.
Let L_1 = (L \otimes _ K K_1)_{red} and M_1 = (M \otimes _ K K_1)_{red} and let B_1 \subset L_1 and C_1 \subset M_1 be the integral closure of B and C. Note that M_1 = (M \otimes _ L L_1)_{red} and that L_1 is a (nonempty) finite product of finite extensions of L. Hence the ring map B_1 \to C_1 is a finite product of ring maps of the form discussed in Remark 15.114.1. In particular, the map \mathop{\mathrm{Spec}}(C_1) \to \mathop{\mathrm{Spec}}(B_1) is surjective. Choose a maximal ideal \mathfrak m \subset C_1 and consider the extensions of discrete valuation rings
(A_1)_{A_1 \cap \mathfrak m} \to (B_1)_{B_1 \cap \mathfrak m} \to (C_1)_\mathfrak m
If the composition is weakly unramified, so is the map (A_1)_{A_1 \cap \mathfrak m} \to (B_1)_{B_1 \cap \mathfrak m}. If the residue field extension \kappa _{A_1 \cap \mathfrak m} \to \kappa _\mathfrak m is separable, so is the subextension \kappa _{A_1 \cap \mathfrak m} \to \kappa _{B_1 \cap \mathfrak m}. Taking into account Lemma 15.111.5 this proves (1). A similar argument works for (2).
\square
Comments (0)
There are also: