Lemma 15.114.3. Let A \to B be an extension of discrete valuation rings with fraction fields K \subset L. Assume that A \to B is formally smooth in the \mathfrak m_ B-adic topology. Then for any finite extension K_1/K we have L_1 = L \otimes _ K K_1, B_1 = B \otimes _ A A_1, and each extension (A_1)_{\mathfrak m_ i} \subset (B_1)_{\mathfrak m_{ij}} (see Remark 15.114.1) is formally smooth in the \mathfrak m_{ij}-adic topology.
Proof. We will use the equivalence of Lemma 15.111.5 without further mention. Let \pi \in A and \pi _ i \in (A_1)_{\mathfrak m_ i} be uniformizers. As \kappa _ A \subset \kappa _ B is separable, the ring
(B \otimes _ A (A_1)_{\mathfrak m_ i})/\pi _ i (B \otimes _ A (A_1)_{\mathfrak m_ i}) = B/\pi B \otimes _{A/\pi A} (A_1)_{\mathfrak m_ i}/\pi _ i (A_1)_{\mathfrak m_ i}
is a product of fields each separable over \kappa _{\mathfrak m_ i}. Hence the element \pi _ i in B \otimes _ A (A_1)_{\mathfrak m_ i} is a nonzerodivisor and the quotient by this element is a product of fields. It follows that B \otimes _ A A_1 is a Dedekind domain in particular reduced. Thus B \otimes _ A A_1 \subset B_1 is an equality. \square
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