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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