Proposition 15.116.8. Let A \to B be an extension of discrete valuation rings with fraction fields K \subset L. If B is essentially of finite type over A, then there exists a finite extension K_1/K which is a solution for A \to B as defined in Definition 15.115.1.
See [Lemma 2.13, alterations] for a special case.
Proof. Observe that a weak solution is a solution if the residue field of A is perfect, see Lemma 15.111.5. Thus the proposition follows immediately from Theorem 15.115.18 if the residue characteristic of A is 0 (and in fact we do not need the assumption that A \to B is essentially of finite type). If the residue characteristic of A is p > 0 we will also deduce it from Epp's theorem.
Let x_ i \in A, i \in I be a set of elements mapping to a p-base of the residue field \kappa of A. Set
where the transition maps send t_{i, n + 1} to t_{i, n}^ p. Observe that A' is a filtered colimit of weakly unramified finite extensions of discrete valuation rings over A. Thus A' is a discrete valuation ring and A \to A' is weakly unramified. By construction the residue field \kappa ' = A'/\mathfrak m_ A A' is the perfection of \kappa .
Let K' be the fraction field of A'. We may apply Lemma 15.116.7 to the extension K'/K. Thus B' is a finite product of Dedekind domains. Let \mathfrak m_1, \ldots , \mathfrak m_ n be the maximal ideals of B'. Using Epp's theorem (Theorem 15.115.18) we find a weak solution K'_ i/K' for each of the extensions A' \subset B'_{\mathfrak m_ i}. Since the residue field of A' is perfect, these are actually solutions. Let K'_1/K' be a finite extension which contains each K'_ i. Then K'_1/K' is still a solution for each A' \subset B'_{\mathfrak m_ i} by Lemma 15.116.1.
Let A'_1 be the integral closure of A in K'_1. Note that A'_1 is a Dedekind domain by the discussion in Remark 15.114.1 applied to K' \subset K'_1. Thus Lemma 15.116.7 applies to K'_1/K. Therefore the integral closure B'_1 of B in L'_1 = (L \otimes _ K K'_1)_{red} is a Dedekind domain and because K'_1/K' is a solution for each A' \subset B'_{\mathfrak m_ i} we see that (A'_1)_{A'_1 \cap \mathfrak m} \to (B'_1)_{\mathfrak m} is formally smooth in the \mathfrak m-adic topology for each maximal ideal \mathfrak m \subset B'_1.
By construction, the field K'_1 is a filtered colimit of finite extensions of K. Say K'_1 = \mathop{\mathrm{colim}}\nolimits _{i \in I} K_ i. For each i let A_ i, resp. B_ i be the integral closure of A, resp. B in K_ i, resp. L_ i = (L \otimes _ K K_ i)_{red}. Then it is clear that
Since the ring maps A_ i \to A'_1 and B_ i \to B'_1 are injective integral ring maps and since A'_1 and B'_1 have finite spectra, we see that for all i large enough the ring maps A_ i \to A'_1 and B_ i \to B'_1 are bijective on spectra. Once this is true, for all i large enough the maps A_ i \to A'_1 and B_ i \to B'_1 will be weakly unramified (once the uniformizer is in the image). It follows from multiplicativity of ramification indices that A_ i \to B_ i induces weakly unramified maps on all localizations at maximal ideals of B_ i for such i. Increasing i a bit more we see that
induces surjective maps on residue fields (because the residue fields of B'_1 are finitely generated over those of A'_1 by Lemma 15.116.7). Picture of residue fields at maximal ideals lying under a chosen maximal ideal of B'_1:
Thus \kappa _{B_ i} is a finitely generated extension of \kappa _{A_ i} such that the compositum of \kappa _{B_ i} and \kappa _{A'_1} in \kappa _{B'_1} is separable over \kappa _{A'_1}. Then that happens already at a finite stage: for example, say \kappa _{B'_1} is finite separable over \kappa _{A'_1}(x_1, \ldots , x_ n), then just increase i such that x_1, \ldots , x_ n are in \kappa _{B_ i} and such that all generators satisfy separable polynomial equations over \kappa _{A_ i}(x_1, \ldots , x_ n). This means that A_ i \to (B_ i)_\mathfrak m is formally smooth in the \mathfrak m-adic topology for all maximal ideals \mathfrak m of B_ i and the proof is complete. \square
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