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Theorem 15.115.18 (Epp). Let A \subset B be an extension of discrete valuation rings with fraction fields K \subset L. If the characteristic of \kappa _ A is p > 0, assume that every element of

\bigcap \nolimits _{n \geq 1} \kappa _ B^{p^ n}

is separable algebraic over \kappa _ A. Then there exists a finite extension K_1/K which is a weak solution for A \to B as defined in Definition 15.115.1.

Proof. If the characteristic of \kappa _ A is zero or if the residue characteristic is p, the ramification index is prime to p, and the residue field extension is separable, then this follows from Abhyankar's lemma (Lemma 15.114.4). Namely, suppose the ramification index is e. Choose a uniformizer \pi \in A. Let K_1/K be the extension obtained by adjoining an eth root of \pi . By Lemma 15.114.2 we see that the integral closure A_1 of A in K_1 is a discrete valuation ring with ramification index over A. Thus A_1 \to (B_1)_\mathfrak m is formally smooth in the \mathfrak m-adic topology for all maximal ideals \mathfrak m of B_1 by Lemma 15.114.4 and a fortiori these are weakly unramified extensions of discrete valuation rings.

From now on we let p be a prime number and we assume that \kappa _ A has characteristic p. We first apply Lemma 15.115.5 to reduce to the case that A and B have separably closed residue fields. Since \kappa _ A and \kappa _ B are replaced by their separable algebraic closures by this procedure we see that we obtain

\kappa _ A \supset \bigcap \nolimits _{n \geq 1} \kappa _ B^{p^ n}

from the condition of the theorem.

Let \pi \in A be a uniformizer. Let A^\wedge and B^\wedge be the completions of A and B. We have a commutative diagram

\xymatrix{ B \ar[r] & B^\wedge \\ A \ar[u] \ar[r] & A^\wedge \ar[u] }

of extensions of discrete valuation rings. Let K^\wedge be the fraction field of A^\wedge . Suppose that we can find a finite extension M/K^\wedge which is (a) a weak solution for A^\wedge \to B^\wedge and (b) a compositum of a separable extension and an extension obtained by adjoining a p-power root of \pi . Then by Lemma 15.113.2 we can find a finite extension K_1/K such that K^\wedge \otimes _ K K_1 = M. Let A_1, resp. A_1^\wedge be the integral closure of A, resp. A^\wedge in K_1, resp. M. Since A \to A^\wedge is formally smooth in the \mathfrak m^\wedge -adic topology (Lemma 15.111.5) we see that A_1 \to A_1^\wedge is formally smooth in the \mathfrak m_1^\wedge -adic topology (Lemma 15.114.3 and A_1 and A_1^\wedge are discrete valuation rings by discussion in Remark 15.114.1). We conclude from Lemma 15.115.4 part (2) that K_1/K is a weak solution for A \to B^\wedge . Applying Lemma 15.115.4 part (1) we see that K_1/K is a weak solution for A \to B.

Thus we may assume A and B are complete discrete valuation rings with separably closed residue fields of characteristic p and with \kappa _ A \supset \bigcap \nolimits _{n \geq 1} \kappa _ B^{p^ n}. We are also given a uniformizer \pi \in A and we have to find a weak solution for A \to B which is a compositum of a separable extension and a field obtained by taking p-power roots of \pi . Note that the second condition is automatic if A has mixed characteristic.

Set k = \bigcap \nolimits _{n \geq 1} \kappa _ B^{p^ n}. Observe that k is an algebraically closed field of characteristic p. If A has mixed characteristic let \Lambda be a Cohen ring for k and in the equicharacteristic case set \Lambda = k[[t]]. We can choose a ring map \Lambda \to A which maps t to \pi in the equicharacteristic case. In the equicharacteristic case this follows from the Cohen structure theorem (Algebra, Theorem 10.160.8) and in the mixed characteristic case this follows as \mathbf{Z}_ p \to \Lambda is formally smooth in the adic topology (Lemmas 15.111.5 and 15.37.5). Applying Lemma 15.115.4 we see that it suffices to prove the existence of a weak solution for \Lambda \to B which in the equicharacteristic p case is a compositum of a separable extension and a field obtained by taking p-power roots of t. However, since \Lambda = k[[t]] in the equicharacteristic case and any extension of k((t)) is such a compositum, we can now drop this requirement!

Thus we arrive at the situation where A and B are complete, the residue field k of A is algebraically closed of characteristic p > 0, we have k = \bigcap \kappa _ B^{p^ n}, and in the mixed characteristic case p is a uniformizer of A (i.e., A is a Cohen ring for k). If A has mixed characteristic choose a Cohen ring \Lambda for \kappa _ B and in the equicharacteristic case set \Lambda = \kappa _ B[[t]]. Arguing as above we may choose a ring map A \to \Lambda lifting k \to \kappa _ B and mapping a uniformizer to a uniformizer. Since k \subset \kappa _ B is separable the ring map A \to \Lambda is formally smooth in the adic topology (Lemma 15.111.5). Hence we can find a ring map \Lambda \to B such that the composition A \to \Lambda \to B is the given ring map A \to B (see Lemma 15.37.5). Since \Lambda and B are complete discrete valuation rings with the same residue field, B is finite over \Lambda (Algebra, Lemma 10.96.12). This reduces us to the special case discussed in Lemma 15.115.17. \square


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