Proof.
Let us use notation as in Remark 15.114.1; we will use all the observations made there. Since L/K is separable, the algebra L_1 = L \otimes _ K K_1 is reduced (Algebra, Lemma 10.43.6). Since B is Nagata, the ring extension B \subset B_1 is finite where B_1 is the integral closure of B in L_1 and B_1 is a Nagata ring. Similarly, the ring A is Nagata by Lemma 15.116.2 hence A \subset A_1 is finite and A_1 is a Nagata ring too. Moreover, the same assertions are true for K_2, i.e., L_2 = L \otimes _ K K_2 is reduced, the ring extensions A_1 \subset A_2 and B_1 \subset B_2 are finite where A_2, resp. B_2 is the integral closure of A, resp. B in K_2, resp. L_2.
Let \pi \in A be a uniformizer. Observe that \pi is a nonzerodivisor on K_1, K_2, A_1, A_2, L_1, L_2, B_1, and B_2 and we have K_1 = (A_1)_\pi , K_2 = (A_2)_\pi , L_1 = (B_1)_\pi , and L_2 = (B_2)_\pi . We may write K_2 = K_1(\alpha ) where \alpha ^ p = a_1 \in K_1, see Fields, Lemma 9.14.5. After multiplying \alpha by a power of \pi we may and do assume a_1 \in A_1. For the rest of the proof it is convenient to write K_2 = K_1[x]/(x^ p - a_1) and L_2 = L_1[x]/(x^ p - a_1). Consider the extensions of rings
A'_2 = A_1[x]/(x^ p - a_1) \subset A_2 \quad \text{and}\quad B'_2 = B_1[x]/(x^ p - a_1) \subset B_2
We may apply Lemma 15.116.3 to A'_2 \subset A_2 and f = \pi ^2 and to B'_2 \subset B_2 and f = \pi ^2. Choose an integer n large enough which works for both of these.
Consider the algebras
K_3 = K_1[x]/(x^ p - \pi ^{2n} x - a_1) \quad \text{and}\quad L_3 = L_1[x]/(x^ p - \pi ^{2n} x - a_1)
Observe that K_3/K_1 and L_3/L_1 are finite étale algebra extensions of degree p. Consider the subrings
A'_3 = A_1[x]/(x^ p - \pi ^ n x - a_1) \quad \text{and}\quad B'_3 = B_1[x]/(x^ p - \pi ^ n x - a_1)
of K_3 = (A'_2)_\pi and L_3 = (B'_3)_\pi . We are going to construct a commutative diagram
\xymatrix{ B'_2/\pi ^{2n} B'_2 \ar[r]_{\psi '} & B'_3/\pi ^{2n} B'_3 \\ A'_2/\pi ^{2n} A'_2 \ar[r]^{\varphi '} \ar[u] & A'_3/\pi ^{2n} A'_3 \ar[u] }
Namely, \varphi ' is the unique A_1-algebra isomorphism sending the class of x to the class of x. Similarly, \psi ' is the unique B_1-algebra isomorphism sending the class of x to the class of x. By our choice of n we obtain, via Lemma 15.116.3 and Remark 15.116.4 finite ring extensions A'_3 \subset A_3 and B'_3 \subset B_3 such that A'_3 \to B'_3 extends to a ring map A_3 \to B_3 and a commutative diagram
\xymatrix{ B_2/\pi ^2 B_2 \ar[r]_\psi & B_3/\pi ^2B_3 \\ A_2/\pi ^2 A_2 \ar[r]^\varphi \ar[u] & A_3/\pi ^2A_3 \ar[u] }
with all the properties asserted in the references mentioned above (in particular \varphi and \psi are isomorphisms).
With all of this data in hand, we can finish the proof. Namely, we first observe that A_3 and B_3 are finite products of Dedekind domains with \pi contained in all of the maximal ideals. Namely, if \mathfrak p \subset A_3 is a maximal ideal, then \pi \in \mathfrak p as A \to A_3 is finite. Then \mathfrak p/\pi ^2 A_3 corresponds via \varphi to a maximal ideal in A_2 / \pi ^2A_2 which is principal as A_2 is a finite product of Dedekind domains. We conclude that \mathfrak p/\pi ^2 A_3 is principal and hence by Nakayama we see that \mathfrak p (A_3)_\mathfrak p is principal. The same argument works for B_3. We conclude that A_3 is the integral closure of A in K_3 and that B_3 is the integral closure of B in L_3. Let \mathfrak q \subset B_3 be a maximal ideal lying over \mathfrak p \subset A_3. To finish the proof we have to show that (A_3)_\mathfrak p \to (B_3)_\mathfrak q is formally smooth in the \mathfrak q-adic topology. By the criterion of Lemma 15.111.5 it suffices to show that \mathfrak p (B_3)_\mathfrak q = \mathfrak q (B_3)_\mathfrak q and that the field extension \kappa (\mathfrak q)/\kappa (\mathfrak p) is separable. This is true because we may check both assertions by looking at the ring map A_3/\pi ^2 A_3 \to B_3/\pi ^2 B_3 and this is isomorphic to the ring map A_2/\pi ^2 A_2 \to B_2/\pi ^2 B_2 where the corresponding statement holds by our assumption that K_2 is a solution for A \subset B. Some details omitted.
\square
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