**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 $x$. Simiarly, $\psi '$ 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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