The Stacks project

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. $\square$