The Stacks project

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 $e$th 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

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

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. $\square$