The Stacks project

Proof. Since the characteristic of $L$ is $p$ we know that $M$ is an Artin-Schreier extension of $L$ (Fields, Lemma 9.25.1). Thus we may pick $z \in M$, $z \not\in L$ such that $\xi = z^ p - z \in L$. Choose $n \geq 0$ such that $\pi ^ n\xi \in B$. We pick $z$ such that $n$ is minimal. If $n = 0$, then $M/L$ is unramified with respect to $B$ (Lemma 15.115.11) and we are done. Thus we have $n > 0$.

Assumption (4) implies that $\kappa _ A$ is perfect. Thus we may choose compatible ring maps $\overline{\sigma } : \kappa _ A \to A/\pi ^ n A$ and $\overline{\sigma } : \kappa _ B \to B/\pi ^ n B$ as in Lemma 15.115.10. We lift the second of these to a map of sets $\sigma : \kappa _ B \to B$¹. Then we can write

for some $\lambda _ i \in \kappa _ B$ and $b \in B$. Let

and

We will argue by induction on the size of the finite set $J$.

The case $J = \emptyset$. Here for all $i \in \{ n, \ldots , 1\}$ we have $\sigma (\lambda _ i) = a_ i + \pi ^ n b_ i$ for some $a_ i \in A$ and $b_ i \in B$ by our choice of $\sigma$. Thus $\xi = \pi ^{-n} a + b$ for some $a \in A$ and $b \in B$. If $p | n$, then we write $a = a_0^ p + \pi a_1$ for some $a_0, a_1 \in A$ (as the residue field of $A$ is perfect). We compute

for some $b' \in B$. This would contradict the minimality of $n$. Thus $p$ does not divide $n$. Consider the degree $p$ extension $K_1$ of $K$ given by $w^ p - w = \pi ^{-n}a$. By Lemma 15.115.11 this extension is Galois and totally ramified with respect to $A$. Thus $L_1 = L \otimes _ K K_1$ is a field and $A_1 \subset B_1$ is weakly unramified (Lemma 15.115.3). By Lemma 15.115.11 the ring $M_1 = M \otimes _ K K_1$ is either a product of $p$ copies of $L_1$ (in which case we are done) or a field extension of $L_1$ of degree $p$. Moreover, in the second case, either $C_1$ is weakly unramified over $B_1$ (in which case we are done) or $M_1/L_1$ is degree $p$, Galois, and totally ramified with respect to $B_1$. In this last case the extension $M_1/L_1$ is generated by the element $z - w$ and

with $b \in B$ (see above). Thus by Lemma 15.115.11 once more the extension $M_1/L_1$ is unramified with respect to $B_1$ and we conclude that $K_1$ is a weak solution for $A \to C$. From now on we assume $J \not= \emptyset$.

Suppose that $j', j \in J$ such that $j' = p^ r j$ for some $r > 0$. Then we change our choice of $z$ into

Then $\xi$ changes into $\xi ' = (z')^ p - (z')$ as follows

Writing $\xi ' = \sum \nolimits _{i = n, \ldots , 1} \sigma (\lambda '_ i) \pi ^{-i} + b'$ as before we find that $\lambda '_ i = \lambda _ i$ for $i \not= j, j'$ and $\lambda '_ j = 0$. Thus the set $J$ has gotten smaller. By induction on the size of $J$ we may assume no such pair $j, j'$ exists. (Please observe that in this procedure we may get thrown back into the case that $J = \emptyset$ we treated above.)

For $j \in J$ write $\lambda _ j = \mu _ j^{p^{r_ j}}$ for some $r_ j \geq 0$ and $\mu _ j \in \kappa _ B$ which is not a $p$th power. This is possible by our assumption (4). Let $j \in J$ be the unique index such that $j p^{-r_ j}$ is maximal. (The index is unique by the result of the preceding paragraph.) Choose $r > \max (r_ j + 1)$ and such that $j p^{r - r_ j} > n$ for $j \in J$. Choose a separable extension $K_1/K$ totally ramified with respect to $A$ of degree $p^ r$ such that the corresponding discrete valuation ring $A_1 \subset K_1$ has uniformizer $\pi '$ with $(\pi ')^{p^ r} = \pi + \pi ^{n + 1}a$ for some $a \in A_1$ (Lemma 15.115.7). Observe that $L_1 = L \otimes _ K K_1$ is a field and that $L_1/L$ is totally ramified with respect to $B$ (Lemma 15.115.3). Computing in the integral closure $B_1$ we get

for some $b_1 \in B_1$. Note that $\sigma (\lambda _ i)$ for $i \in I$ is a $q$th power modulo $\pi ^ n$, i.e., modulo $(\pi ')^{n p^ r}$. Hence we can rewrite the above as

As in the previous paragraph we change our choice of $z$ into

to obtain

for some $b'_1 \in B_1$. Since there is a unique $j$ such that $j p^{r - r_ j}$ is maximal and since $j p^{r - r_ j}$ is bigger than $i \in I$ and divisible by $p$, we see that $M_1 / L_1$ falls into case (C) of Lemma 15.115.11. This finishes the proof. $\square$