Lemma 15.115.12. Let $A \subset B \subset C$ be extensions of discrete valuation rings with fractions fields $K \subset L \subset M$. Assume

1. $A \subset B$ weakly unramified,

2. the characteristic of $K$ is $p$,

3. $M$ is a degree $p$ Galois extension of $L$, and

4. $\kappa _ A = \bigcap _{n \geq 1} \kappa _ B^{p^ n}$.

Then there exists a finite Galois extension $K_1/K$ totally ramified with respect to $A$ which is a weak solution for $A \to C$.

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

$\xi = \sum \nolimits _{i = n, \ldots , 1} \sigma (\lambda _ i) \pi ^{-i} + b$

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

$I = \{ i \in \{ n, \ldots , 1\} \mid \lambda _ i \in \kappa _ A\}$

and

$J = \{ j \in \{ n, \ldots , 1\} \mid \lambda _ i \not\in \kappa _ A\}$

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

$(z - \pi ^{-n/p}a_0)^ p - (z - \pi ^{-n/p}a_0) = \pi ^{-(n - 1)}(a_1 + \pi ^{n - 1 - n/p}a_0) + b'$

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

$(z - w)^ p - (z - w) = z^ p - z - (w^ p - w) = b$

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

$z' = z - (\sigma (\lambda _ j) \pi ^{-j} + \sigma (\lambda _ j^ p) \pi ^{-pj} + \ldots + \sigma (\lambda _ j^{p^{r - 1}}) \pi ^{-p^{r - 1}j})$

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

$\xi ' = \xi - \sigma (\lambda _ j) \pi ^{-j} + \sigma (\lambda _ j^{p^ r}) \pi ^{-j'} + \text{something in }B$

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

$\xi = \sum \nolimits _{i \in I} \sigma (\lambda _ i) (\pi ')^{-i p^ r} + \sum \nolimits _{j \in J} \sigma (\mu _ j)^{p^{r_ j}} (\pi ')^{-j p^ r} + b_1$

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

$\xi = \sum \nolimits _{i \in I} x_ i^{p^ r} (\pi ')^{-i p^ r} + \sum \nolimits _{j \in J} \sigma (\mu _ j)^{p^{r_ j}} (\pi ')^{- j p^ r} + b_1$

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

\begin{align*} z' & = z \\ & - \sum \nolimits _{i \in I} \left(x_ i (\pi ')^{-i} + \ldots + x_ i^{p^{r - 1}} (\pi ')^{-i p^{r - 1}}\right) \\ & - \sum \nolimits _{j \in J} \left( \sigma (\mu _ j) (\pi ')^{- j p^{r - r_ j}} + \ldots + \sigma (\mu _ j)^{p^{r_ j - 1}} (\pi ')^{- j p^{r - 1}} \right) \end{align*}

to obtain

$(z')^ p - z' = \sum \nolimits _{i \in I} x_ i (\pi ')^{-i} + \sum \nolimits _{j \in J} \sigma (\mu _ j) (\pi ')^{- j p^{r - r_ j}} + b_1'$

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. $\square$

[1] If $B$ is complete, then we can choose $\sigma$ to be a ring map. If $A$ is also complete and $\sigma$ is a ring map, then $\sigma$ maps $\kappa _ A$ into $A$.

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