**Proof.**
Let $\pi \in A$ be a uniformizer. Let $w \in B$ and $P \in B[t]$ be as in Lemma 15.115.13 (for $B$). Set $e_1 = \text{ord}_ B(w)$, so that $w$ and $\pi ^{e_1}$ are associates in $B$. Pick $z \in M$ generating $M$ over $L$ with $\xi = P(z) \in K$ and $n$ such that $\pi ^ n\xi \in B$ as in the definition of the level of $M$ over $L$, i.e., $l = n/e_1$.

The proof of this lemma is completely similar to the proof of Lemma 15.115.12. To explain what is going on, observe that

15.115.16.1
\begin{equation} \label{more-algebra-equation-first-congruence} P(z) \equiv z^ p - z \bmod \pi ^{-n + e_1}B \end{equation}

for any $z \in L$ such that $\pi ^{-n} P(z) \in B$ (use that $z$ has valuation at worst $-n/p$ and the shape of the polynomial $P$). Moreover, we have

15.115.16.2
\begin{equation} \label{more-algebra-equation-second-congruence} \xi _1 + \xi _2 + w^ p \xi _1 \xi _2 \equiv \xi _1 + \xi _2 \bmod \pi ^{-2n + pe_1}B \end{equation}

for $\xi _1, \xi _2 \in \pi ^{-n}B$. Finally, observe that $n - e_1 = (l - 1)/e_1$ and $-2n + pe_1 = -(2l - p)e_1$. Write $m = n - e_1 \max (0, l - 1, 2l - p)$. The above shows that doing calculations in $\pi ^{-n}B / \pi ^{-n + m}B$ the polynomial $P$ behaves exactly as the polynomial $z^ p - z$. This explains why the lemma is true but we also give the details below.

Assumption (4) implies that $\kappa _ A$ is perfect. Observe that $m \leq e_1$ and hence $A/\pi ^ m$ is annihilated by $w$ and hence $p$. Thus we may choose compatible ring maps $\overline{\sigma } : \kappa _ A \to A/\pi ^ mA$ and $\overline{\sigma } : \kappa _ B \to B/\pi ^ mB$ as in Lemma 15.115.10. We lift the second of these to a map of sets $\sigma : \kappa _ B \to B$. Then we can write

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

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

\[ I = \{ i \in \{ n, \ldots , n - m + 1\} \mid \lambda _ i \in \kappa _ A\} \]

and

\[ J = \{ j \in \{ n, \ldots , n - m + 1\} \mid \lambda _ i \not\in \kappa _ A\} \]

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

The case $J = \emptyset $. Here for all $i \in \{ n, \ldots , n - m + 1\} $ we have $\sigma (\lambda _ i) = a_ i + \pi ^{n - m}b_ i$ for some $a_ i \in A$ and $b_ i \in B$ by our choice of $\overline{\sigma }$. Thus $\xi = \pi ^{-n} a + \pi ^{-n + m} b$ 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). Set $z_1 = - \pi ^{-n/p} a_0$. Note that $P(z_1) \in \pi ^{-n}B$ and that $z + z_1 + w z z_1$ is an element generating $M$ over $L$ (note that $wz_1 \not= -1$ as $n < pe_1$). Moreover, by Lemma 15.115.13 we have

\[ P(z + z_1 + w z z_1) = P(z) + P(z_1) + w^ p P(z) P(z_1) \in K \]

and by equations (15.115.16.1) and (15.115.16.2) we have

\[ P(z) + P(z_1) + w^ p P(z) P(z_1) \equiv \xi + z_1^ p - z_1 \bmod \pi ^{-n + m}B \]

for some $b' \in B$. This contradict the minimality of $n$! Thus $p$ does not divide $n$. Consider the degree $p$ extension $K_1$ of $K$ given by $P(y) = -\pi ^{-n}a$. By Lemma 15.115.14 this extension is separable 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 15.115.14 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, totally ramified with respect to $B_1$. In this last case the extension $M_1/L_1$ is generated by the element $z + y + wzy$ and we see that $P(z + y + wzy) \in L_1$ and

\begin{align*} P(z + y + wzy) & = P(z) + P(y) + w^ p P(z) P(y) \\ & \equiv \xi - \pi ^{-n}a \bmod \pi ^{-n + m}B_1 \\ & \equiv 0 \bmod \pi ^{-n + m}B_1 \end{align*}

in exactly the same manner as above. By our choice of $m$ this means exactly that $M_1/L_1$ has level at most $\max (0, l - 1, 2l - p)$. From now on we assume that $J \not= \emptyset $.

Suppose that $j', j \in J$ such that $j' = p^ r j$ for some $r > 0$. Then we set

\[ z_1 = - \sigma (\lambda _ j) \pi ^{-j} - \sigma (\lambda _ j^ p) \pi ^{-pj} - \ldots - \sigma (\lambda _ j^{p^{r - 1}}) \pi ^{-p^{r - 1}j} \]

and we change $z$ into $z' = z + z_1 + wzz_1$. Observe that $z' \in M$ generates $M$ over $L$ and that we have $\xi ' = P(z') = P(z) + P(z_1) + wP(z)P(z_1) \in L$ with

\[ \xi ' \equiv \xi - \sigma (\lambda _ j) \pi ^{-j} + \sigma (\lambda _ j^{p^ r}) \pi ^{-j'} \bmod \pi ^{-n + m}B \]

by using equations (15.115.16.1) and (15.115.16.2) as above. Writing

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

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 there is no pair $j, j'$ of $J$ such that $j'/j$ is a power of $p$. (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 $j \in J$. Let $K_1/K$ be the extension of degree $p^ r$, totally ramified with respect to $A$, defined by $(\pi ')^{p^ r} = \pi $. Observe that $\pi '$ is the uniformizer of the corresponding discrete valuation ring $A_1 \subset K_1$. Observe that $L_1 = L \otimes _ K K_1$ is a field and $L_1/L$ is totally ramified with respect to $B$ (Lemma 15.115.3). Computing in the integral closure $B_1$ we get

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

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

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

Similar to our choice in the previous paragraph we set

\begin{align*} z_1 & - \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*}

and we change our choice of $z$ into $z' = z + z_1 + wzz_1$. Then $z'$ generates $M_1$ over $L_1$ and $\xi ' = P(z') = P(z) + P(z_1) + w^ p P(z) P(z_1) \in L_1$ and a calculation shows that

\[ \xi ' \equiv \sum \nolimits _{i \in I} x_ i (\pi ')^{-i} + \sum \nolimits _{j \in J} \sigma (\mu _ j) (\pi ')^{- j p^{r - r_ j}} + (\pi ')^{(-n + m)p^ r}b'_1 \]

for some $b'_1 \in B_1$. There is a unique $j$ such that $j p^{r - r_ j}$ is maximal and $j p^{r - r_ j}$ is bigger than $i \in I$. If $j p^{r - r_ j} \leq (n - m)p^ r$ then the level of the extension $M_1/L_1$ is less than $\max (0, l - 1, 2l - p)$. If not, then, as $p$ divides $j p^{r - r_ j}$, we see that $M_1 / L_1$ falls into case (C) of Lemma 15.115.14. This finishes the proof.
$\square$

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