**Proof.**
Let $e$ be the ramification index of $C$ over $B$. If $e = 1$, then we are done. If not, then $e = p$ by Lemmas 15.111.2 and 15.111.4. This in turn implies that the residue fields of $B$ and $C$ agree. Choose a uniformizer $\pi _ C$ of $C$. Write $\pi _ C^ p = u \pi $ for some unit $u$ of $C$. Since $\pi _ C^ p \in L$, we see that $u \in B^*$. Also $M = L[\pi _ C]$.

Suppose there exists an integer $m \geq 0$ such that

\[ u = \sum \nolimits _{0 \leq i < m} b_ i^ p \pi ^ i + b \pi ^ m \]

with $b_ i \in B$ and with $b \in B$ an element whose image in $\kappa _ B$ is not a $p$th power. Choose an extension $K_1/K$ as in Lemma 15.115.7 with $n = m + 2$ and denote $\pi '$ the uniformizer of the integral closure $A_1$ of $A$ in $K_1$ such that $\pi = (\pi ')^ p + (\pi ')^{np} a$ for some $a \in A_1$. Let $B_1$ be the integral closure of $B$ in $L \otimes _ K K_1$. Observe that $A_1 \to B_1$ is weakly unramified by Lemma 15.115.3. In $B_1$ we have

\[ u \pi = \left(\sum \nolimits _{0 \leq i < m} b_ i (\pi ')^{i + 1}\right)^ p + b (\pi ')^{(m + 1)p} + (\pi ')^{np} b_1 \]

for some $b_1 \in B_1$ (computation omitted). We conclude that $M_1$ is obtained from $L_1$ by adjoining a $p$th root of

\[ b + (\pi ')^{n - m - 1} b_1 \]

Since the residue field of $B_1$ equals the residue field of $B$ we see from Lemma 15.115.8 that $M_1/L_1$ has degree $p$ and the integral closure $C_1$ of $B_1$ is weakly unramified over $B_1$. Thus we conclude in this case.

If there does not exist an integer $m$ as in the preceding paragraph, then $u$ is a $p$th power in the $\pi $-adic completion of $B_1$. Since $B$ is Nagata, this means that $u$ is a $p$th power in $B_1$ by Algebra, Lemma 10.162.18. Whence the second case of the statement of the lemma holds.
$\square$

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