**Proof.**
Let $M'$ be any finite extension of $L$ and consider the integral closure $C'$ of $B$ in $M'$. Then $C'$ is finite over $B$ as $B$ is Nagata by Algebra, Lemma 10.162.8. Moreover, $C'$ is a discrete valuation ring, see discussion in Remark 15.114.1. Moreover $C'$ is complete as a $B$-module, hence complete as a discrete valuation ring, see Algebra, Section 10.96. It follows in particular that $C$ is the integral closure of $B$ in $M$ (by definition of valuation rings as maximal for the relation of domination).

Let $M \subset M'$ be a finite extension and let $C' \subset M'$ be the integral closure of $B$ as above. By Lemma 15.115.4 it suffices to prove the result for $A \to B \to C'$. Hence we may assume that $M/L$ is normal, see Fields, Lemma 9.16.3.

If $M / L$ is normal, we can find a chain of finite extensions

\[ L = L^0 \subset L^1 \subset L^2 \subset \ldots \subset L^ r = M \]

such that each extension $L^{j + 1}/L^ j$ is either:

purely inseparable of degree $p$,

totally ramified with respect to $B^ j$ and Galois of degree $p$,

totally ramified with respect to $B^ j$ and Galois cyclic of order prime to $p$,

Galois and unramified with respect to $B^ j$.

Here $B^ j$ is the integral closure of $B$ in $L^ j$. Namely, since $M/L$ is normal we can write it as a compositum of a Galois extension and a purely inseparable extension (Fields, Lemma 9.27.3). For the purely inseparable extension the existence of the filtration is clear. In the Galois case, note that $G$ is “the” decomposition group and let $I \subset G$ be the inertia group. Then on the one hand $I$ is solvable by Lemma 15.112.5 and on the other hand the extension $M^ I/L$ is unramified with respect to $B$ by Lemma 15.112.8. This proves we have a filtration as stated.

We are going to argue by induction on the integer $r$. Suppose that we can find a finite extension $K_1/K$ which is a weak solution for $A \to B^1$ where $B^1$ is the integral closure of $B$ in $L^1$. Let $K'_1$ be the normal closure of $K_1/K$ (Fields, Lemma 9.16.3). Since $A$ is complete and the residue field of $A$ is algebraically closed we see that $K'_1/K_1$ is separable and totally ramified with respect to $A_1$ (some details omitted). Hence $K'_1/K$ is a weak solution for $A \to B^1$ as well by Lemma 15.115.3. In other words, we may and do assume that $K_1$ is a normal extension of $K$. Having done so we consider the sequence

\[ L^0_1 = (L^0 \otimes _ K K_1)_{red} \subset L^1_1 = (L^1 \otimes _ K K_1)_{red} \subset \ldots \subset L^ r_1 = (L^ r \otimes _ K K_1)_{red} \]

and the corresponding integral closures $B^ i_1$. Note that $C_1 = B^ r_1$ is a product of discrete valuation rings which are transitively permuted by $G = \text{Aut}(K_1/K)$ by Lemma 15.115.6. In particular all the extensions of discrete valuation rings $A_1 \to (C_1)_\mathfrak m$ are isomorphic and a weak solution for one will be a weak solution for all of them. We can apply the induction hypothesis to the sequence

\[ A_1 \to (B^1_1)_{B^1_1 \cap \mathfrak m} \to (B^2_1)_{B^2_1 \cap \mathfrak m} \to \ldots \to (B^ r_1)_{B^ r_1 \cap \mathfrak m} = (C_1)_\mathfrak m \]

to get a weak solution $K_2/K_1$ for $A_1 \to (C_1)_\mathfrak m$. The extension $K_2/K$ will then be a weak solution for $A \to C$ by what we said before. Note that the induction hypothesis applies: the ring map $A_1 \to (B^1_1)_{B^1_1 \cap \mathfrak m}$ is weakly unramified by our choice of $K_1$ and the sequence of fraction field extensions each still have one of the properties (a), (b), (c), or (d) listed above. Moreover, observe that for any finite extension $\kappa _ B \subset \kappa $ we still have $k = \bigcap \kappa ^{p^ n}$.

Thus everything boils down to finding a weak solution for $A \subset C$ when the field extension $M/L$ satisfies one of the properties (a), (b), (c), or (d).

Case (d). This case is trivial as here $B \to C$ is unramified already.

Case (c). Say $M/L$ is cyclic of order $n$ prime to $p$. Because $M/L$ is totally ramified with respect to $B$ we see that the ramification index of $B \subset C$ is $n$ and hence the ramification index of $A \subset C$ is $n$ as well. Choose a uniformizer $\pi \in A$ and set $K_1 = K[\pi ^{1/n}]$. Then $K_1/K$ is a solution for $A \subset C$ by Abhyankar's lemma (Lemma 15.114.4).

Case (b). We divide this case into the mixed characteristic case and the equicharacteristic case. In the equicharacteristic case this is Lemma 15.115.12. In the mixed characteristic case, we first replace $K$ by a finite extension to get to the situation where $M/L$ is a degree $p$ extension of finite level using Lemma 15.115.15. Then the level is a rational number $l \in [0, p)$, see discussion preceding Lemma 15.115.16. If the level is $0$, then $B \to C$ is weakly unramified and we're done. If not, then we can replacing the field $K$ by a finite extension to obtain a new situation with level $l' \leq \max (0, l - 1, 2l - p)$ by Lemma 15.115.16. If $l = p - \epsilon $ for $\epsilon < 1$ then we see that $l' \leq p - 2\epsilon $. Hence after a finite number of replacements we obtain a case with level $\leq p - 1$. Then after at most $p - 1$ more such replacements we reach the situation where the level is zero.

Case (a) is Lemma 15.115.9. This is the only case where we possibly need a purely inseparable extension of $K$, namely, in case (2) of the statement of the lemma we win by adjoining a $p$th power of the element $\pi $. This finishes the proof of the lemma.
$\square$

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