The Stacks project

Proof. Let $\pi \in A$ be a uniformizer. By Kummer theory (Fields, Lemma 9.24.1) $M$ is obtained from $L$ by adjoining the root of $y^ p = b$ for some $b \in L$.

If $\text{ord}_ B(b)$ is prime to $p$, then we choose a degree $p$ separable extension $K_1/K$ totally ramified with respect to $A$ (for example using Lemma 15.115.7). Let $A_1$ be the integral closure of $A$ in $K_1$. By Lemma 15.115.3 the integral closure $B_1$ of $B$ in $L_1 = L \otimes _ K K_1$ is a discrete valuation ring weakly unramified over $A_1$. If $K_1/K$ is not a weak solution for $A \to C$, then the integral closure $C_1$ of $C$ in $M_1 = M \otimes _ K K_1$ is a discrete valuation ring and $B_1 \to C_1$ has ramification index $p$. In this case, the field $M_1$ is obtained from $L_1$ by adjoining the $p$th root of $b$ with $\text{ord}_{B_1}(b)$ divisible by $p$. Replacing $A$ by $A_1$, etc we may assume that $b = \pi ^ n u$ where $u \in B$ is a unit and $n$ is divisible by $p$. Of course, in this case the extension $M$ is obtained from $L$ by adjoining the $p$th root of a unit.

Suppose $M$ is obtained from $L$ by adjoining the root of $y^ p = u$ for some unit $u$ of $B$. If the residue class of $u$ in $\kappa _ B$ is not a $p$th power, then $B \subset C$ is weakly unramified (Lemma 15.115.8) and we are done. Otherwise, we can replace our choice of $y$ by $y/v$ where $v^ p$ and $u$ have the same image in $\kappa _ B$. After such a replacement we have

for some $b \in B$. Then we see that $P(z) = \pi b/ w^ p$ where $z = (y - 1)/w$. Thus we see that the extension is a degree $p$ extension of finite level with $\xi = \pi b / w^ p$. $\square$