Lemma 15.115.5. Let $A \to B$ be an extension of discrete valuation rings. There exists a commutative diagram

$\xymatrix{ B \ar[r] & B' \\ A \ar[r] \ar[u] & A' \ar[u] }$

of extensions of discrete valuation rings such that

1. the extensions $K'/K$ and $L'/L$ of fraction fields are separable algebraic,

2. the residue fields of $A'$ and $B'$ are separable algebraic closures of the residue fields of $A$ and $B$, and

3. if a solution, weak solution, or separable solution exists for $A' \to B'$, then a solution, weak solution, or separable solution exists for $A \to B$.

Proof. By Algebra, Lemma 10.159.2 there exists an extension $A \subset A'$ which is a filtered colimit of finite étale extensions such that the residue field of $A'$ is a separable algebraic closure of the residue field of $A$. Then $A \subset A'$ is an extension of discrete valuation rings such that the induced extension $K'/K$ of fraction fields is separable algebraic.

Let $B \subset B'$ be a strict henselization of $B$. Then $B \subset B'$ is an extension of discrete valuation rings whose fraction field extension is separable algebraic. By Algebra, Lemma 10.155.9 there exists a commutative diagram as in the statement of the lemma. Parts (1) and (2) of the lemma are clear.

Let $K'_1/K'$ be a (weak) solution for $A' \to B'$. Since $A'$ is a colimit, we can find a finite étale extension $A \subset A_1'$ and a finite extension $K_1$ of the fraction field $F$ of $A_1'$ such that $K'_1 = K' \otimes _ F K_1$. As $A \subset A_1'$ is finite étale and $B'$ strictly henselian, it follows that $B' \otimes _ A A_1'$ is a finite product of rings isomorphic to $B'$. Hence

$L' \otimes _ K K_1 = L' \otimes _ K F \otimes _ F K_1$

is a finite product of rings isomorphic to $L' \otimes _{K'} K'_1$. Thus we see that $K_1/K$ is a (weak) solution for $A \to B'$. Hence it is also a (weak) solution for $A \to B$ by Lemma 15.115.4. $\square$

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