Lemma 15.114.9. Let $A \subset B$ be an extension of discrete valuation rings. Denote $L/K$ the corresponding extension of fraction fields. Let $K'/K$ be a finite separable extension. Then

\[ K' \otimes _ K L = \prod L'_ i \]

is a finite product of fields and the following is true

If $K'$ is unramified with respect to $A$, then each $L'_ i$ is unramified with respect to $B$.

If $K'$ is tamely ramified with respect to $A$, then each $L'_ i$ is tamely ramified with respect to $B$.

**Proof.**
The algebra $K' \otimes _ K L$ is a finite product of fields as it is a finite étale algebra over $L$. Let $A'$ be the integral closure of $A$ in $K'$.

In case (1) the ring map $A \to A'$ is finite étale. Hence $B' = B \otimes _ A A'$ is finite étale over $B$ and is a finite product of Dedekind domains (Lemma 15.44.4). Hence $B'$ is the integral closure of $B$ in $K' \otimes _ K L$. It follows immediately that each $L'_ i$ is unramified with respect to $B$.

Choose a uniformizer $\pi \in A$. To prove (2) we may replace $K'$ by a larger extension tame ramified with respect to $A$ (details omitted; hint: use Lemma 15.114.6). Thus by Lemma 15.114.7 we may assume there exists some $e \geq 1$ invertible in $\kappa _ A$ such that $K'$ contains $K[\pi ^{1/e}]$ and such that $K'$ is unramified with respect to $A[\pi ^{1/e}]$. Choose a product decomposition

\[ K[\pi ^{1/e}] \otimes _ K L = \prod L_{e, j} \]

For every $i$ there exists a $j_ i$ such that $L'_ i/L_{e, j_ i}$ is a finite separable extension. Let $B_{e, j}$ be the integral closure of $B$ in $L_{e, j}$. By (1) applied to $K'/K[\pi ^{1/e}]$ and $A[\pi ^{1/e}] \subset (B_{e, j_ i})_\mathfrak m$ we see that $L'_ i$ is unramified with respect to $(B_{e, j_ i})_\mathfrak m$ for every maximal ideal $\mathfrak m \subset B_{e, j_ i}$. Hence the proof will be complete if we can show that $L_{e, j}$ is tamely ramified with respect to $B$, see Lemma 15.114.5.

Choose a uniformizer $\theta $ in $B$. Write $\pi = u \theta ^ t$ where $u$ is a unit of $B$ and $t \geq 1$. Then we have

\[ A[\pi ^{1/e}] \otimes _ A B = B[x]/(x^ e - u \theta ^ t) \subset B[y, z]/(y^{e'} - \theta , z^ e - u) \]

where $e' = e/\gcd (e, t)$. The map sends $x$ to $z y^{t/\gcd (e, t)}$. Since the right hand side is a product of Dedekind domains each tamely ramified over $B$ the proof is complete (details omitted).
$\square$

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