Lemma 15.114.8. Let $A$ be a discrete valuation ring with fraction field $K$.

1. If $L/K$ is a finite separable extension which is tamely ramified with respect to $A$, then there exists a Galois extension $M/K$ containing $L$ which is tamely ramified with respect to $A$.

2. If $L_1/K$, $L_2/K$ are finite separable extensions which are tamely ramified with respect to $A$, then there exists a a finite separable extension $L/K$ which is tamely ramified with respect to $A$ containing $L_1$ and $L_2$.

Proof. Proof of (2). Choose a uniformizer $\pi \in A$. We can choose an integer $e$ invertible in $\kappa _ A$ and extensions $L_ i'/K' = K[\pi ^{1/e}]$ unramified with respect to $A' = A[\pi ^{1/e}]$ with $L'_ i/L_ i$ as extensions of $K$, see Lemma 15.114.7. By Lemma 15.111.8 we can find an extension $L'/K'$ which is unramified with respect to $A'$ such that $L'_ i/K$ is isomorphic to a subextension of $L'/K'$ for $i = 1, 2$. This finishes the proof of (3) as $L'/K$ is tamely ramified (use same lemma as above).

Proof of (1). We may first replace $L$ by a larger extension and assume that $L$ is an extension of $K' = K[\pi ^{1/e}]$ unramified with respect to $A' = A[\pi ^{1/e}]$ where $e$ is invertible in $\kappa _ A$, see Lemma 15.114.7. Let $M$ be the normal closure of $L$ over $K$, see Fields, Definition 9.16.4. Then $M/K$ is Galois by Fields, Lemma 9.21.5. On the other hand, there is a surjection

$L \otimes _ K \ldots \otimes _ K L \longrightarrow M$

of $K$-algebras, see Fields, Lemma 9.16.6. Let $B$ be the integral closure of $A$ in $L$ as in Remark 15.111.6. The condition that $L$ is unramified with respect to $A' = A[\pi ^{1/e}]$ exactly means that $A' \to B$ is an étale ring map, see Algebra, Lemma 10.143.7. Claim:

$K' \otimes _ K \ldots \otimes _ K K' = \prod K'_ i$

is a product of field extensions $K'_ i/K$ tamely ramified with respect to $A$. Then if $A'_ i$ is the integral closure of $A$ in $K'_ i$ we see that

$\prod A'_ i \otimes _{(A' \otimes _ A \ldots \otimes _ A A')} (B \otimes _ A \ldots \otimes _ A B)$

is finite étale over $\prod A'_ i$ and hence a product of Dedekind domains (Lemma 15.44.4). We conclude that $M$ is the fraction field of one of these Dedekind domains which is finite étale over $A'_ i$ for some $i$. It follows that $M/K'_ i$ is unramified with respect to every maximal ideal of $A'_ i$ and hence $M/K$ is tamely ramified by Lemma 15.114.5.

It remains the prove the claim. For this we write $A' = A[x]/(x^ e - \pi )$ and we see that

$A' \otimes _ A \ldots \otimes _ A A' = A'[x_1, \ldots , x_ r]/(x_1^ e - \pi , \ldots , x_ r^ e - \pi )$

The normalization of this ring certainly contains the elements $y_ i = x_ i/x_1$ for $i = 2, \ldots , r$ subject to the relations $y_ i^ e - 1 = 0$ and we obtain

$A[x_1, y_2, \ldots , y_ r]/(x_1^ e - \pi , y_2^ e - 1, \ldots , y_ r - 1) = A'[y_2, \ldots , y_ r]/(y_2^ e - 1, \ldots , y_ r^ e - 1)$

This ring is finite étale over $A'$ because $e$ is invertible in $A'$. Hence it is a product of Dedekind domains each unramified over $A'$ as desired (see references given above in case of confusion). $\square$

There are also:

• 2 comment(s) on Section 15.114: Abhyankar's lemma and tame ramification

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).