Lemma 15.114.7. Let $A$ be a discrete valuation ring with fraction field $K$. Let $\pi \in A$ be a uniformizer. Let $L/K$ be a finite separable extension. The following are equivalent

1. $L$ is tamely ramified with respect to $A$,

2. there exists an $e \geq 1$ invertible in $\kappa _ A$ and an extension $L'/K' = K[\pi ^{1/e}]$ unramified with respect to $A' = A[\pi ^{1/e}]$ such that $L$ is contained in $L'$, and

3. there exists an $e_0 \geq 1$ invertible in $\kappa _ A$ such that for every $d \geq 1$ invertible in $\kappa _ A$ (2) holds with $e = de_0$.

Proof. Observe that $A'$ is a discrete valuation ring with fraction field $K'$, see Lemma 15.114.2. Of course the ramification index of $A'$ over $A$ is $e$. Thus if (2) holds, then $L'$ is tamely ramified with respect to $A$ by Lemma 15.114.5. Hence $L$ is tamely ramified with respect to $A$ by Lemma 15.114.6.

The implication (3) $\Rightarrow$ (2) is immediate.

Assume that (1) holds. Let $B$ be the integral closure of $A$ in $L$ and let $\mathfrak m_1, \ldots , \mathfrak m_ n$ be its maximal ideals. Denote $e_ i$ the ramification index of $A \to B_{\mathfrak m_ i}$. Let $e_0$ be the least common multiple of $e_1, \ldots , e_ r$. This is invertible in $\kappa _ A$ by our assumption (1). Let $e = de_0$ as in (3). Set $A' = A[\pi ^{1/e}]$. Then $A \to A'$ is an extension of discrete valuation rings with fraction field $K' = K[\pi ^{1/e}]$, see Lemma 15.114.2. Choose a product decomposition

$L \otimes _ K K' = \prod L'_ j$

where $L'_ j$ are fields. Let $B'_ j$ be the integral closure of $A$ in $L'_ j$. Let $\mathfrak m_{ijk}$ be the maximal ideals of $B'_ j$ lying over $\mathfrak m_ i$. Observe that $(B'_ j)_{\mathfrak m_ i}$ is the integral closure of $B_{\mathfrak m_ i}$ in $L'_ j$. By Abhyankar's lemma (Lemma 15.114.4) applied to $A \subset B_{\mathfrak m_ i}$ and the extension $K'/K$ we see that $A' \to (B'_ j)_{\mathfrak m_{ijk}}$ is formally smooth in the $\mathfrak m_{ijk}$-adic topology. This implies that the ramification index is $1$ and that the residue field extension is separable (Lemma 15.111.5). In this way we see that $L'_ j$ is unramified with respect to $A'$. This finishes the proof: we take $L' = L'_ j$ for some $j$. $\square$

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).