The Stacks project

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 $j$. $\square$

Comments (0)

There are also:

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

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EXW. Beware of the difference between the letter 'O' and the digit '0'.