The Stacks project

Lemma 15.114.6. Let $A$ be a discrete valuation ring with fraction field $K$. If $M/L/K$ are finite separable extensions and $M$ is tamely ramified with respect to $A$, then $L$ is tamely ramified with respect to $A$.

Proof. We will use the results of the discussion in Remark 15.111.6 without further mention. Let $C/B/A$ be the integral closures of $A$ in $M/L/K$. Since $C$ is a finite ring extension of $B$, we see that $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(B)$ is surjective. Hence for ever maximal ideal $\mathfrak m \subset B$ there is a maximal ideal $\mathfrak m' \subset C$ lying over $\mathfrak m$. By the multiplicativity of ramification indices (Lemma 15.111.3) and the assumption, we conclude that the ramification index of $B_\mathfrak m$ over $A$ is prime to the residue characteristic. Since $\kappa (\mathfrak m')/\kappa _ A$ is finite separable, the same is true for $\kappa (\mathfrak m)/\kappa _ A$. $\square$

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Your email address will not be published.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 0EXV. Beware of the difference between the letter 'O' and the digit '0'.