The Stacks project

Lemma 15.114.5. Let $A$ be a discrete valuation ring with fraction field $K$. Let $M/L/K$ be finite separable extensions. Let $B$ be the integral closure of $A$ in $L$. If $L/K$ is tamely ramified with respect to $A$ and $M/L$ is tamely ramified with respect to $B_\mathfrak m$ for every maximal ideal $\mathfrak m$ of $B$, then $M/K$ is tamely ramified with respect to $A$.

Proof. Let $C$ be the integral closure of $A$ in $M$. Every maximal ideal $\mathfrak m'$ of $C$ lies over a maximal ideal $\mathfrak m$ of $B$. Then the lemma follows from the multiplicativity of ramification indices (Lemma 15.111.3) and the fact that we have the tower $\kappa (\mathfrak m')/\kappa (\mathfrak m)/\kappa _ A$ of finite extensions of fields. $\square$

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Your email address will not be published.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



View Lemma 15.114.5 as pdf