The Stacks project

Lemma 15.115.3. Let $A \to B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. Assume that $A \to B$ is weakly unramified. Then for any finite separable extension $K_1/K$ totally ramified with respect to $A$ we have that $L_1 = L \otimes _ K K_1$ is a field, $A_1$ and $B_1 = B \otimes _ A A_1$ are discrete valuation rings, and the extension $A_1 \subset B_1$ (see Remark 15.114.1) is weakly unramified.

Proof. Let $\pi \in A$ and $\pi _1 \in A_1$ be uniformizers. As $K_1/K$ is totally ramified with respect to $A$ we have $\pi _1^ e = u_1 \pi $ for some unit $u_1$ in $A_1$. Hence $A_1$ is generated by $\pi _1$ over $A$ and the minimal polynomial $P(t)$ of $\pi _1$ over $K$ has the form

\[ P(t) = t^ e + a_{e - 1} t^{e - 1} + \ldots + a_0 \]

with $a_ i \in (\pi )$ and $a_0 = u\pi $ for some unit $u$ of $A$. Note that $e = [K_1 : K]$ as well. Since $A \to B$ is weakly unramified we see that $\pi $ is a uniformizer of $B$ and hence $B_1 = B[t]/(P(t))$ is a discrete valuation ring with uniformizer the class of $t$. Thus the lemma is clear. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 15.115: Eliminating 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Your email address will not be published.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 09ER. Beware of the difference between the letter 'O' and the digit '0'.