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. $\square$

There are also:

• 2 comment(s) on Section 15.115: Eliminating 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).