Lemma 15.115.7. Let $A$ be a discrete valuation ring with uniformizer $\pi$. If the residue characteristic of $A$ is $p > 0$, then for every $n > 1$ and $p$-power $q$ there exists a degree $q$ separable extension $L/K$ totally ramified with respect to $A$ such that the integral closure $B$ of $A$ in $L$ has ramification index $q$ and a uniformizer $\pi _ B$ such that $\pi _ B^ q = \pi + \pi ^ n b$ and $\pi _ B^ q = \pi + (\pi _ B)^{nq}b'$ for some $b, b' \in B$.

Proof. If the characteristic of $K$ is zero, then we can take the extension given by $\pi _ B^ q = \pi$, see Lemma 15.114.2. If the characteristic of $K$ is $p > 0$, then we can take the extension of $K$ given by $z^ q - \pi ^ n z = \pi ^{1 - q}$. Namely, then we see that $y^ q - \pi ^{n + q - 1} y = \pi$ where $y = \pi z$. Taking $\pi _ B = y$ we obtain the desired result. $\square$

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).