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. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: