Lemma 15.115.11. Let $A$ be a discrete valuation ring with fraction field $K$ of characteristic $p > 0$. Let $\xi \in K$. Let $L$ be an extension of $K$ obtained by adjoining a root of $z^ p - z = \xi$. Then $L/K$ is Galois and one of the following happens

1. $L = K$,

2. $L/K$ is unramified with respect to $A$ of degree $p$,

3. $L/K$ is totally ramified with respect to $A$ with ramification index $p$, and

4. the integral closure $B$ of $A$ in $L$ is a discrete valuation ring, $A \subset B$ is weakly unramified, and $A \to B$ induces a purely inseparable residue field extension of degree $p$.

Let $\pi$ be a uniformizer of $A$. We have the following implications:

1. If $\xi \in A$, then we are in case (1) or (2).

2. If $\xi = \pi ^{-n}a$ where $n > 0$ is not divisible by $p$ and $a$ is a unit in $A$, then we are in case (3)

3. If $\xi = \pi ^{-n} a$ where $n > 0$ is divisible by $p$ and the image of $a$ in $\kappa _ A$ is not a $p$th power, then we are in case (4).

Proof. The extension is Galois of order dividing $p$ by the discussion in Fields, Section 9.25. It immediately follows from the discussion in Section 15.112 that we are in one of the cases (1) – (4) listed in the lemma.

Case (A). Here we see that $A \to A[x]/(x^ p - x - \xi )$ is a finite étale ring extension. Hence we are in cases (1) or (2).

Case (B). Write $\xi = \pi ^{-n}a$ where $p$ does not divide $n$. Let $B \subset L$ be the integral closure of $A$ in $L$. If $C = B_\mathfrak m$ for some maximal ideal $\mathfrak m$, then it is clear that $p \text{ord}_ C(z) = -n \text{ord}_ C(\pi )$. In particular $A \subset C$ has ramification index divisible by $p$. It follows that it is $p$ and that $B = C$.

Case (C). Set $k = n/p$. Then we can rewrite the equation as

$(\pi ^ kz)^ p - \pi ^{n - k} (\pi ^ kz) = a$

Since $A[y]/(y^ p - \pi ^{n - k}y - a)$ is a discrete valuation ring weakly unramified over $A$, the lemma follows. $\square$

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).