Lemma 15.115.14. Let $A$ be a discrete valuation ring of mixed characteristic $(0, p)$ which contains a primitive $p$th root of $1$. Let $P(t) \in A[t]$ be the polynomial of Lemma 15.115.13. Let $\xi \in K$. Let $L$ be an extension of $K$ obtained by adjoining a root of $P(z) = \xi$. 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. Adjoining a root of $P(z) = \xi$ is the same thing as adjoining a root of $y^ p = w^ p(1 + \xi )$. Since $K$ contains a primitive $p$th root of $1$ the extension is Galois of order dividing $p$ by the discussion in Fields, Section 9.24. 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 $A \to A[x]/(P(x) - \xi )$ 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

$(\pi ^ kz)^ p - \pi ^{n - k} (\pi ^ kz) + \sum a_ i \pi ^{n - ik} (\pi ^ kz)^ i = a$

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

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