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Let $R$ be a DVR with $K = Quot(R)$ and residue field $k$. Let $k'/k$ be a finite field extension. I would like to have a reference for the following statement (or to see, that it is not true):

There exists a finite field extension $K'/K$ s.t. the residue field of the integral closure $R'$ of $R$ in $K'$ is $k'$.

(In my concrete situation: $K/\mathbb Q_p$ finite field extension, $R = \mathcal O_K$ local number field of $K$, and $k \cong \mathbb F_q$ for some prime power $q$ of $p$.)

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Do you mean for $R'$ to be the integral closure of $R$ in $K'$, i.e. that $R' = \mathcal{O}_{K'}$? – Brandon Carter May 25 '12 at 16:05
@BrandonCarter: Yes, I mixed it up with the algebro-geometric term of normalization. (edited now) – boxdot May 25 '12 at 16:08
up vote 4 down vote accepted

Write $k^\prime=k[X]/(\bar{p})$ where $\bar{p}$ is a monic irreducible in $k[X]$. Let $p$ be a monic lift of $\bar{p}$ in $R[X]$. Consider $R^\prime=R[X]/(p(x))$. This is a local ring with residue field $\bar{k}$. It is a finite $R$-algebra, so it's also Noetherian. The maximal ideal of $R^\prime$ is generated by the image of $\pi$, where $\pi$ is a uniformizer for $R$. The ring map $R\rightarrow R^\prime$ is injective, so $\pi$ is not nilpotent in $R^\prime$. This is enough to conclude that $R^\prime$ is a discrete valuation ring. Take $K^\prime$ to be the field of fractions of $R^\prime$. Then $R^\prime$ is integral over $R$ and integrally closed in $K^\prime$, so it must be the integral closure of $R$ in $K^\prime$. One has $K^\prime=K[X]/(p(X))$, which is finite over $K$.

For more details see the first chapter of Serre's Local Fields.

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I like this way too. And it's unramified, which is nice. – Dylan Moreland May 25 '12 at 16:42
This is beautiful. Could it be generalized to an algebraic closure of the residue field? I mean, is it again possible to find a discrete valuation ring whose residue field is the algebraic closure of $k$? – Simone Apr 28 '15 at 7:19

Yes, this is true in your case. Write $K = L(\alpha)$ where $L$ is the maximal subextension of $K$ unramified over $\mathbb{Q}_p$ and $\alpha$ satisfies an Eisenstein polynomial (so $L$ is the field of fractions of the ring of Witt vectors on $k$). Then you can take $K' = L'(\alpha)$ where $L'$ is the field of fractions of the ring of Witt vectors of $k'$. Since $\alpha$ satisfies an Eisenstein polynomial, $L'$ will be the maximal unramified subextension of $K'$ and so the residue field of $K'$ will be $k'$ and not a proper extension thereof.

If we don't require the residue field to be perfect, I believe it's still true by the theory of Cohen rings, but the construction is not as nice. If we don't require the ring of integers to be complete, then I don't know.

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