Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $p$.)

share|improve this question
    
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) –  finite May 25 '12 at 16:08
add comment

2 Answers 2

up vote 3 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.

share|improve this answer
    
I like this way too. And it's unramified, which is nice. –  Dylan Moreland May 25 '12 at 16:42
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.