Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If $S$ is the Zariski-Riemann space of a noetherian subring $k$ of a field $K$, Zariski-Samuel prove that $S$ is quasi-compact. If $S'$ is the subspace of valuations that are discrete (i.e. that valuation group is isomorphic to $\mathbb{Z}^n$ with the lexicographical ordering), is $S'$ still quasi-compact?

Is $S'$ dense in $S$?

share|improve this question
    
Your definition of 'discrete valuation' is not the usual definition. Usually if one says 'DVR', then one means a valuation with its value group isomorphic to $\mathbb{Z}$. If one says 'discrete valuation' in the general sense, then one means the value group is discrete as an ordered group, i.e. every element has a predecessor and a successor; in this case the value group could also be $\mathbb{Z}\times\mathbb{Q}\times\cdots$. That being said, S′ could be empty if all the valuations are with divisible value groups, as in the case of separably/real/radically closed fields. –  Jizhan Hong Sep 25 '11 at 15:53
    
Hi, thanks for your comment! As is often the case, I forgot one of my hypotheses: $K$ is finitely generated as a field over $k$. Also, as an aside, I was using Samuel-Zariski's definition of discrete page 48-49. –  name Sep 26 '11 at 17:42
add comment

1 Answer 1

up vote 1 down vote accepted

This is only a partial answer: $S'$ is dense in $S$. (As mentioned in the comments, assuming that $K$ is finitely generated over $k$ as a field.)

In fact, a stronger result is true: Let $S_{\mathrm{DVR}}$ be the subspace of all DVRs (i.e. value group isomorphic to $\mathbb{Z}$), then $S_{\mathrm{DVR}}$ is dense in $S$.

This is a direct consequence of the following lemma (Page 487) from the book Algebraic geometry I, schemes with examples and exercises by Ulrich Görtz and Torsten Wedhorn:

Lemma 15.6 Let $A$ be a local integral domain, $K=\mathrm{Frac{A}}$ and let $K'$ be an extension of $K$. There exists a valuation ring $A'$ with $\mathrm{Frac}(A')=K'$ that dominates $A$. If $A$ is in addition noetherian and $K'\supseteq K$ is finitely generated, then we can find a discrete valuation ring $A'$ with $\mathrm{Frac}(A')=K'$ that dominates $A$.

I don't know if $S'$ (or $S_{\mathrm{DVR}}$) is quasi-compact at this moment. But I suspect that it's obvious to experts.

share|improve this answer
    
Cool thanks. Density is actually enough for what I want to do. By the way, that Lemma also appears in EGA II 7.1.2/7.1.7. I imagine it has been reproduced from Bourbaki somewhere. –  name Sep 29 '11 at 17:01
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.