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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 $S$?

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  • $\begingroup$ 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. $\endgroup$
    – Jizhan Hong
    Commented Sep 25, 2011 at 15:53
  • $\begingroup$ 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. $\endgroup$
    – name
    Commented Sep 26, 2011 at 17:42

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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.

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  • $\begingroup$ 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. $\endgroup$
    – name
    Commented Sep 29, 2011 at 17:01

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