Is the subspace of DVR's of the Zariski-Riemann space still quasi-compact? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T00:29:36Z http://mathoverflow.net/feeds/question/75734 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75734/is-the-subspace-of-dvrs-of-the-zariski-riemann-space-still-quasi-compact Is the subspace of DVR's of the Zariski-Riemann space still quasi-compact? name 2011-09-18T07:06:48Z 2011-09-27T18:52:54Z <p>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?</p> <p>Is $S'$ dense in $S$?</p> http://mathoverflow.net/questions/75734/is-the-subspace-of-dvrs-of-the-zariski-riemann-space-still-quasi-compact/76544#76544 Answer by Jizhan Hong for Is the subspace of DVR's of the Zariski-Riemann space still quasi-compact? Jizhan Hong 2011-09-27T18:52:54Z 2011-09-27T18:52:54Z <p>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.)</p> <p>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 $S$. </p> <p>This is a direct consequence of the following lemma (Page 487) from the book <em>Algebraic geometry I, schemes with examples and exercises</em> by Ulrich G&ouml;rtz and Torsten Wedhorn:</p> <blockquote> <p><strong>Lemma 15.6</strong> 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 $A$.</p> </blockquote> <p>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. </p>