Is the subspace of DVR's of the Zariski-Riemann space still quasi-compact?

2011-09-18T07:06:48Z

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 $S$?

Answer by Jizhan Hong for Is the subspace of DVR's of the Zariski-Riemann space still quasi-compact?

2011-09-27T18:52:54Z

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.

Is the subspace of DVR's of the Zariski-Riemann space still quasi-compact? - MathOverflow

Is the subspace of DVR's of the Zariski-Riemann space still quasi-compact?

Answer by Jizhan Hong for Is the subspace of DVR's of the Zariski-Riemann space still quasi-compact?