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 $S$?



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



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> **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 $A$.
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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. 


