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Let $B \subset C$ be Noetherian integral domains, and $g \in B$. Thus, $\mathrm{Spec} B \to \mathrm{Spec} B_g$ is an open immersion.

If furthermore $C \subset B_g$, does it follow that $\mathrm{Spec} C \to \mathrm{Spec} B$ is an open immersion?

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1 Answer 1

up vote 6 down vote accepted

No. $B=k[x,y]$, $g=x$, $C=k[x, x^{-1} y]$.

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After a few computations, Spec C -> Spec B appears to be the blowing-down morphism in disguise. Is this correct? – Charles Staats Sep 3 '10 at 20:47
One of the two charts of it, yes. – David Speyer Sep 3 '10 at 21:01

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