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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asked Sep 3, 2010 at 20:32
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1 Answer
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No. $B=k[x,y]$, $g=x$, $C=k[x, x^{-1} y]$.
answered Sep 3, 2010 at 20:38
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$\begingroup$ After a few computations, Spec C -> Spec B appears to be the blowing-down morphism in disguise. Is this correct? $\endgroup$ Commented Sep 3, 2010 at 20:47
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$\begingroup$ One of the two charts of it, yes. $\endgroup$ Commented Sep 3, 2010 at 21:01