MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 6 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.