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Let $X$ be an irreducible Noetherian scheme. Consider some flat closed immersion into it. I want to show that it is also open, so that the morphism is surjective.

I have a few thoughts, but I can't seem to finish. Fixing some affine chart, we have a map

$$\operatorname{Spec} A/I \rightarrow \operatorname{Spec} A.$$

Associated to it we have a ring map $A \rightarrow A/I$. It is flat at every prime ideal by hypothesis, so it is flat. One can show this implies that $I=I^2$, and using Nakayama's lemma we see that $I=eA$ for some $e$ with $e^2=e$. One can also show that $A=(1-e)A\oplus eA$.

Why does this imply that the map is open? And does this result hold if we weaken the hypotheses so that $X$ is only assumed to be locally Noetherian?

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$X$ doesn't need to be irreducible. You can remove the noetherian condition if you assume just that the closed immersion is of finite presentation. Then one gets the equality $A = (e) + (1-e)$ as you stated. Now it is immediate that $V(I) = V(e) = D(1-e)$ is open.

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    $\begingroup$ The Noetherian assumption on the scheme insures the ideal is in fact finitely generated, right? Thanks for your answer! $\endgroup$ – Potato Jun 27 '13 at 7:05
  • $\begingroup$ @Potato: ah yes that's right. $\endgroup$ – user314 Jun 27 '13 at 7:08

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