0
$\begingroup$

I come across this question which I have trouble understanding. It says

(7.3F) Suppose $Z$ is a closed subset of an affine scheme $Spec A$ locally cut out by one equation, (In other words, $Spec A$ can be covered by smaller open sets and on each such set $Z$ is cut out by one equation...

Question: Does he mean to say that there exists a smaller affine open set $U=Spec\ C$ (for example) such that $U\cap Z$ is something of the form $Spec\ (C/(g))$ for some $g\in C$. Otherwise what can this possibly mean if we do not assume that $U$ is affine?

Thanks!

$\endgroup$
  • $\begingroup$ You can assume that his "smaller open sets" are affine if you like. But a function on any locally ringed space cuts out a closed set. $\endgroup$ – Hoot Nov 28 '14 at 19:25
  • $\begingroup$ To be more precise what does this exactly mean? i.e. If $U$ is an open set (arbitrary), how do I capture the meaning that $Z\cap U$ is cut out by a function (because then I don't think it makes sense unless we pass to the stalk? For example, there exists $f\in \Gamma(U, \mathcal{O}_{X})$ such that $Z\cap U=V_{f}$ where $V_{f}=\{u\in U: [f]=0\in \mathcal{O}_{U,u}/\mathfrak{m}_{u}\}$. Does this make sense? $\endgroup$ – enoughsaid05 Nov 28 '14 at 19:53
  • $\begingroup$ That looks fine to me. As you say, we're using the stalk and in particular that the stalks are local rings. $\endgroup$ – Hoot Nov 29 '14 at 3:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.