# Closed subscheme intersected with open affine: Irreducible?

(In the following we suppose that $$X$$ is a smooth variety over an algebraically closed field $$k$$ (in particular smooth $$\iff$$ regular). Here a variety is an integral separated scheme of finite type.)

I am currently reading through a paper, where the author is making the following statement:

Let $$Z$$ be a smooth closed subvariety of $$X$$ of codimension $$p$$. Choose a finite affine open covering $$U_\alpha$$ of $$X$$ and a regular sequence $$f_{\alpha 1}, \dots, f_{\alpha p} \in \mathcal O_X(U_\alpha)$$ such that $$Z \cap U_\alpha$$ is defined by the ideal $$(f_{\alpha 1}, \dots, f_{\alpha p})$$. We may assume that $$Z \cap U_\alpha$$ is irreducible.

I am wondering about the last sentence. First of all, $$Z \cap U_\alpha$$ doesn't have to be irreducible. However I don't see how we can reduce to that case while at the same time keeping the property of having the regular sequence. Maybe I'm not seeing something obvious here.

Can anybody help out maybe? Thanks in advance!

• Write $Z\cap U_\alpha$ as $F\cup G$ with $F,G$ closed in $U_\alpha$, therefore $\overline{F}\cup \overline{G}=Z$ as $Z\cap U_\alpha$ is dense in $Z$, thus we have say $\overline{F}=Z$ and $F=Z\cap U_\alpha$.
– Ahr
Commented Jun 5, 2019 at 12:26
• If $Z$ is an irreducible topological space, then each non-empty open subset of $Z$ is also irreducible: this is pure topology and has nothing to do with algebraic geometry. So in fact your phrase "First of all, $Z \cap U_\alpha$ doesn't have to be irreducible" is not correct. The best definition for "$Z$ is irreducible" is: $Z\neq \emptyset$ and any two non-empty open subsets of $Z$ have non-empty intersection. Commented Jun 6, 2019 at 11:29
• Of course, to both. +1 Commented Jun 6, 2019 at 11:47

This is presented perhaps a little out of order - the aside about irreducibility maybe is more natural to prove first. First we show that given any covering, one may find a refinement with $$U_\alpha$$ affine and $$U_\alpha\cap Z$$ irreducible, then we show that one may find appropriate regular sequences (after perhaps refining our covering).
An important point is that for smooth varieties, irreducible components are connected components. So as $$Z$$ has finitely many irreducible/connected components there's a finite covering of $$X$$ by open sets so that each open set only intersects one component of $$Z$$: if $$Z=\bigcup Z_i$$, then $$\hat{Z}_i := X\setminus \left(\bigcup_{i\neq j} Z_i\right)$$ is such an open cover. By taking common refinements with the $$\hat{Z}_i$$, any finite open cover may be assumed to only intersect each component of $$Z$$ once. From here, we may further refine this cover by taking a finite affine open cover of each $$\hat{Z}_i$$, so we are in the situation of having an affine open cover of $$X$$ so that each set in our covering intersects at most one irreducible component of $$Z$$. Now as any open subset of an irreducible variety is irreducible, we see that $$U_\alpha\cap Z = U_\alpha\cap Z_i$$ is also irreducible.
Now we deal with the regular sequence. The point here is that on any scheme, an irreducible closed subscheme of codimension $$p$$ may not be cut out by regular sequence of the right length, but this is true on the level of stalks (by definition of regularity, essentially). So we can pick an affine covering of our scheme so that this is true on each element of the cover. Now we're done.