# Intersection of closed sub schemes

For a family of closed sub schemes, $$\phi_ i: Z_ i\rightarrow X$$ the scheme intersection $$\phi:\cap_ i Z_ i \rightarrow X$$ is defined to be the closed sub scheme of X such that for each affine open $$U \subset X$$, the kernel of $$\mathscr{O}_ X(U)\rightarrow \mathscr{O}_{\cap_ i Z_ i} (\phi^{-1}( U))$$ is the ideal generated by the kernels of all the $$\mathscr{O}_X(U)\rightarrow \mathscr{O}_{Z_ i} (\phi_ i^{-1}( U))$$ . My question is whether the same is true when U is is not necessarily affine. That is for all open subsets of X.

• Try the case when $X=\mathbb{P}^n$ and $Z$ any proper closed subscheme and $U=X$. Jan 5, 2020 at 17:25
• @Mohan I’m sorry I don’t understand. Do you want me to intersect Z with X or intersect Z with some other proper closed Z? In the cases I know $Z\subset P^n_k$ Will have Global ring of functions equal to k.
– bart
Jan 5, 2020 at 23:14
• So the map from $O_X(X)\to O_Z(Z\cap U)$ has kernel just zero. Jan 5, 2020 at 23:17
• @Mohan right, if you take U=X, it seems like the statement is always true since all kernels from $O_X(X)=k$ are zero... but i thought you were gesturing at a counterexample. have i misunderstood?
– bart
Jan 6, 2020 at 11:52
• @Mohan +1 request to elaborate. Jan 6, 2020 at 13:19

1. $$Z_1, Z_2$$ disjoint closed subschemes of $$U=X=\mathbb{P}^n$$ e.g. $$Z_1=\{{0\}}, Z_2=\{{\infty\}}, U=X=\mathbb{P}^1$$.
2. $$U=X=\mathbb{A}^2-\{{(0,0)\}}, Z_1=(y=0), Z_2=(y-x^2-x=0)$$.