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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.

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    $\begingroup$ Try the case when $X=\mathbb{P}^n$ and $Z$ any proper closed subscheme and $U=X$. $\endgroup$
    – Mohan
    Jan 5, 2020 at 17:25
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    $\begingroup$ @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. $\endgroup$
    – bart
    Jan 5, 2020 at 23:14
  • $\begingroup$ So the map from $O_X(X)\to O_Z(Z\cap U)$ has kernel just zero. $\endgroup$
    – Mohan
    Jan 5, 2020 at 23:17
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    $\begingroup$ @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? $\endgroup$
    – bart
    Jan 6, 2020 at 11:52
  • $\begingroup$ @Mohan +1 request to elaborate. $\endgroup$
    – Latimer
    Jan 6, 2020 at 13:19

1 Answer 1

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Two counterexamples:

  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. $U=X=\mathbb{A}^2-\{{(0,0)\}}, Z_1=(y=0), Z_2=(y-x^2-x=0)$.

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