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How to define the union of closed subschemes in an affine scheme? Suppose $I$ and $J$ define closed subschemes of $\operatorname{Spec}R$, how should we define their intersection?

Eisenbud and Harris (GTM 197, p24) defined it by $I\cap J$ and used it to derive the "double points"(p60). We can also define it by $IJ$, which has the same underluing space.

Which one is more useful and why?

  • I assume "Jarris" is meant to be "Harris". (I would fix it, but the silly lower bound on edits won't let me.) –  Jan 10 '14 at 09:54

1 Answers1

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My default choice would for the scheme-theoretic union would be $I\cap J$.

There is a natural injection $R/I\cap J \to R/I \times R/J$, which you can interpret as saying that Spec $R/I\cap J$ is the scheme-theoretic image of the natural map from Spec $R/I \coprod $ Spec $R/J$, and this scheme-theoretic image is a natural choice for what you would call the scheme-theoretic union of the two closed subschemes.

Also, one would expect union to be idempotent, i.e. the union of Spec $R/I$ with itself should just be Spec $R/I$ again.

Considering Spec $R/IJ$ can also have its uses (especially if $I = J$), but I wouldn't call it a union.

Matt E
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  • Also, one would like to have that to write a variety as the union of its irreducible component is the same as to find the primary decomposition of the associated ideal. – nowhere dense Jul 30 '18 at 19:09