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Define the Euler characteristic of a scheme to be the Euler characteristic of its structure sheaf. I remember being told that for curves, this invariant satisfies inclusion-exclusion. That is, if $C_1, C_2$ are curves , then

$$\chi(C_1 \cup C_2) + \chi(C_1\cap C_2) = \chi(C_1) + \chi(C_2)$$

The intersection is scheme theoretic.Does any one knows a proof or a reference for this ?

flag	asked Dec 12 2011 at 5:12 Tmonk 173●1●5

Sándor Kovács · Accepted Answer · 2011-12-12 10:19:39Z UTC

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$$ 0\to \mathscr O_{C_1\cup C_2} \to \mathscr O_{C_1}\oplus \mathscr O_{C_2} \to \mathscr O_{C_1\cap C_2} \to 0 $$ with maps $a\mapsto (a,a)$ and $(a,b)\mapsto a-b$
is exact and $\chi$ is additive.

edited Dec 12 2011 at 10:19

answered Dec 12 2011 at 5:27

Sándor Kovács
23.7k●2●52●73

Why is the latter map surjective? – Martin Brandenburg Dec 12 2011 at 9:30

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It's surjective individually already. $C_1\cap C_2$ is a closed subscheme of $C_i$ for $i=1,2$, so $\mathscr O_{C_i}\to \mathscr O_{C_1\cap C_2}$ is surjective. So if $a$ in $\mathscr O_{C_1\cap C_2}$ lifts to $a$ in $\mathscr O_{C_1}$, then it is the image of $(a,0)$ by the above map. – Sándor Kovács Dec 12 2011 at 10:17

Oh yes, sorry for the stupid question. The left exactness is equivalent to the generalized Chinese Remainder Theorem which I have mentioned here mathoverflow.net/questions/10014. So in fact $C_1,C_2$ don't have to be curves, but they can be arbitrary closed subschemes. – Martin Brandenburg Dec 12 2011 at 11:26

Yes, right. I meant to mention that it works in any dimension (and they don't have to be equi-dimensional either). – Sándor Kovács Dec 12 2011 at 11:31

Euler characteristic and inclusion-exclusion

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