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Let $X$ be projective surface and $C$ be an irreducible nonsingular curve on $X$. For any curve $D$, I know that

$$0 \rightarrow \mathcal{I}_D \rightarrow \mathcal{O}_X \rightarrow \mathcal{O}_D \rightarrow 0$$

Tensoring with $\mathcal{O}_C$ I saw in Hartshorne book (p.358) that

$$0 \rightarrow \mathcal{I}_D\otimes\mathcal{O}_C \rightarrow \mathcal{O}_C \rightarrow \mathcal{O}_{C\cap D} \rightarrow 0$$ is exact.

But I can't understand injectivity of this sequence.

Sang Cheol Lee
  • 3,119

1 Answers1

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This is pretty terse stuff from Hartshorne! Just to be clear, he is talking about the case when $C$ and $D$ intersect transversally.

To show that the second sequence above is exact on the left, we have to show that the sheaf $\operatorname{Tor}_1(\mathcal O_D, \mathcal O_C)$ vanishes.

It's enough to check this on stalks.

First let $p$ be a point of $C \cap D$, and let $R=\mathcal O_{X,\, p}$. Then the stalks of $\mathcal O_D$ and $\mathcal O_C$ at $p$ have the form $R/(f)$ and $R/(g)$, for some irreducible elements $f, \, g$ of $R$.

Now for any $R$-module $B$, we have (according to Wikipedia)

$$\operatorname{Tor}_1^R(R/(f),B) = \{ b \in B : fb = 0 \} ;$$

applying this with $B=R/(g)$, and using the fact that $f$ and $g$ generate the maximal ideal of $R$ (by definition of transversal intersection), we get the vanishing we want.

Finally, if $p$ is a point that doesn't lie on both $C$ and $D$, then the stalk of either $\mathcal O_C$ or $\mathcal O_D$ will be zero at $p$, so again the stalk of the Tor sheaf will vanish there.

  • Dear Asal, I don't understand the nullity of $\operatorname{Tor}_1^R(R/(f),B)$ for $B=R/(g)$: why does $fr\in gR$ imply $r\in g$ under the hypothesis that $f,g$ generate the maximal ideal ? – Georges Elencwajg Nov 20 '14 at 10:04
  • @GeorgesElencwajg: the missing ingredient is that $X$ is nonsingular (a standing assumption at this point in Hartshorne, although the OP didn't day it.) Hence $R$ is a UFD, so irreducible elements (in particular $g$) are prime. So if $g$ divides $fr$ then $g$ divides $f$ or $r$. But if $g$ divides $f$ then $\langle f,g\rangle = gR$, contradicting maximality. –  Nov 20 '14 at 12:11
  • Dear Asal, thank you for your very clear explanation . – Georges Elencwajg Nov 20 '14 at 13:13