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Hartshorne, Algebraic Geometry, Chapter V, Lemma 1.3, reads (in part):

Throughout this chapter, a surface will mean a nonsingular projective surface over an algebraically closed field $k$. [...] Let $X$ be a surface.

[...]

Lemma 1.3. Let $C$ be an irreducible nonsingular curve on $X$ , and let $D$ be any curve meeting $C$ transversally. Then $$\#(C \cap D) = \deg_C(\mathscr{L}(D) \otimes \mathcal{O}_C).$$ Proof. [...] We use the fact that $\mathscr{L}(D)$ is the ideal sheaf of $D$ on $X$. Therefore, tensoring with $\mathcal{O}_X$, we have an exact sequence $$0 \to \mathscr{L}(-D) \otimes \mathcal{O}_C \to \mathcal{O}_C \to \mathcal{O}_{C \cap D} \to 0$$ where now $C \cap D$ denotes the scheme-theoretic intersection. [...]

So apparently we've taken the exact sequence $$0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X \to \mathcal{O}_D \to 0$$ of coherent sheaves on $X$, tensored by the coherent sheaf $\mathcal{O}_C$, and the resulting sequence is still exact.

Now, as I understand it, $\mathcal{O}_C$ is not flat: Proposition III.9.2(e) says that a coherent sheaf on a noetherian scheme is flat iff it is locally free, and $\mathcal{O}_C$ is clearly not locally free since it's supported on a curve. (Of course I guess I don't actually need flatness here.)

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  • $\begingroup$ I apologize if the below is not helpful! $\endgroup$ – Alex Youcis Nov 19 '14 at 6:37
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    $\begingroup$ I think there's another version of the same question here: math.stackexchange.com/questions/688247/… $\endgroup$ – user64687 Nov 19 '14 at 17:14
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Not that $$0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X \to \mathcal{O}_D \to 0$$ tensor by $\mathcal{O}_C$ is just

$$0\to {\mathcal Tor}_1( \mathcal{O}_D,\mathcal{O}_C)\to \mathscr{L}(-D) \otimes \mathcal{O}_C \to \mathcal{O}_C \to \mathcal{O}_{C \cap D} \to 0$$ since $\mathcal{O}_X$ is flat.

Now ${\mathcal Tor}_1( \mathcal{O}_D,\mathcal{O}_C)$ is a sheaf supported on ${C \cap D}$ and $\mathscr{L}(-D) \otimes \mathcal{O}_C$ is locally free on $C$.

Note that the subsheaf of a locally free sheaf can not be torsion, hence $${\mathcal Tor}_1( \mathcal{O}_D,\mathcal{O}_C)=0$$.

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Just to answer your question as posed, the sequence remains exact because $C$ and $D$ meet transversally, in particular, $C \cap D \subsetneq C$. In particular, the local equation for $D$ remains nonzero when restricted to $C$. Since $C$ is integral, the map $\mathcal{O}_X(-D) \to \mathcal{O}_X$ (multiplication by the local equation for $D$) remains injective after tensoring with $\mathcal{O}_C$.

If you want to see this algebraically: the sequence

$$0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X \to \mathcal{O}_D \to 0$$

corresponds (in a sufficiently small affine chart of $X$, say $X = \mathrm{Spec}(R)$, where $D$ has local equation $f$ and $C$ has local equation $g$,) to the sequence

$$0 \to R \to R \to R/(f) \to 0,$$

where the first map is multiplication by $f$. Tensoring with $C$ means tensoring with $R/g$, as you said:

$$0 \to R/(g) \to R/(g) \to R/(f,g) \to 0,$$

where the first map is again multiplication by $f$ and not, a priori, injective anymore. But by our hypothesis, $C \cap D \subsetneq C$, so the image of $f$ is not zero in $R/g$. Since $R/g$ is a domain, this means multiplication by $f$ is still injective.

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I actually was confused by large portions of V.I, in particular, this part. Hopefully someone will come by and tell us why we're being silly.

I'd like to give you an alternate path to proving the various basic theorems here, that I personally found more helpful. I would have made this a comment, but, unfortunately, it's too long.

Hopefully it's of some help for you, it's the only way I remember how the stuff in V.I works!


In the usual way, define the multiplicity of the intersection of two non-singular curves $C$ and $D$ on a surface $X$ at a point $x$ to be $m_x(C,D):=\ell_{\mathcal{O}_{X,x}}(\mathcal{O}_{X,x}/(f,g))$ where $f$ and $g$ are local equations defining $C$ and $D$ at $x$. Define then $\langle C,D\rangle$ to be $\displaystyle \sum_{x\in C\cap D}m_x(C,D)$. Clearly then $m_x(C,D)=\#(C\cap D)$ if $C$ and $D$ meet transversely. Moreover, it's clear that $m_x(C,D)=\dim_k H^0(X,\mathcal{O}_{C\cap D})$.

Now, for any two line bundles $\mathscr{L},\mathscr{L}'$ define $\langle\mathscr{L},\mathscr{L}'\rangle$ by

$$\chi_X(\mathcal{O}_X)-\chi_X(\mathscr{L}^{-1})-\chi_X(\mathscr{L}'^{-1})+\chi_X(\mathscr{L}^{-1}\otimes\mathscr{L}'^{-1})$$ I claim then that:

For $C$ and $D$ as above, $\langle C,D\rangle=\langle\mathcal{O}(C),\mathcal{O}(D)\rangle$.

To prove this, merely check that the following sequence is exact

$$0\to\mathcal{O}(-D-C)\to\mathcal{O}(-C)\oplus\mathcal{O}(-D)\to \mathcal{O}_X\to i_\ast\mathcal{O}_{C\cap D}\to 0$$

where $i:C\cap D\to X$ is the inclusion, and the first two maps are $x\mapsto s'x-sx$, and $(x,y)\mapsto sx+s'y$, where $s$ and $s'$ global sections of $\mathcal{O}(C)$ and $\mathcal{O}(D)$ respectively. This is easy to check affine locally.

Then, taking $\chi$ we get

$$\chi_X(i_\ast(\mathcal{O}_{C\cap D}))=\chi_X(\mathcal{O}_S)-\chi_X(\mathcal{O}(-D))-\chi_X(\mathcal{O}(-C))+\chi_X(\mathcal{O}(-C-D))$$ which gives the desired result after noting that

$$\chi_X(i_\ast\mathcal{O}_{C\cap D})=\chi_{C\cap D}(\mathcal{O}_{C\cap D})=H^0(X,\mathcal{O}_{C\cap D})$$ since $C\cap D$ is zero dimensional $\blacksquare$

Now, we want to prove that for a smooth curve $C\subseteq X$, and a line bundle $\mathscr{L}$ that $$\langle\mathcal{O}(C),\mathscr{L}\rangle=\deg(i^\ast\mathscr{L})$$ where $i$ is the embedding $C\to X$. Indeed, if this were true then, for two smooth transversely interesting curves $C$ and $D$ we have that

$$\#(C\cap D)=\langle C,D\rangle=\langle \mathcal{O}(C),\mathcal{O}(D)\rangle=\deg(i^\ast\mathcal{O}(D))$$ which is what you were after.

To prove this one just considers the closed subscheme SES $$0\to\mathcal{O}(-C)\to\mathcal{O}_X\to i_\ast\mathcal{O}_C\to 0$$ Tensor this with $\mathscr{L}$ to get $$0\to\mathcal{O}(-C)\otimes\mathscr{L}\to\mathscr{L}\to i_\ast\mathcal{O}_C\otimes\mathscr{L}\to 0$$ Taking $\chi_X$ gives $$\chi_X(\mathscr{L}^{-1})-\chi_X(\mathscr{L}^{-1}\otimes \mathcal{O}(-C))=\chi_C(i^\ast\mathscr{L}^{-1})$$ where I used the projection formula to see that $$\mathscr{L}^{-1}\otimes i_\ast\mathcal{O}_C=i_\ast(i^\ast\mathscr{L})$$ and the fact that $\chi_X(i_\ast(i^\ast\mathscr{L}^{-1}))=\chi_C(i^\ast\mathscr{L}^{-1})$.

So,

$$\begin{aligned}\langle\mathcal{O}(C),\mathscr{L}\rangle &=\chi_X(\mathcal{O}_X)-\chi_X(\mathscr{L}^{-1})-\chi_X(\mathcal{O}(-C))-\chi_X(\mathcal{O}(-C)\otimes\mathscr{L}^{-1})\\ &=\chi_C(\mathcal{O}_C)-\chi_C(i^\ast\mathscr{L}^{-1})\\ &=-\deg(\mathscr{L}^{-1})\\ &=\deg(\mathscr{L})\end{aligned}$$ where I have used "Riemman-Roch" (""=no need for Serre duality) at the obvious place.

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  • $\begingroup$ Dear Alex, very neat answer! :) $\endgroup$ – Brenin Nov 19 '14 at 17:23

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