5

I've been trying to make sense of the exact sequence in Lemma 1.3 chapter 5.

The Lemma is the following:

Let $C$ be a smooth irreducible curve on a smooth projective surface X, and let $D$ be any curve meeting $C$ transversally. Then $$\# (C \cap D) = \text{deg}_C(\mathcal{L}(D) \otimes \mathcal{O}_C)$$

Hartshorne claims the result is deduced from the exact sequence $$0 \to \mathcal{L}(-D) \otimes \mathcal{O}_C \to \mathcal{O}_C \to \mathcal{O}_{C \cap D} \to 0.$$

I'm trying to get the details down for how he gets this exact sequence. Let $i: D \to X$ be the inclusion map. Then we get the exact sequence $$0 \to \mathcal{L}(-D) \to \mathcal{O}_X \to i_*\mathcal{O}_D \to 0.$$

Let $j: C \to X$ be another inclusion map. Then by tensoring with $j_*\mathcal{O}_C$, we get $$0 \to \mathcal{L}(-D) \otimes j_*\mathcal{O}_C \to j_*\mathcal{O}_C \to i_*\mathcal{O}_D \otimes j_*\mathcal{O}_C \to 0.$$

The first map is injective since $C$ and $D$ intersect transversally, so this is indeed an exact sequence. In order to get the result, we need to apply $j^*$ to the exact sequence. Then we get $$0 \to j^*(\mathcal{L}(-D) \otimes j_*\mathcal{O}_C) \to \mathcal{O}_C \to j^*(i_*\mathcal{O}_D \otimes j_*\mathcal{O}_C) \to 0.$$

We can look at the stalks to see that this is exact as well.

If $p : C \times_X D \to C$ is the canonical map coming from the fiber product, how do I see that $j^*(i_*\mathcal{O}_D \otimes j_*\mathcal{O}_C) \cong p_*\mathcal{O}_{C \times_X D}$ and $j^*(\mathcal{L}(-D) \otimes j_*\mathcal{O}_C) \cong \mathcal{L}(-D) \otimes \mathcal{O}_C$? Does he mean that $\mathcal{L}(-D) \otimes \mathcal{O}_C = j^*\mathcal{L}(-D)$? If so, I see that the stalks are the same by just dealing with the presheaves, but is there some slick way to see this other than actually writing the maps down? Are there some universal property tricks that will yield the results?

Daniel Levine
  • 311

1 Answers1

2

Just take the exact sequence on $X$:

$$0 \to \mathcal L(-D) \to \mathcal O_X \to i_* \mathcal O_D \to 0$$

and pull it back along $j: C \hookrightarrow X$ to obtain (As you pointed out, exactness is preserved by assumption):

$$0 \to j^*\mathcal L(-D) \to \mathcal O_C \to j^*i_* \mathcal O_D \to 0.$$

Indeed we have $j^* \mathcal L(-D) = \mathcal L(-D) \otimes \mathcal O_C$.

Furthermore, we have the following Cartesian diagram:

\begin{array}{ccc} C \cap D& \xrightarrow{\hat j} & D \\[3pt] \downarrow {\hat i} & & \downarrow{i} \\ C& \xrightarrow{j} & X \end{array}

This shows that $j^*i_* \mathcal O_D = \hat i_*\hat j^*\mathcal O_D = \hat i_* \mathcal O_{C \cap D}$, i.e. we have the exact sequence on $C$:

$$0 \to \mathcal L(-D) \otimes \mathcal O_C \to \mathcal O_C \to \hat i_* \mathcal O_{C \cap D} \to 0.$$

As usual, the $\hat i_*$ is omitted, so it is written as

$$0 \to \mathcal L(-D) \otimes \mathcal O_C \to \mathcal O_C \to \mathcal O_{C \cap D} \to 0.$$

MooS
  • 31,390
  • Are those equalities true because all the maps are inclusions, or is this true when you have a commuting diagram in general? – Daniel Levine Mar 29 '16 at 16:17