Numerically Trivial Invertible Sheaf

Question

Vakil's Algebraic Geometry gives the following definition of numerically trivial invertible sheaf:

Suppose $X$ is a proper k-variety, and $\mathscr{L}$ is an invertible sheaf on $X .$ If $i: C \hookrightarrow X$ is a one-dimensional closed reduced subscheme of $X$, define the degree of $\mathscr{L}$ on $C$ by

$$ \deg_{C} \mathscr{L}:=\deg_{C} i^{*} \mathscr{L} $$

If $\deg_{C} \mathscr{L}=0$ for all $C,$ we say that $\mathscr{L}$ is numerically trivial.

Then there is one exercise:

Show that $\mathscr{L}$ is numerically trivial if and only if $\deg_{C} \mathscr{L}=0$ for all integral curves $C$ in $X$.

One part is trivial. For the other part, I want to solve it as followings:

First suppose $C$ has two irreducible components $C_{1}$ and $C_{2}$. Suppose we have $i_{1}: C_{1} \hookrightarrow C$ and $i_{2}: C_{2} \hookrightarrow C$. Then I want to prove that $\deg_{C} \mathscr{L}$=$\deg_{C_{1}} \mathscr{L}+\deg_{C_{2}} \mathscr{L}$. Since $\deg_{C} \mathscr{L}=\deg_{C} i^{*} \mathscr{L}=\chi(C,i^{*}\mathscr{L})-\chi(C,\mathscr{O}_{C})$, $\deg_{C_{1}} \mathscr{L}=\deg_{C_{1}} i_{1}^{*}i^{*} \mathscr{L}=\chi(C_{1},i_{1}^{*}i^{*}\mathscr{L})-\chi(C_{1},\mathscr{O}_{C_{1}})$, $\deg_{C_{2}} \mathscr{L}=\deg_{C_{2}} i_{2}^{*}i^{*} \mathscr{L}=\chi(C_{2},i_{2}^{*}i^{*}\mathscr{L})-\chi(C_{2},\mathscr{O}_{C_{2}})$,

I want to prove that $\chi(C,i^{*}\mathscr{L})$=$\chi(C_{1},i_{1}^{*}i^{*}\mathscr{L})$+$\chi(C_{2},i_{2}^{*}i^{*}\mathscr{L})$, and $\chi(C,\mathscr{O}_{C})=\chi(C,\mathscr{O}_{C_{1}})+\chi(C,\mathscr{O}_{C_{2}})$.

But I do not know how to prove it. Could you help me figure it out? Thank you very much.

Answer 1

Let's take a look at the morphism $f:X'\to X$, where $X'=C_1\sqcup C_2$ and $X=C_1\cup C_2$. Our goal is to prove that $\deg(f^*\mathcal{L})=\deg(\mathcal{L})$. In fact, we'll prove something stronger: degree is preserved by pullback along any birational morphism of proper curves over a field.

Here are some preliminary results:

Lemma 1: Let $f:X\to Y$ be an affine morphism. Then for any quasicoherent $\mathcal{O}_X$-module $\mathcal{F}$, we have $H^i(X,\mathcal{F})=H^i(Y,f_*\mathcal{F})$ based on the vanishing of the higher direct image and the Leray spectral sequence.

Lemma 2 (Projection Formula): For a morphism $f:X\to Y$, an $\mathcal{O}_X$-module $\mathcal{F}$ and a finite locally free $\mathcal{O}_Y$-module $\mathcal{E}$, we have $$\mathcal{E}\otimes_{\mathcal{O}_Y} R^pf_*\mathcal{F} \cong R^pf_*(f^*\mathcal{E}\otimes_{\mathcal{O}_X} \mathcal{F}).$$

Lemma 3: Euler characteristic adds over exact sequences.

Lemma 4: Euler characteristic for sheaves $\mathcal{F}$ with support of dimension zero multiplies by $\operatorname{rank}\mathcal{E}$ when you tensor $\mathcal{F}$ by a locally free sheaf $\mathcal{E}$.

First, we notice that $f$ is quasifinite and proper, thus finite and therefore affine. By lemma 1, we get that $$ \chi(X,\mathcal{L})-\chi(X',f^*\mathcal{L}) = \chi(X,\mathcal{L}) - \chi(X,f_*f^*\mathcal{L}).$$

By the projection formula, the RHS is equal to $$\chi(X,\mathcal{L}) - \chi(X,\mathcal{L} \otimes f_*\mathcal{O}_{X'}).$$

Next, complete the natural morphism $\mathcal{O}_{X}\to f_*\mathcal{O}_{X'}$ to an exact sequence of quasicoherent $\mathcal{O}_X$ modules: $$ 0\to \mathcal{K}\to \mathcal{O}_X\to f_*\mathcal{O}_{X'} \to \mathcal{C}\to 0,$$ where $\mathcal{K}$ and $\mathcal{C}$ are supported on a finite collection of points. Since $\mathcal{L}$ is locally free, tensoring with it is an exact functor, so$$ 0\to \mathcal{K}\otimes\mathcal{L}\to \mathcal{L}\to f_*\mathcal{O}_{X'}\otimes\mathcal{L} \to \mathcal{C}\otimes\mathcal{L}\to 0$$ is also exact.

By lemma 3 applied to this exact sequence, we see that $$\chi(X,\mathcal{L}) - \chi(X,\mathcal{L} \otimes f_*\mathcal{O}_{X'}) = \chi(X,\mathcal{K}\otimes\mathcal{L}) - \chi(X,\mathcal{C}\otimes\mathcal{L})$$ and then applying lemma 4, we get $$ \chi(X,\mathcal{K}\otimes\mathcal{L}) - \chi(X,\mathcal{C}\otimes\mathcal{L}) = (\operatorname{rank}\mathcal{L})\chi(X,\mathcal{K}) - (\operatorname{rank}\mathcal{L})\chi(X,\mathcal{C}).$$ Using additivity again, we see $$(\operatorname{rank}\mathcal{L})\chi(X,\mathcal{K}) - (\operatorname{rank}\mathcal{L})\chi(X,\mathcal{C}) = (\operatorname{rank}\mathcal{L})\chi(X,\mathcal{O}_X) - (\operatorname{rank}\mathcal{L})\chi(X,f_*\mathcal{O}_{X'}).$$

Putting it all together, we have $$\chi(X,\mathcal{L}) - \chi(X',f^*\mathcal{L}) = (\operatorname{rank}\mathcal{L})\chi(X,\mathcal{O}_X) - (\operatorname{rank}\mathcal{L})\chi(X,f_*\mathcal{O}_{X'})$$ and after swapping around the terms with the negative signs, we see that this is exactly the statement that $\deg(\mathcal{L})=\deg(f^*\mathcal{L})$.