The Stacks project

Lemma 33.44.12. Let $k$ be a field. Let $X$ be a proper curve over $k$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module.

  1. If $\mathcal{L}$ has a nonzero section, then $\deg (\mathcal{L}) \geq 0$.

  2. If $\mathcal{L}$ has a nonzero section $s$ which vanishes at a point, then $\deg (\mathcal{L}) > 0$.

  3. If $\mathcal{L}$ and $\mathcal{L}^{-1}$ have nonzero sections, then $\mathcal{L} \cong \mathcal{O}_ X$.

  4. If $\deg (\mathcal{L}) \leq 0$ and $\mathcal{L}$ has a nonzero section, then $\mathcal{L} \cong \mathcal{O}_ X$.

  5. If $\mathcal{N} \to \mathcal{L}$ is a nonzero map of invertible $\mathcal{O}_ X$-modules, then $\deg (\mathcal{L}) \geq \deg (\mathcal{N})$ and if equality holds then it is an isomorphism.

Proof. Let $s$ be a nonzero section of $\mathcal{L}$. Since $X$ is a curve, we see that $s$ is a regular section. Hence there is an effective Cartier divisor $D \subset X$ and an isomorphism $\mathcal{L} \to \mathcal{O}_ X(D)$ mapping $s$ the canonical section $1$ of $\mathcal{O}_ X(D)$, see Divisors, Lemma 31.14.10. Then $\deg (\mathcal{L}) = \deg (D)$ by Lemma 33.44.9. As $\deg (D) \geq 0$ and $= 0$ if and only if $D = \emptyset $, this proves (1) and (2). In case (3) we see that $\deg (\mathcal{L}) = 0$ and $D = \emptyset $. Similarly for (4). To see (5) apply (1) and (4) to the invertible sheaf

\[ \mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{N}^{\otimes -1} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{N}, \mathcal{L}) \]

which has degree $\deg (\mathcal{L}) - \deg (\mathcal{N})$ by Lemma 33.44.7. $\square$

Comments (2)

Comment #8769 by Paola on

In order to prove (5), if you apply (4) as it is stated, I think you only get that and are isomorphic, not that the given morphism is an isomorphism. However, from the proof of (4) one sees that the following stronger claim holds: if and is a nonzero section of , then the morphism mapping to is an isomorphism. Am I right?

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  • 4 comment(s) on Section 33.44: Degrees on curves

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