(From Vakil's notes, Exercise 18.4.K) If $C$ is an integral projective curve over a field $k$, and $\mathscr{L}$ is an ample line bundle on $C$, why is the degree of $\mathscr{L}>0$?

If $C$ is also a regular curve, then I think I can do this, as it can be quickly checked that ${\rm deg}(\mathscr{L}^{\otimes n}) = n {\rm deg}(\mathscr{L})$ and, if $\mathscr{L}$ is very ample, there exists a global section (and that section must vanish somewhere or else our very ample bundle is trivial).

However, I don't see how to extend this argument to the case where the curve might not be regular. The degree of a coherent sheaf is defined in the notes as \begin{align*} {\rm deg}(\mathscr{F}):=\chi(C,\mathscr{F}) - ({\rm rank} \mathscr{F})\chi(C,\mathscr{O}_C) \end{align*}

As one possible useful fact, Exercise 18.4.S seems to say that we can express $\mathscr{L}$ as $\mathscr{O}_C(\sum{n_ip_i})$ where $p_i$ are regular points that are not associated points. I don't know why this is true, but it would at least allow us to view $\mathscr{L}$ more concretely (in particular, repeating the argument for Riemann Roch for regular projective curves tells us the degree of $\mathscr{L}$). However, even assuming this fact, I would need to know be able to compute the degree of $\mathscr{L}$ using the zeros and poles of a global section to use the argument above.

  • $\begingroup$ I was wondering about this as well, but there is probably a reason why the exercise appeared right after he introduced the definition of degree for a curve that is not regular. $\endgroup$
    – DCT
    Jun 18, 2014 at 20:17

2 Answers 2


The result and your argument (which is the first I would think of) are still valid for integral projective curves, but the theory is more delicate. For example, you can find the result as 7.3.2 Proposition 3.25(b) of Liu's wonderful book, in which he develops Riemann-Roch results with minimal assumptions on what constitutes a "curve" (e.g. possibly singular, nonreduced).

  • $\begingroup$ All right thanks (upvote). I'll accept this once I understand it fully. It might be a while because I'll be away and because I packed Liu's book in storage where I can't get it... $\endgroup$
    – DCT
    Jun 19, 2014 at 4:46
  • $\begingroup$ Dear @Dtseng, you're welcome. I'm glad to hear that you own a copy of the book, so can look up the reference (somewhat) easily! $\endgroup$
    – Andrew
    Jun 19, 2014 at 4:54
  • 1
    $\begingroup$ Okay, understanding it fully was a bad term to use, but I tried to take a look. I think the gist is that we can use cartier divisors instead of weil divisors in Riemann-Roch because the idea of length works even if the stalks aren't DVRs. Given this, the proof of Riemann Roch isn't changed much. The argument seems to work in exactly the same way as above. The identity ${\rm deg}(\mathscr{L}^{\otimes n})= n{\rm deg}(\mathscr{L})$ still holds, and the fact that very ample bundles have of lots of sections (more than 1-dimensional) plus the proposition you cited finishes it. $\endgroup$
    – DCT
    Jun 19, 2014 at 6:25

In the newly updated Rising Sea (August 2022), this exercise is now 19.4.M. And in this version, there is a new exercise preceding it,

19.4.K. EXERCISE. If $\mathcal{F}$ is a torsion sheaf on $C$, show that $\deg(\mathcal{F})\geq 0$, with equality if and only $\mathcal{F}=0$.

Here $C$ is an integral projective curve. From this, one can deduce that if a line bundle $\mathcal{N}$ has a nonzero global section, and $\mathcal{N}$ is not trivial, then $\deg(\mathcal{N})>0$. This is done by examining the exact sequence $$0\to \mathcal{O}_C\to \mathcal{N}\to \mathcal{F}\to 0$$ where the first map is the one induced by the global section of $\mathcal{N}$, and $\mathcal{F}$ is defined to be the cokernel. In this section, it is shown that both rank and degree are additive on exact sequences. The first gives that the rank of $\mathcal{F}$ is $0$, and the second gives that the degrees of $\mathcal{F}$ and $\mathcal{N}$ are equal. By 19.4.K, using the fact that $\mathcal{N}$ is not trivial, we have $\deg(\mathcal{N})>0$.

This then shows that for $\mathcal{N}$ very ample, we have $\deg(\mathcal{N})>0$. But we want this for ample line bundles, not very ample line bundles. As is mentioned in the comments to Andrew's answer, $\deg(\mathcal{L}^{\otimes n})=n\deg(\mathcal{L})$ (even better, degree is additive across tensor products), even for singular curves. But it seems like this may take a lot of effort, so we will go around this slightly.

Suppose $\mathcal{L}$ is ample, with $\mathcal{L}^{\otimes n}$ very ample (or even just has basepoint-free) for $n\geq N$, and $\mathcal{M}$ is any line bundle. Set $\mathcal{N}:=\mathcal{L}^{\otimes n}$ for some $n\geq N$. Tensoring the above exact sequence with $\mathcal{M}$ gives $$0\to \mathcal{M} \to \mathcal{N}\otimes \mathcal{M}\to \mathcal{F}\otimes \mathcal{M}\cong \mathcal{F}\to 0$$ exact, from which we get $$\deg(\mathcal{N}\otimes \mathcal{M})=\deg(\mathcal{M})+\deg(\mathcal{F})=\deg(\mathcal{M})+\deg(\mathcal{N})$$

Plugging in $\mathcal{N}=\mathcal{L}^{\otimes n}$, and setting $\mathcal{M}=\mathcal{L}^{\otimes a}$, we see that $$\deg(\mathcal{L}^{\otimes n}\otimes \mathcal{L}^{\otimes a})=\deg(\mathcal{L}^{\otimes n})+\deg(\mathcal{L}^{\otimes a})$$ for $n>N$, and any $a\in\mathbb{Z}$. And then, for any $a,b\in\mathbb{Z}$, choosing $n$ so that $n,n+a\geq N$, we have

$$\deg(\mathcal{L}^{\otimes n})+\deg(\mathcal{L}^{\otimes a})+\deg(\mathcal{L}^{\otimes b})=\deg(\mathcal{L}^{\otimes (n+a+b)})=\deg(\mathcal{L}^{\otimes n})+\deg(\mathcal{L}^{\otimes (a+b)})$$ which implies $\deg(\mathcal{L}^{\otimes a})+\deg(\mathcal{L}^{\otimes b})=\deg(\mathcal{L}^{\otimes (a+b)})$, for any $a,b\in\mathbb{Z}$, and so $0<\deg(\mathcal{L}^{\otimes N})=N\deg(\mathcal{L})\implies \deg(\mathcal{L})>0$.

  • $\begingroup$ I have realized that my argument at the end can extended to show that degree is additive for all vector bundles on projective curves, using that every line bundle is the difference of very ample line bundles. $\endgroup$ Jan 18 at 16:53

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