2
$\begingroup$

This should not be hard, but I am stuck on it nonetheless, so I would much appreciate a solution.

Suppose $C$ is a projective non-singular curve of genus $g\geq 2$ and $P,Q$ are distinct points on $C$.

The notes of my algebraic geometry course now claim: "since $g\geq 2$, the dimension of the linear system $|P+Q|$ must be equal to 0 or 1."

Why is this?

$\endgroup$
5
  • $\begingroup$ Have you checked how far Riemann-Roch alone could take you? $\endgroup$ – Hoot May 12 '15 at 2:21
  • $\begingroup$ Yes. I don't see how you can use it to get an answer. $\endgroup$ – Misja May 12 '15 at 7:42
  • $\begingroup$ @DanieleA: the hyperelliptic case is dimension 1. ("Linear system dimension" is vector space dimension minus one.) $\endgroup$ – user64687 May 12 '15 at 15:04
  • $\begingroup$ @DanieleA: For a justification of the hyperelliptic case giving dimension 1, see the proof of prop. IV.5.2 in Hartshorne. $\endgroup$ – Misja May 12 '15 at 16:40
  • $\begingroup$ Ah ok, sorry, I thought it was the dimension of the vector space, not of the projective one, I will delete my comment. $\endgroup$ – Daniele A May 13 '15 at 9:25
1
$\begingroup$

Assertion:
If $D$ is an effective divisor of degree $d\geq 1$ on a smooth projective curve $C$ of genus $g\geq 1$, then $l(D)\leq d$.

The proof is by induction on $d$:

Initial step d=1
If $d=1$, then $D=1. P$ for some $P\in C$ and $L(1.P)=k$ (= base field) so that the assertion is true. [Here we have used that $g\geq 1$, i.e. that $C$ is not isomorphic to $\mathbb P^1$]

Inductive step $d-1\to d$
Choose some $P\in \operatorname {supp} D$ and consider the exact sequence of $k$-vector spaces $$0\to L(D-P)\to L(D)\to k$$ It immediately implies $l(D)\leq l(D-P)+1\leq (d-1)+1=d$

The assertion, which is much stronger than what you ask about, is thus proved.
Note carefully that this proof is from first principles and does definitely not involve grandiose concepts like Riemann-Roch, sheaf cohomology or Serre duality...

$\endgroup$
4
  • $\begingroup$ Very clear - thanks! $\endgroup$ – Misja Aug 1 '15 at 8:57
  • $\begingroup$ You are welcome, dear Misja (or Миша ?) $\endgroup$ – Georges Elencwajg Aug 1 '15 at 10:30
  • $\begingroup$ @GeorgesElencwajg Excuse me, why $L(1\cdot P)=k$? And which is the map $L(D)\rightarrow k$? (I am sorry if this is obvious, I am starting to study divisors) $\endgroup$ – H. Jackson Dec 16 '15 at 9:39
  • 1
    $\begingroup$ @H.Jackson. $L(1.P)=k$ because the only rational functions with at most a pole at $P$ are the constants: if there were a nonconstant rational function with a pole of order one at $P$ and regular elsewhere it would give an isomorphism with $\mathbb P^1$: contradiction with $g\geq 1$. The map $L(D)\to k$ is for $f\in L(D)$ evaluation at $P$ of $f(z)\cdot z^{D(P)}$, where $z$ is a uniformizing parameter (=local coordinate). This map is unfortunately non canonical. And no need to apologize: your questions are not trivial at all! $\endgroup$ – Georges Elencwajg Dec 16 '15 at 10:31
2
$\begingroup$

I think that you can reason like this: you know in general that (in Hartshorne notation) $\ell(P+Q)\leq \deg(P+Q)+1 = 3$ so that you just have to rule out the case $\ell(P+Q)=3$. Suppose that this is the case, then $\ell(P)\geq \ell(P+Q)-1 \geq 2$, so that by the previous inequality, $\ell(P)=2$. But since this divisor has degree $1$, the induced map to $\mathbb{P}^1$ is an isomorphism.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.