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Let $C$ be an elliptic curve in $\mathbb{P}^3$. By Riemann-Roch we have $\ell(D)=\deg D$ for a divisor $D$, since the genus $g$ equals 1. So if $a$ is a point on the curve, then the complete linear system of $D=1\cdot a$ will only contain $D$ itself. Apart from Riemann-Roch, one can decude this by translating rational functions on $C$ to elliptic functions on the corresponding torus. And one sees that such elliptic functions can't have poles of order one, and so again the complete linear system of $D=1\cdot a$ just contains itself. I've learned most of this watching Borcherds' youtube series.

My question is, is there a more direct or elementary way of understanding this fact about the complete linear system of $1\cdot a$? Let's consider a rational function $f/g$ with $\deg f=\deg g$. Why is it not possible to choose $g$ to have a zero at $a$, $f$ to be non-zero at $a$ and that $g$ and $f$ otherwise have the same zeros on $C$?

Cellardoor
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  • Because then $f/g$ is a rational function with a single zero and a single pole, implying it defines an isomorphism to $\Bbb P^1$. But elliptic curves are not isomorphic to the projective line. – Hank Scorpio Aug 09 '22 at 12:55
  • Does this answer your question? Show that function gives a bijection from curve to projective line but $\textrm{div}(f) = [P] - [Q] \implies P = Q$? – Hank Scorpio Aug 09 '22 at 12:56
  • The usual definition of elliptic curve is a nonsingular projective curve of genus 1; the "of genus 1" part presumes Riemann-Roch. So you would have to specify what your definition of elliptic curve is in order for your question to make sense. If you define an elliptic curve using a Weierstrass equation, then your question is answered at https://math.stackexchange.com/questions/408115/is-there-a-more-elementary-proof-of-this-special-case-of-riemann-roch/40854 – djao Aug 09 '22 at 13:10
  • Thank you, what is still missing is why $f/g$ must be injective. What about $f(x)/g(x)=f(y)/g(y)$ for $x\neq y$ guarantees that $f/g$ cannot be rational with a single zero and single pole? Again, I'd appreciate an elementary approach. – Cellardoor Aug 10 '22 at 09:43

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