Elliptic curves as cubics as discussed in Ravi Vakil's notes

Question

I was reading the section of Ravi Vakil's Algebraic Geometry notes where he discusses elliptic curves.

If we let an elliptic curve be $(E,p)$ (Where $p$ is the distinguished point), we have $\mathcal{O}(3p)$, which has $3$ sections, which gives us a closed embedding into $\mathbb{P}^2$. In the next sentence Ravi Vakil writes

Thus we have a closed embedding $E\hookrightarrow \mathbb{P}^2_k$ as a cubic curve.

My question is how does he know that it must be cubic, and what does this depend on (like in what way if any does it depend on the fact that $k$ is a field)?

Thank you for any help or advice.

Answer 1

The degree of the line bundle is $3$. This means that a generic section of $\mathcal O(3p)$ has $3$ zeroes. This means that a generic line in $\mathbb P^2$ intersects the image of the curve in $3$ points, which means the image has degree $3$.

This kind of computation is part of the basic package related to using line bundles to embed varieties into projective space, and if you're not familiar with it, I would suggest spending some time to make sure you understand it. If you read some of Hartshorne Ch. IV you can find many examples and applications.

Answer 2

A nonsingular curve $C\subset\mathbb{P}^2$ of degree $d$ always has genus $g = \frac{(d-1)(d-2)}{2}$ (this can be computed explicitly), so $g=1$ implies $d=3$.