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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.

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    $\begingroup$ I don't know the tome in question but if my guess about the terms you use is correct then the following might work. Let $1,x$ span the sections of $\mathcal{O}(2p)$, and $1,x,y$ those of $\mathcal{O}(3p)$. Then $1,x,x^2,x^3,y,xy,y^2$ all belong to $\mathcal{O}(6p)$. But that has a six-dimensional space of sections, so those seven elements must be linearly dependent. The non-trivial linear dependency relation is your cubic equation. $\endgroup$ – Jyrki Lahtonen Dec 8 '14 at 18:36
  • $\begingroup$ Jyrki, this was what I was looking for. Eventually while poking around books, I found this in Silverman's book on elliptic curves. Thank you much. $\endgroup$ – user45150 Dec 9 '14 at 0:41
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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.

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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 $d=3$.

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  • $\begingroup$ Is this true for instance if we replace $k$ by a ring $A$? $\endgroup$ – user45150 Dec 7 '14 at 22:22
  • $\begingroup$ I am pretty confident the answer is no, so excepting that the answer is actually no, what do we need about $\mathbb{P}^n_A$ for this to be true? $\endgroup$ – user45150 Dec 7 '14 at 22:23
  • $\begingroup$ @user45150 What do you propose to call the genus of a curve over something that isn't a field? If $X\to S$ is an $S$-scheme, then I think of genus as a function taking some point $s$ of $S$ to the genus of the curve $X_s$ over $k(s)$. This function is constant if $X\to S$ has nice properties, but various things can go wrong if it doesn't. To take a simple example, if $A=k\times k$, then $S=\operatorname{Spec} A$ is a discrete space with two points, and a curve over $S$ is just two curves over $k$, with no condition on their genera. $\endgroup$ – Slade Dec 7 '14 at 23:26
  • $\begingroup$ @user45150 If we are discussing elliptic curves, we are probably interested the case where $A$ is a Dedekind domain, or at least an integral domain. In this case, we can ask for the genus at the generic point, but I don't know offhand exactly what happens to the divisor argument when we are no longer working over a field. $\endgroup$ – Slade Dec 7 '14 at 23:34

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