# Is the following definition of an elliptic curve correct?

Im new to algebraic geometry so I want to make sure im getting my definitions right. I know there are a few ways to state what an elliptic curve is (ex a smooth projective curve of genus one with distinguished $$K$$-rational point). But I am wondering if the following is equivalent. For simplicity lets just work over $$\mathbb{C}$$.:

$$\textbf{Definition:}$$ An elliptic curve $$E$$ is a non-singular projective curve in $$\mathbb{P}^2$$ of the form $$E: Y^2Z + a_1XYZ + a_3YZ^2 = X^3 + a_2X^2Z + a_4XZ^2 + a_6Z^3$$

I am wondering if this is sufficient to define an elliptic curve?

• Yes, this is enough. The equivalence is shown in Silverman "The arithmetic of ellipic curves" chapter 3. Your first definition needs the added statement "with a fixed $K$-rational point" Jan 21, 2021 at 1:02
• A truly pedantic quibble, you should say "in" $\mathbb{P}^2$ and not over $\mathbb{P}^2$, which means something else in algebraic geometry. Jan 21, 2021 at 17:19

This is subtle. It is true that every elliptic curve can be written in this form, and that every smooth projective curve of this form has genus $$1$$.

But an elliptic curve is not a smooth projective curve of genus $$1$$. An elliptic curve (over a field $$K$$) is a smooth projective curve of genus $$1$$ together with a distinguished $$K$$-rational point, and this is key to the theory because it's the point that forms the identity for the group law. You need to say something about what this distinguished point is: with an equation of this form it's conventional to take it to be the point at infinity, with coordinates $$(X : Y : Z) = (0 : 1 : 0)$$. But this needs to be said explicitly in the definition; it is in fact possible to pick another point, which changes the group law. This is important for the following reasons among others:

1. There are smooth projective curves of genus $$1$$ over non-algebraically closed fields $$K$$ which have no $$K$$-rational points, and hence there is no choice of point which turns them into elliptic curves.

2. Elliptic curves have endomorphisms and automorphisms, and these are required to preserve the distinguished point (which turns out to imply that they preserve the group law). If you don't keep this in mind you will be confused when you read statements about endomorphisms and automorphisms of elliptic curves in the literature (which in fact happened recently on MO). For example, endomorphisms of an elliptic curve form a ring, but only if you require that endomorphisms preserve the distinguished point.

• It's worth pointing out that by the rigidity lemma, the isomorphism class of elliptic curve doesn't actually depend on the chosen base point, assuming one exists. Namely, if $C$ is a smooth connected curve of genus $1$ over $k$ and $p_1,p_2$ in $C(k)$ then there exists unique group multiplications $\mu_i$ on $C$ such that $p_i$ is the identity element. But, the map $C\to C:x\mapsto x-p_1+p_2$ is an isomorphism between them. Jan 21, 2021 at 5:34
• @Alex: thanks. Can you also comment on smooth projective curves of genus $1$ with no $K$-rational points? I'm guessing the correct statement here is that they are all torsors over elliptic curves, and even more specifically torsors over their Jacobians, but I wasn't able to track down a resource saying this explicitly. (Maybe it just follows straightforwardly from Galois descent?) Jan 21, 2021 at 5:39
• Since quasi-projective varieties are a stack for the etale topology one knows that twists of an elliptic curve $(E,0)$, thought of as an undecorated curve, can be identified with $\mathrm{Aut}(E)$-torsors. But, $\mathrm{Aut}(E)=E\rtimes \mathrm{Aut}(E,0)$ and so we see that one has an inclusion of $E$-torsors into $\mathrm{Aut}(E)$-torsors=twists of $E$. But, since $E$ is itself nice (e.g. smooth and proper) one knows from abstract nonsense that $E$-torsors correspond to $E$-bundles--curves $C$ with a simply transitive action of $E$ as a scheme. Since such $C$ are evidently twists of $E$ Jan 21, 2021 at 6:05
• one obtains the natural inclusion of $E$-torsors into twists for $E$. One can easily see that if $C$ is a smooth proper geometrically connected curve with no rational point, then it's a twist of $\mathrm{Pic}^0(C)$ and, in fact, can be easily seen to be a $\mathrm{Pic}^0(C)$-bundle. This is all covered, in less fancy language, in Section X.2 of Silverman's first book. Jan 21, 2021 at 6:08
• I'm glad this got accepted now. It's a much better answer than mine :-). Jan 21, 2021 at 17:18

Yes, this will do. The abstract definition of an elliptic curve is that it is nonsingular of genus $$1$$ as you mentioned. If your curve is cut out by a degree $$d$$ equation in $$\Bbb{P}^2$$, the degree-genus formula gives $$g=\frac{(d-1)(d-2)}{2}.$$ Hence, since you have a cubic $$C$$, $$g(C)=\frac{2}{2}=1$$. Conversely, you can show that any abstract genus $$1$$ curve over an algebraically closed field $$k$$ has an embedding into projective space $$\Bbb{P}^2$$ which realizes it as a degree $$3$$ curve.