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?
 A: 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:

*

*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.


*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.
A: 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.
