Why does an elliptic curve have genus one?

Question

I read that one definition of an elliptic curve goes as follows: Let $k$ be a field. We define the elliptic curve over $k$ be a smooth projective curve $E$ over $k$, isomorphic to a closed subvariety of $\mathbb P_k^2$ defined by a homogeneous polynomial $F(u,v,w)$ of the form $$F(u,v,w)=v^2w+(a_1u+a_3w)vw-(u^3+a_2u^2w+a_4uw^2+a_6w^3),$$ where $a_i\in k$ for $k\in\{1,2,3,4,6\}$, with privileged rational point $o=(0,1,0)$. How can I prove from this definition that the arithmetic and geometric genus of any elliptic curve over a given field $k$ is 1 if $\operatorname{char}(k)\ne 2$? This problem was popped on my mind when I was reading Takeshi Saito's book Fermat's last theorem: Basic tools lemma 1.10. (2) as well as definition of an elliptic curve from Qing Liu's book Algebraic Geometry and Arithmetic curves.

Answer 1

If you define it to be a projective curve given by a degree 3 polynomial, you just need a formula linking the degree of a plane curve with the genus. This you can find:

http://en.wikipedia.org/wiki/Genus%E2%80%93degree_formula

(By smoothness, which you can check by hand, it doesn't matter if you want to refer to arithmetic or geometric genus).

It's a bit more natural to define it to be a (smooth projective) genus one curve with a marked point, and then prove every such object can be given an equation like the one you wrote, but it can be easier to define everything in terms of equations.