# Why does an elliptic curve have genus one?

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.

• It's better to define as a proper smooth geometrically connected curve of genus 1 and then embed in projective space. – user40276 Feb 23 '15 at 22:15