Genus of the product of two elliptic curves

In trying to understand the trichotomy of the genus of algebraic curves, I first consider the following two elliptic curves (over $\mathbb{Q}$), well-known to be of rank $2$,

$y^2 = x^3+17$ and $y^2 = x^3+15$.

Then I define the curve $C$ defined by $(y^2z - x^3-17z^3)(y^2z - x^3-15z^3) = 0$. Now the rational points on $C$ obviously is the union of the points of the two curves above and according to Sage the genus of $C$ is $1$.

If it is correct that the genus is $1$ then I suppose that $C$ itself is an elliptic curve.

Can one find the Weierstrass form of $C$ or is there an error in the reasoning?

• The zero locus of that product polynomial is not irreducible, i.e. not a variety. It most certainly is not an elliptic curve, but rather the union of the two elliptic curves. I don't know how the genus of a reducible curve is defined, so I cannot comment on how Sage handles this. Anyway, $C$ is not an elliptic curve, and therefore has no Weierstrass form. – Jyrki Lahtonen Apr 29 '15 at 7:03
• Also, this is not what a person would typically mean by "the product of two elliptic curves." If $E_1$ and $E_2$ are two elliptic curves then this should mean the product variety $E_1 \times E_2$, which is not a curve but a surface. – Qiaochu Yuan Apr 29 '15 at 7:26
• Thanks for both answers. I see that I forgot about reducibility. And that a better terminology would be the union of two elliptic curves. – Jesper Petersen Apr 29 '15 at 7:50
• (You have a typo: your two factors have $z^2$ where they should have $z^3$.) There's a perfectly good definition of genus for any connected 1-dimensional scheme, namely the dimension of $H^1(\mathcal O_X)$. This is constant in flat families, so your curve must have the same genus as a smooth plane sextic, namely $(6-1)(6-2)/2=10$. I don't know if you made an input error or if Sage uses a definition I am not familiar with (or has been on the sauce), but the answer 1 seems wildly off to me. – user64687 Apr 29 '15 at 12:55