how is a cubic curve topologically equivalent to a torus? Can someone explain this in simple terms?
What I don't get is, how can a 1-D curve be topologically equivalent to a 2-D surface?  I guess isomorphism cannot hold between the 2 sets, and I don't see the homeomorphism.
 A: If your curve is given by a polynomial $p(x,y)=0$, you need to consider all complex $x,y$ satisfying the equation (so it's not surprising that you're getting something 2-dimensional), and you also need to add points at infinity to get a closed surface.
If the equation if $y^2=q(x)$, where $q$ is a cubic polynomial: for every $x\in\mathbb C$ s.t. $q(x)\neq0$ you have 2 solutions for $y$, and if $q(x)=0$ then you have just one solution (namely $y=0$). So you can imagine the complex cubic curve as lying over the plane of (complex) $x$-es, in two layers everywhere, except for the 3 roots of $q$. There is just one "point at infinity" to be added, we can call it $x=\infty$, $y=\infty$. Now we see the surface as lying over the (Riemann) sphere of $x$-es. 
To understand that the surface is actually a torus requires some work: we slit the $x$-sphere along two cuts, one connecting 2 roots of $q$ and the second connecting the third root with $\infty$. As we remove these arcs, the surface above splits to two spheres with the same cuts. To get the surface back in whole, we need to re-glue the cuts: imagine two spheres with two holes each, when we glue the corresponding holes, we get a torus:

A: An elliptic curve is a $1$-dimensional smooth variety of genus $1$ over some field. If the field is $\mathbb{C}$, the complex dimension is $1$, hence the real dimension is $2$. Using Weierstrass's elliptic functions one can prove that the group of $\mathbb{C}$-valued points of a complex elliptic curve is a torus, i.e. isomorphic to $\mathbb{C}/\Lambda$ for some lattice $\Lambda$.
