Why can't elliptic curves be parameterized with rational functions?

Solution 1:

Isn't $\Bbb Q(t,\sqrt{t^3-t})$ just isomorphic to $$ \Bbb Q(t)[X]/(X^2-t^3+t), $$ so it is an algebraic extension of a purely transcendental field of transcendence degree 1?

Anyway, if $E$ is an elliptic curve (non-singuar complex plane cubic) Weierstrass' theory shows that $$ E\simeq\Bbb C/\Lambda $$ as complex variety, where $\Lambda\subset\Bbb C$ is a lattice, i.e. a discrete subgroup of maximal rank ($=2$). Thus the set of complex points of $E$ is a torus, a topological space with non-trivial fundamental group.

On the other hand, any smooth algebraic curve admitting a rational parametrization is isomorphic to the projective line $\Bbb P^1$ and the set of complex points $\Bbb P^1(\Bbb C)$ (the projective complex line) is topologically equivalent to the sphere $S^2$ which has trivial fundamental group.

Solution 2:

Assume an elliptic curve $E$ has a rational parametrization by a holomorphic map

$$f: \mathbb P^1 \longrightarrow E.$$

The formula of Riemann-Hurwitz implies

$$g(\mathbb P^1) = b/2 + (deg \ f)(g(E) -1) + 1$$

with genus $g(\mathbb P^1) = 0$, $g(E) = 1$ and $b \ge 0$ the total branching order of $f$, a contradiction, q.e.d