Background: For our abstract algebra class, we were asked to prove that $\mathbb{Q}(t, \sqrt{t^3 - t})$ is not purely transcendental. It clearly has transcendence degree $1$, so if it is purely transcendental, there is a transcendental $u$ and rational functions $f$ and $g$ such that $f(u) = t$ and $g(u) = \sqrt{t^3 - t}$. Therefore, $f(u)^3 - f(u) = g(u)^2$. But since $u$ is transcendental, $f(x)^3 - f(x) = g(x)^2$ as polynomials. This would be a rational parameterization of the elliptic curve $y^2 = x^3 - x$.
Since I'm not very familiar with elliptic curves, I couldn't show directly that such a parameterization cannot exist. So I showed that it would give rise to an integer solution to $pq(p + q)(p - q) = r^2$. The pathway to get there is really neat, but long, so unless someone asks, I'll omit it. Using a vaguely geometric argument from Fermat, I showed there are no integer solutions.
But this was a very 1) lengthy 2) tricky-to-motivate 3) bizarre proof, and it would have been much easier if I could have proved that elliptic curves do not admit a rational parameterization. Internet searches have mentioned all sorts of things about the topology of the curve, and parameterizations in the Weierstrass $\wp$ function, but they seemed to take the fact for granted, because I never saw a proof.
Could anyone show me a proof for this statement? Also, $y^2 = x^3 + 0x + 0$ does have a rational parameterization; is this some kind of degenerate case that can be kicked out?
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.