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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?

Henry Swanson
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    I like this question. I can think of a couple of proofs/reasons, but all of them are much more powerful than "should" be necessary. I don't expect your solution method for $y^2 = x^3 - x$ will work for a general elliptic curve. Note, for example, that there are elliptic curves with infinitely many rational points ( $y^2 = x^3 - x$ is not one of them ). I think Andrea's proof below is probably the first one that most geometer's will think of, but the Weierstrass' theory is not obvious ( to me at least! ). – Callus - Reinstate Monica Jun 09 '14 at 09:02
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    I'm wrong above, your proof does generalize in a really neat way! See Corollary 1.8 in these notes: http://math.mit.edu/~ebelmont/ec-notes.pdf – Callus - Reinstate Monica Jun 09 '14 at 09:17
  • Oh wow, that is really neat! I didn't expect such a nice change of coordinates. And if I'm not mistaken, it also explains why $y^2 = x^3$ is an exception: it has a multiple root. Thanks! – Henry Swanson Jun 10 '14 at 00:31
  • +1. It's not important, but there is some conflict between $x$'s and $t$'s in the first paragraph. –  Jun 10 '14 at 07:33
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    Dear Henry, There are some related questions posted here. E.g. this one, this one, and this one. The last one in particular is directly related to your question (although perhaps not quite obviously). Regards, – Matt E Jun 10 '14 at 11:19

2 Answers2

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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.

Andrea Mori
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    Yeah, it is isomorphic to that, but algebraic extensions of purely transcendental extensions can still be purely transcendental. Consider $\mathbb{Q}(t)[X]/(X^2 - t) \cong \mathbb{Q}(t, \sqrt{t})$, which has a transcendence base of $\sqrt t$, which generates the field. As for the proof, that's the kind of thing I'm having a lot of trouble following. In particular, I was looking for something more algebraic (mostly because I've never run into the Weierstrass $\mathfrak P$ before, and while I know what tori, $\mathbb{P}^1$, and $S^2$ are, I don't know how to manipulate them in proofs yet). – Henry Swanson Jun 09 '14 at 09:00
  • Oh, ok, I see what the point of the question was. I don't think that you can prove that the generic plane cubic has no rational parametrization by purely algebraic methods. Maybe the specific curve you're dealing with may be attacked via algebra as your eventually-Fermat strategy shows. – Andrea Mori Jun 09 '14 at 09:17
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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

Jo Wehler
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