7
$\begingroup$

See David Speyer's answer here.

I saw Brian Conrad give an excellent one hour talk to undergraduates where he proved that there do not exist nonconstant, relatively prime, polynomials $a(t)$, $b(t)$ and $c(t) \in \mathbb{C}[t]$ such that $$a(t)^3 + b(t)^3 = c(t)^3.$$ He gave an elementary proof, then outlined the better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve.

Could anyone expand a bit on the "outlined the better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve" part? That's not so clear to me. Thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ AFAICT this follows for the ABC theorem for polynomials. Yours will contradict it. $\endgroup$ – joro Nov 21 '15 at 15:07
  • $\begingroup$ You are right, and both proofs of "Fermat-3 for polynomials" generalize to Mason's theorem (= polynomial ABC). $\endgroup$ – Noam D. Elkies Nov 21 '15 at 15:23
  • 1
    $\begingroup$ Check the answers here: mathoverflow.net/questions/116583/… Can FLT fail with a parametrization over some extension of Z? $\endgroup$ – joro Nov 21 '15 at 15:37
12
$\begingroup$

(With corrections noted by @GH from MO): The map $t\to (a(t),b(t),c(t))$, if at least one of the ratios $a(t)/c(t)$ or $b(t)/c(t)$ is non-constant, would extend to a non-constant map from a lower-genus (compact connected) curve ($\mathbb P^1$) to a higher-genus such, the elliptic curve defined by $a^3+b^3=c^3$. (Maybe open mapping easily shows surjectivity in the complex case, for example.) Impossible, by Riemann-Hurwitz formula.

Also, in the complex case, since elliptic curves are quotients $\mathbb C/L$ by lattices, and $\mathbb C\mathbb P^1$ is simply-connected, a map to the elliptic curve would lift (by homotopy lifting...) to a map to $\mathbb C$. This would contradict Liouville's theorem.

$\endgroup$
  • 4
    $\begingroup$ Perhaps you can add that a nonconstant holomorphic map from the Riemann sphere $\mathbb{C}\mathbb{P}^1$ to an elliptic curve $E=\mathbb{C}/\Lambda$ would lift to a nonconstant holomorphic map $\mathbb{C}\mathbb{P}^1\to\mathbb{C}$, which is impossible by Liouville's theorem (using that $\mathbb{C}\mathbb{P}^1$ is compact). $\endgroup$ – GH from MO Nov 21 '15 at 15:21
4
$\begingroup$

Let $a(t)$, $b(t)$, $c(t)$ have degree $n_1$, $n_2$, $n_3$, respectively, and $n = \text{max}(n_1, n_2, n_3)$. Then we can define $A(u, v)$, $B(u, v)$, $C(u, v)$ as homogeneous polynomials of degree $n$ such that$$A(u, 1) = a(u),\text{ }B(u, 1) = b(u),\text{ }C(u, 1) = c(u).$$Then, by construction,$$A(u, v)^3 + B(u, v)^3 = C(u, v)^3.$$Let$$E = \{(x, y, z) \in \mathbb{P}^2 : x^3 + y^3 = z^3\}.$$$E$ is a smooth curve of genus $1$, i.e. an elliptic curve. Now, define a map$$\varphi: \mathbb{P}^1 \to E,\text{ }(u, v) \mapsto (A(u, v), B(u, v), C(u, v)).$$This map is well-defined since $A(u, v)$, $B(u, v)$, $C(u, v)$ are homogeneous polynomials of the same degree which do not vanish simultaneously. Moreover, this map is nonconstant and proper since its source is projective. As the image of a proper map is closed and it is not a point, and $E$ is an irreducible $1$-dimensional variety follows that $\varphi$ is a surjective morphism. $E$ is a topologically a torus, i..e it has genus $1$ and there is a one up to scaling differential form $\tau$ on it. This will imply that its pullback $\varphi^*\tau$ is a differential form on $\mathbb{P}^1$, which is impossible since it has genus $0$. In more formal terms, a surjective map $\mathbb{P}^1 \to E$ gives rise to the injection $H^0(E, \Omega^1) \to H^0(\mathbb{P}^1, \Omega^1)$. But this is absurd, since the former is a vector space of dimension $1$ and the latter is a vector space of dimension $0$.

Perhaps you might want to say something about why the pullback of a non-zero holomorphic differential is non-zero.

$X$ and $Y$ are Riemann surfaces and $f: X \to Y$ holomorphic differential form looks like $g(z)\,dz$, where $g$ is a holomorphic function. Then its pullback is locally given by $g(f(z))\,df$. It is obviously nonzero as long as $f$ is nonconstant.

$\endgroup$
  • 1
    $\begingroup$ Perhaps you might want to say something about why the pullback of a non-zero holomorphic differential is non-zero. $\endgroup$ – Ben Lim Nov 21 '15 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy