David Speyer commented the following 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.
Here is my elementary proof, which I gave here.
Suppose there are some solutions of$$a(t)^3 + b(t)^3 = c(t)^3$$in $\mathbb{C}[t]$. Choose a solution $(a(t), b(t), c(t))$ such that the maximum $m > 0$ of the degrees of $a$, $b$, $c$ is minimal possible among all solutions. Clearly, this choice ensures that $a(t)$, $b(t)$, $c(t)$ are coprime. Then we have$$a(t)^3 = c(t)^3 – b(t)^3 = (c(t) – b(t))(c(t) – \omega b(t)) (c(t) – \omega^2 b(t)),$$where $\omega$ is a third primitive root of unity. Now, we look at the factors $c(t) – b(t)$, $c(t) – \omega b(t)$. Suppose that they have a common factor $q(t)$. Considering their sum and difference, we conclude that $c(t)$ and $b(t)$ have a common factor too. Moreover, $q(t)$ is a factor of $a(t)$. Thus, $a$, $b$, $c$ are not relatively prime, which is a contradiction. Repeating the same game with other pairs of factors, we see that all three factors $c(t) – b(t)$, $c(t) – \omega b(t)$, $c(t) – \omega^2 b(t)$ are pairwise coprime. Therefore,$$c(t) – b(t) = d_1(t)^3,\text{ }c(t) – \omega b(t) = d_2(t)^3,\text{ }c(t) – \omega^2b(t) = d_3(t)^3,\text{ where }d_1,\,d_2,\,d_3 \in \mathbb{C}[t].$$Note that$$\omega^2 + \omega + 1 = 0.$$Multiplying the second equation by $\omega$ and the third equation by $\omega^2$ and adding all three, we arrive at $$d_1(t)^3 + \omega d_2(t)^3 + \omega^2d_3(t)^3 = 0.$$Choosing $\eta_1$ and $\eta_2$ as any third roots of $-\omega$ and $-\omega^2$, respectively, and letting$$a_1 = d_1^3,\text{ }b_1 = \eta_1d_2^3,\text{ }c_1 = \eta_2d_2^3,$$we get$$a_1^3 = b_1^3 + c_1^3.$$By construction, at least one of $a_1$, $b_1$, $c_1$ is a nonconstant polynomial, and the maximum of their degrees is smaller than that of $a$, $b$, $c$. This is a contradiction to the choice of $a$, $b$, $c$.
Here is my better motivated "algebraic geometric" proof, which I gave here.
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.
My question is, is there anyone who knows more algebraic geometry than me who can comment on the precise/possibly deep relationship between these two different proofs, i.e. why they are fundamentally the same or different?
Best Answer
This answer is basically a longer version of Felipe Voloch's, but maybe it will be useful. Both proofs take a class in $H^1(E)$, pull it back to $H^1(\mathbb{P}^1)$ and note that $H^1(\mathbb{P}^1)$ is zero to conclude that the map $\mathbb{P}^1 \to E$ was constant. The difference is what form of $H^1$ we work with.
The topological version is that a continuous map from a sphere to a torus must have degree zero; then holomorphic maps of degree zero are constant.
Proof with differentials We take the class $dy/x$ in $H^0(E, \Omega^1)$ and pull it back to $H^0(\mathbb{P}^1, \Omega^1)=0$. This gives some equations on derivatives which imply that we are pulling back by the zero map. By Hodge theory, for any curve $X$, the sheaf cohomology $H^0(X, \Omega^1)$ is half of $H^1(X, \mathbb{C})$.
The proof by descent The map $$ \phi(x:y:z) = (x+\omega y + \omega^2 z : x+\omega^2 y + \omega z : - 3 x y z)$$ is a degree $3$ unbranched cover of $E$, giving a class in $H^1(E, \mathbb{Z}/3)$. If we pull this cover back to $\mathbb{P}^1$, then it must become trivial. Your computations are explicitly showing that, if $a(t)^3+b(t)^3+c(t)^3=0$, then we can find some $x(t)$, $y(t)$, $z(t)$ such that $(a:b:c)=\phi(x:y:z)$. So you have shown that this $\mathbb{Z}/3$ cover has a section, and is thus trivial. Abstractly, the reason there must be a section is that $H^1(\mathbb{P}^1, \mathbb{Z}/3)=0$.
Here is an abstract proof that $H^1(\mathbb{P}^1, \mathbb{Z}/3)=0$. Look at the short exact sequence $1 \to \mu_3 \to \mathcal{O}^{\ast} \to \mathcal{O}^{\ast} \to 1$, where $\mu_3$ is the group of cube roots of unity, $\mathcal{O}^{\ast}$ is the sheaf of nonvanishing holomorphic functions and the second map is cubing. (The topology is either analytic or etale.) Looking at the long exact sequence in cohomology, we see that $H^1(\mathbb{P}^1, \mu_3)$ is the $3$-torsion in $Pic(\mathbb{P}^1)$. But $\mathbb{C}[t]$ is a PID, so $Pic(\mathbb{A}^1)=0$ and $Pic(\mathbb{P}^1) \cong \mathbb{Z}$. You wind up using the exact same fact -- that $\mathbb{C}[t]$ is a PID -- when you write down the proof in an elementary way.
I remark that the first proof using a "de Rham" description of $H^1$ (by differential forms) and the second proof uses a "Betti" description (by covering maps). Relating de Rham and Betti descriptions is always complicated.