Understanding proof by algebraic geometry, Fermat's last theorem for polynomials when $n = 3$. This is a followup to my question here.
See here. The question is as follows.

How do we see that there do not exist nonconstant, relatively prime, polynomials $a(t)$, $b(t)$, and $c(t) \in \mathbb{C}[t]$ such that$$a(t)^3 + b(t)^3 = c(t)^3?$$

There is the following answer, the "better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve."

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.

Unfortunately, I do not know any algebraic geometry and do not really understand what is going on. Can anyone summarize/explain the motivation/explain what is going on/explain the key steps/help me understand the proof? Thanks.
 A: It seems to me that there is no better "explanation" than your introducing sentence about "the better motivated proof that CP1 can't map holomorphically to a genus 1 curve". But of course you need to know the notion of genus and its functorial properties. One can try to explain by "hand waving" that the genus of an algebraic surface is the number of "holes" in it, and is a topological invariant. The complex projective line can be viewed as a sphere (the Riemann sphere), of genus 0 (no hole!). The 3rd degree Fermat equation can be viewed as the equation, in homogeneous coordinates, of a smooth plane cubic. Such curves are called "elliptic curves" (not ellipses, which are curves of the 2nd degree; the name comes actually from the "elliptic functions" which intervene in the calculation of the length of an ellipse arc). Elliptic curves are remarkable because they own a natural structure of abelian groups. By the Riemann-Roch theorem,they are known to have genus 1 (think of a torus, or a bretzel!). The existence of polynomial solutions of the 3rd degree Fermat equation, roughly speaking, would give a "parametrization" of the Fermat curve by rational functions, so that the key point in the proof by algebraic geometry that you mention lies in the Riemann-Hurwitz formula which relates the genera our two holomorphic curves (surfaces) .
Coming back to Fermat's theorem for polynomials (and for any exponent n not less than 3), it has been proved by Liouville. Nowadays, an elementary demonstration can be easily derived from a not less elementary theorem of Mason (1984):   "For any polynomial P(t), denote by N(P) the number of distinct roots of P. If a, b, c are 3 polynomials in t such that a + b = c, then N(abc) - 1 is not less than max(deg(a), deg(b), deg(c))". Note that this inspired the formulation of the famous "abc conjecture" (Masser-Oesterlé) which - if true - would allow to solve many important diophantine problems. 
