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My question is, what are some of the different ways we can see that there do not exist nonconstant, relatively prime, polynomials $a(t)$, $b(t)$, and $c(t) \in \mathbb{C}[t]$ such that$$a(t)^3 + b(t)^3 = c(t)^3?$$In particular, could anyone outline/supply a proof the "better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve" mentioned by David Speyer in the link? Thanks.
We give a completely elementary proof simply working in the ring $\mathbb{C}[t]$ and using that it is a UFD. 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 $a$, $b$, $c$.
In particular, could anyone outline/supply a proof the "better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve" mentioned by David Speyer in the link? Thanks.
See my answer here.
The key lemma for the proof of Fermat's last theorem for polynomials usually is the so-called Mason-Stothers Theorem:
Theorem: Let $K$ be a field and $A,B,C$ nonzero polynomials in $K[T]$ with $A+B+C = 0$ and $gcd(A,B,C) = 1$. If ${\rm deg}\, A ≥ deg \, rad \, ABC$, then $A′ = B′ = C′ = 0$.
Here $rad \, f$ is the radical of $f$, i.e., if $f=cp_1^{e_1}\cdots f_n^{e_n}$ with distinct prime factors $p_i$, then $rad \, f=p_1\cdots p_n$. Note the analogy with the abc-conjecture for integers.
How this implies Fermat for polynomials is explained very nicely in Lemmermayer's notes. This is elementary, and I do not think, that the cubic case is much easier than the general case for $n\ge 3$.
Edit: Just now the OP has edited, about David Speyer's link. For this see here.
Fermat’s Last Theorem for polynomials can be proved in many different ways; Shanks, in his book Solved and Unsolved Problems in Number Theory, discusses a proof given by Chebyshev using integration.
