Polynomial FLT and Beyond

The classical Fermat's Last Theorem says if n >= 3, then the equation x^n + y^n = z^n admits no positive integer solutions.

Here is the polynomial version of Fermat's Last Theorem:

Theorem(Polynomial FLT):
Suppose a, b, c are in C[t] (i.e. polynomials with complex coefficients) such that gcd(a, b, c) = 1. (simply means that they do not have a common complex root) For n >= 3, then a^n + b^n = c^n for all t in C implies a, b, and c are constant polynomials.

The proof of this theorem usually uses the following so-called ABC conjecture for function fields, whose proof is totally elementary.

Theorem(Mason)
Suppose a, b, c in C[t] such that (a, b, c) = 1. If a + b = c, then we have:
max{deg(a), deg(b), deg(c)} <= N(abc) - 1, where N(f) denotes the number of distinct complex roots of f.

I will not prove this here as you will probably see the proof during the upcoming talk.

However, I shall give a short proof of the polynomial FLT using Mason's theorem:

Proof of Polynomial FLT: Let n >= 3, we have N(a^n*b^n*c^n) = N(abc). Mason's theorem yields:

n deg(a) = deg(a^n) <= max{deg(a^n), deg(b^n), deg(c^n)} <= N(abc) - 1 <= deg(a) + deg(b) + deg(c) - 1.

Therefore, we have:

n(deg(a) + deg(b) + deg(c)) <= 3(deg(a) + deg(b) + deg(c)) - 3 <= n(deg(a) + deg(b) + deg(c)) - 3, where the last inequality follows from the fact that 3 <= n. This is a contradiction and we are done.

Next, we go a step further. It is now known that if a, b, c are polynomials in t with coefficients in C, then a^n + b^n = c^n has no constants solution if n >= 3. We ask ourselves whether the same holds if a, b, c are allowed to be entire functions. That is a, b, c are power series in t with coefficients in C. The answer is no if n = 3, and yes if n >= 4.

To this end, we change notation a little bit and we write f = a/c, and g = b/c, where f, g are now meromorphic functions in C. We ask whether f^n + g^n = 1 forces f, and g to be constant functions.

For n = 3, we can transform the equation f^3 + g^3 = 1 into an equation of the form: Y^2 = X^3 - 432, via setting f = (36 - Y)/6X, g = (36 + Y)/6X.

E: Y^2 = X^3 - 432 can be realized as an elliptic curve, and it's a well known theorem in elliptic curve that we have an analytic isomorphism:

C/L -> E(C),

where L = {aw_1 + bw_2 | a, b in Z, and w_1, w_2 are linearly independent over R}, and E(C) = complex points in E. As an example, L could be Z[i] the Gaussian integers for example.

This isomorphism is given by the map: z |-> (p(z), p'(z)), where p(z) is the Weierstrass p-function, which is meromorphic in C with double poles precisely at the points in L.

Now, via this isomorphism, we will get non-constant f, g in terms of p(z), and p'(z).

As for n >= 4, the theorem is true and here is a quick proof using theory in compact Riemann surface.

First we know that f = (1 - g^n)^{1/n} defines a compact Riemann surface of genus g >= 2 (since n >= 4) [In fact, it can be shown that f^n + g^n = 1 has genus g = (n - 1)(n - 2)/2.

Let S denotes this Riemann surface. Therefore, we have a meromorphic map f : C to S. Next, we resort to the uniformization theorem which asserts that the universal cover of S is D, the open unit disc since g >= 2. Since D is actually a universal cover, and that C is simply connected, we have a meromorphic lift j: C to D. By Liouville's theorem (or by Picard if one wishes), j must be constant and hence the covering map from D to S is also constant and therefore f is constant. This concludes the proof.