Consider the transcendental extension Q(t) of the field of rationals.
To Q(t) adjoin the root of the polynomial x^5+t^5=1. The resulting 
field Q(t)[x] is a radical extension of Q(t). Is it true that the 
only solutions to the equation X^5+Y^5=1 in the field Q(t)[x] are 
(0,1), (1,0), (t,x), (x,t), (1/t,-x/t) and (-x/t, 1/t)?


Comment: Using the ABC theorem one can prove that the Fermat
curve X^n+Y^n=1 does not have a non-trivial solution in Q(t) for n>2.
In particular in Q(t) the equation X^5+Y^5=1 does not have
non-trivial solutions.



a question on function fields
First, replace Q by the complex numbers C. 


Write A = C[x,y]/(x^n+y^n-1). Then the field you write down, call it K, is the fraction field of A, which is the function field of the Fermat Curve.


Finding a solution (X,Y) to x^n + y^n = 1 is equivalent to finding a map A --> K, where one sends x to X and y to Y. Any map to a field factors though A/P for some prime ideal P of A.


Since (x^n+y^n-1) is irreducible, the only primes in A are either (0) or maximal ideals m with A/m = C.


If A/P = C, then the map A --> K factors through C. Maps from A to C correspond to complex points on the Fermat curve.


If P = 0, then A --> K extends to a non-trivial map from K --> K. Giving a map between function fields is equivalent to giving a map of the corresponding smooth projective curves (in the opposite direction). Thus, the question becomes: what are the non-trivial maps from the Fermat curve to itself? For symmetry reasons, write the Fermat curve as x^n + y^n + z^n = 0 (over C).


Assume that n > 3. Since the genus of the Fermat curve is > 1, the Reimann-Hurwitz formula tells us that any non-trivial map must be an automorphism. On the other hand, the isomorphism group of the Fermat curve is the semi-direct product of (Z/nZ)^2 by S`_`3. (Explicitly, this corresponds to multiplying (x,y,z) by n-th roots of unity, and permuting the entries.)
In terms of the affine coordinates the S\_3 action is generated by (x,y)->(y,x) and (x,y)->(1/y,-x/y).


To summarize, the K points are given by the (finitely many) automorphisms, and the C-points of the curve. You have noted all the automorphisms over Q, I would note that you are also invoking the non-trivial fact that the Fermat curve has no non-trivial points over Q, which (even for n = 5) that is harder than everything else in this answer (and is especially harder for general n > 2!). (The "ABC theorem" for polynomials only tells you that there are no *non-constant* solutions in C[t]. In general, a solution in C[t] will correspond to a map from P^1-->X, which can be non-trivial only if X has genus 0.)


Note that this argument used basically nothing about the Fermat curve itself --- any curve of genus > 1 has only finitely many automorphisms.


If g = 1, then there are some non-trivial maps from a curve of genus g to itself, but Riemann-Hurwitz tells us that they are all unramfied. X/C in this case is an elliptic curve, so there will be infinitely many non-trivial solutions in K, corresponding to the rational functions giving the multiplication by [n] map (or more if X has CM). In the Fermat case, this means that for n = 3 there will be many more solutions.


If g = 0, there are buckets of maps from P^1 to itself, as one knows in the Fermat case when n = 2 or 1.


