Not sure if this makes sense, but is it possible Fermat's Last Theorem to fail
with a parametrization over some extension of $\mathbb{Z}$, i.e. are there
not all constant $x(t),y(t),z(t) \in K[t]$ where $K$ is an extension of $\mathbb{Z}$
s.t. $$x(t)^p + y(t)^p=z(t)^p, x(t)y(t)z(t) \ne 0, p > 2,\gcd(x(t),y(t),z(t))=1 $$


I suppose this would mean in an extension of $\mathbb{Q}$ the curve $x^p + y^p = z^p$ will
be of genus $0$.


Tried equating coefficients with $p=3$, got relatively small undetermined system but failed to solve it
or compute groebner basis. The system has solutions like $x(t)=0$.


EDIT: The coprimality condition is to avoid scaling a single solution by a polynomial.



Can FLT fail with a parametrization over some extension of Z?
No; in fact, you can take $K = \mathbb{C}$. This follows from the [Mason-Stothers theorem](http://en.wikipedia.org/wiki/Mason%E2%80%93Stothers_theorem) in the same way that FLT for sufficiently large $n$ follows from the abc conjecture. 


As Chandan indicates in the comments, a more geometric reason this is false is that the Fermat curves $x^n + y^n = z^n$ have positive genus for $n \ge 3$ and hence do not admit rational parameterizations (over fields of characteristic zero or something like that). 


