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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.

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Isn't there a formula for the genus of a plane curve in terms of its degree ? – Chandan Singh Dalawat Dec 17 '12 at 11:03
The genus of a curve is invariant under base change. – François Brunault Dec 17 '12 at 12:01
@François : Voir Genus change in inseparable extensions of function fields par John Tate (…). – Chandan Singh Dalawat Dec 17 '12 at 12:23
@Chandan : Thanks for this reference. I should have said "separable base change" in the above comment. As far as the Fermat curve of degree $n$ is concerned, everything is fine as long as the characteristic of $K$ doesn't divide $n$ : the curve is absolutely irreducible, smooth, and the $K$-vector space of regular differentials on it has dimension $(n-1)(n-2)/2$. – François Brunault Dec 18 '12 at 7:51
up vote 8 down vote accepted

No; in fact, you can take $K = \mathbb{C}$. This follows from the Mason-Stothers 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).

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Thank you Qiaochu – joro Dec 17 '12 at 12:06

Here is a possibly simpler proof than the one given by quid. Suppose that $A^n+B^n=C^n$ where $A,B,C$ are coprime and non-constant. Assume that the characteristic does not divide $n$. Differentiate and eliminate $A,B,C$ in turn, then compare degrees. It is not hard to deduce that $n<3$. I do not need to give the details as I have set this as an exercise after the first lecture of an undergraduate course. The follow-on question is to generalise and show that if $A^{n_1}+B^{n_2}=C^{n_3}$ (with $A,B,C$ coprime and non-constant polynomials as before) then $1/n_1+1/n_2+1/n_3>1$.

Now if only we could differentiate integers...

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Thank you Cremona. – joro Dec 22 '12 at 12:14

In addition to the two proofs already given, it might be interesting to know that there is also a more selfcontained proof (by descent); along the lines of (unsuccesfull or at least only partially succesfull) ideas to prove original FLT.

Working with complex polynomials, and for $n\ge 3$: suppose there is a coprime solution $a^n +b^n = c^n$ and pick one such that the maximum of degrees of $a,b,c$ is minimal; denote this max $N$.

One has the factorization $a^n + b^n = \prod_{j=0}^{n-1} (a + \zeta^j b)$ with $\zeta$ an $n$-th root of unity.

The factors on the right can be seen to be relatively prime. Since their product equals the $n$-th power $c^n$ each of them is an $n$-th power. Write $(a+\zeta^j b) = f_j^n$. Note $a= (f_1^n - \zeta f_0^n)/(1 - \zeta)$ and $b=(f_1^n - f_0^n)/(\zeta -1)$, and $f_2^n = (\zeta + 1) f_1^n - \zeta f_0^n$. (Here the fact $n\ge 3$ is used.)

Now, with $a_1 = \sqrt[n]{\zeta+1}f_1 $ and $b_1 = -\sqrt[n]{\zeta}f_0$ one has a triple $(a_1,b_1,f_2)$ fullfulling the equation.

Then, the degree of these are bounded above by $N/n$ contradicting the minimality of $N$.

This proof is an abridged version of the one given in these lecture notes by Franz Lemmermeyer (it is attributed there to Greenleaf); see section 2.2. The two other proofs via ABC and genus are also discussed there.

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Thank you quid. – joro Dec 17 '12 at 12:13

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