# Fermat over Number Fields

If $n>2$, does the impossibility of solving $x^n +y^n=z^n$ with $x, y, z$ rational integers imply the same with $x, y, z$ algebraic integers?

Rather, If insolvability in algebraic integers does follow, then does it follow from simple considerations, or is it still an interesting question?

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It definitely doesn't follow! Given n>2 and any large integers x,y, just let z be the n'th root of x^n+y^n: that's an algebraic integer. FLT is open for all fields other than Q and Q(sqrt(2)) AFAIK. –  Kevin Buzzard Jan 30 '10 at 14:45

This is mostly an amplification of Kevin Buzzard's comment.

You ask about points on the Fermat curve $F_n: X^n + Y^n = Z^n$ with values in a number field $K$.

First note that since the equation is homogeneous, any nonzero solution with $(x,y,z) \in K^3$ can be rescaled to give a nonzero solution $(Nx,Ny,Nz) \in \mathbb{Z}_K$, the ring of algebraic integers of $K$ -- here $N$ can taken to be an ordinary positive integer.

Thus you have two "parameters": the degree $n$ and the number field $K$.

If you fix $n$ and ask (as you have seemed to) whether there are solutions in some number field $K$, the answer is trivially yes as Kevin says: take $x$ and $y$ to be whatever algebraic integers you want; every algebraic integer has an $n$th root which is another algebraic integer, so you can certainly find a $z$ in some number field which gives a solution. Moreover, if you take $x$ and $y$ in a given number field $K$ (e.g. $\mathbb{Q}$), then you can find infinitely many solutions in varying number fields $L/K$ of degrees at most $n$. But it is interesting to ask over which number fields (or which number fields of a given degree) there is a nontrivial solution.

On the other hand, if you fix the number field $K$ and ask for which $n$ the Fermat curve $F_n$ has a solution $(x,y,z) \in K$ with $xyz \neq 0$, then you're back in business: this is a deep and difficult problem. (You can ask such questions for any algebraic curve, and many people, myself included, have devoted a large portion of their mathematical lives to this kind of problem.) So far as I know / remember at the moment, for a general $K$ there isn't that much which we know about this problem for the family of Fermat curves specifically, and there are other families (modular curves, Shimura curves) that we understand significantly better. But there are some beautiful general results of Faltings and Frey relating the plenitude of solutions (in fact not just over a fixed number field but over all number fields of bounded degree) to geometric properties of the curves, like the least degree of a finite map to the projective line (the "gonality").

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Interesting example: A Aigner looked at $x^4+y^4=z^4$ over quadratic extensions of the rationals and proved that $\mathbb{Q}(\sqrt{-7})$ is the only quadratic extension to contain nontrivial solutions. Note that $(1+\sqrt{-7})^4+(1-\sqrt{-7})^4=2^4$. This was looked at again by Faddeev and then by Mordell (Acta Arith. 14 1967/1968 347--355).

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I think it's natural to look at quadratic extensions and $n=3$, which I believe eliminates obvious solutions. I used $(5-\sqrt 6)^3 + (3\sqrt 6)^3 = (5+\sqrt 6)^3$ for this problem.

This paper in Integers has some useful references, and a new proof that the case $n=3$ has no nontrivial solutions over the Gaussian integers $\mathbb Z[i]$.

"In L.E. Dickson’s History of the Theory of Numbers, one can find that this question has already been answered by R. Feuter within the frame of algebraic number theory. Namely, he has proven that if $a^3+b^3+c^3 = 0$ is solvable by numbers $\ne 0$ of an imaginary quadratic domain $k(\sqrt m)$, where $m \lt 0$, $m = 2 (\mod 3)$, then the class number of $k$ is divisible by 3. Now $k = \mathbb Z$, $m = −1$ and $\mathbb Z$ is a Principal Ideal Domain having class number 1 not divisible by 3."

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The Fermat cubic is an elliptic curve. Using known results in elliptic curve theory (specifically existence of quadratic twists with rank 0 and with rank 1, say), it follows easily that there will be infinitely many quadratic fields over which the Fermat cubic has only the trivial solutions and infinitely many more over which it has nontrivial solutions. –  Pete L. Clark Jan 30 '10 at 8:33
Pete, despite your statement "So this is not so interesting," I maintain that for a fixed n, it is interesting to ask for which number fields there are solutions. –  Douglas Zare Jan 30 '10 at 19:51
@DZ: You're absolutely right: I removed the offending sentence. By the way, if you had made this as a comment to my answer, I would have been notified of it automatically. –  Pete L. Clark Jan 30 '10 at 20:27
Would a converse to the above be true as well? i.e. existence of a class of order 3 implies non trivial solution? If so, then this is quite more interesting than very difficult results in the theory of elliptic curves. I mean in this specific case, of course. –  Dror Speiser Jan 30 '10 at 20:40
@Dror: To the best of my knowledge, no, it is not so easy. That was my point: the question whether F_3 has a nontrivial rational point over Q(\sqrt{m}) is equivalent to whether the quadratic twist of F_3 by m has positive rank. I believe that no explicit condition for the latter is known for any rational elliptic curve (including this one, which is simpler than usual in many respects because it has complex multiplication). –  Pete L. Clark Jan 30 '10 at 21:02

Kolyvagin's "On the First Case of the Fermat Theorem for Cyclotomic Fields" is worth mentioning here. The main point of this paper is that while various case-by-case eliminations (e.g., Kummer's regularity, Sophie Germain primes, Wieferich conditions) are made obsolete by the full solution to FLT over $\mathbb{Q}$, the exact same conditions, or suitable generalizations, should still work over other number fields -- he deals in particular with cyclotomic fields and eliminates the analogous set of prime exponents for the first case of FLT over $\mathbb{Q}(\zeta_\ell)$.

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