# Fermat's Last Theorem For Polynomials

• Jul 12th 2011, 09:49 PM
Drexel28
Fermat's Last Theorem For Polynomials
Presumably anyone who is on this forum is well aware of the famous Fermat's Last Theorem(FLT): "There exists no solutions to the equation $\displaystyle a^n+b^n=c^n$ for $\displaystyle n>2$, $\displaystyle a,b,c\in\mathbb{Z}$ and $\displaystyle abc\ne0$." Moreover, I'm sure it's equally well known that this 'theorem' was really a conjecture for over three hundred years until Sir Andrew Wiles of Princeton finally proved a result which implied FLT in 1994-1995 in one of the most stupendous mathematical feats in the history of the subject (the proof may be found here for those interested).

Despite the infamous difficulty of FLT most students of mathematics can attest that they have, at one point or another, devoted a day, week, or month to trying to prove it--after all, it doesn't look that hard. Well, a more mathematically cultured person may have a different pursuit, namely asking whether an analogue of FLT holds for other fields (rings). Namely, given a ring $\displaystyle R$ does there exist non-trivial solutions to $\displaystyle a^n+b^n=c^n$ for $\displaystyle n>2$? Of course, this isn't true at all for general rings or fields (take $\displaystyle \mathbb{C}$)! So perhaps the question needs to be narrowed from a general ring to a more specific, tame ring, the question is, which ring? Well, besides $\displaystyle \mathbb{Z}$ perhaps every math students favorite ring is $\displaystyle k[x]$ (the ring of polynomials with coefficients in $\displaystyle k$, for those unfamiliar to algebra) for some field $\displaystyle k$, so why not this? I mean, for the sake of simplicity why not restrict to the easiest infinite field $\displaystyle \mathbb{C}$? So, the question now is, does there exist solutions to $\displaystyle a(x)^n+b(x)^n=c(x)^n$ for coprime $\displaystyle a(x),b(x),c(x)\in\mathbb{C}[x]$ for $\displaystyle n>2$? There is clearly two distinct cases to this problem, namely when $\displaystyle a(x),b(x),c(x)$ are constant, and when they are all non-constant (clearly these are the only two cases). The first of these cases, as discussed, is evidently false, and so the crux of the problems occurs when one decides to consider only when $\displaystyle \min\{\deg(a(x)),\deg(b(x)),\deg(c(x))\}\geqslant 1$. So, what's the answer, do there exist non-trivial solutions?

I won't keep you in suspense, the answer is no. So, who won a Fields medal for the proof? Which great mathematician solved this problem? Well, shockingly the person who solved this originally didn't win a Fields medal, isn't famous, and isn't even (at least widely) known. In fact, one can prove this theorem (more generally over fields of characteristic zero) with such elementary methods that a mathematically inclined high school student would have no difficulty understanding it. Particularly, once one proves the so-called Mason-Stothers Theorem (which states that under the conditions of our 'theorem' one has that $\displaystyle \max\{\deg(a(x)),\deg(b(x)),\deg(c(x))\}+1 \leqslant \text{number of complex roots of }a(x)b(x)c(x)$) then one can prove the FLT for $\displaystyle \mathbb{C}[x]-\mathbb{C}$ in less than a page (a proof of both the Mason-Stothers and the FLT for $\displaystyle \mathbb{C}[x]-\mathbb{C}$ here).

So my question, the discussion I'd like to start is, what intuition is there for this version of the FLT to be SO much easier than for $\displaystyle \mathbb{Z}$?
• Jul 13th 2011, 12:03 PM
FernandoRevilla
Re: Fermat's Last Theorem For Polynomials
Quote:

Originally Posted by Drexel28
So my question, the discussion I'd like to start is, what intuition is there for this version of the FLT to be SO much easier than for $\displaystyle \mathbb{Z}$?

At first there is no surprise for me. Although $\displaystyle \mathbb{Z}$ is embedded in $\displaystyle \mathbb{R}[x]$ , we are considering $\displaystyle x(t), y(t),z(t)$ nontrivial relatively prime polynomials i.e. essentially different "objects" from integers.
• Jul 13th 2011, 12:04 PM
Drexel28
Re: Fermat's Last Theorem For Polynomials
Quote:

Originally Posted by FernandoRevilla
At first there is no surprise for me. Although $\displaystyle \mathbb{Z}$ is embedded in $\displaystyle \mathbb{R}[x]$ , we are considering $\displaystyle x(t), y(t),z(t)$ nontrivial relatively prime polynomials i.e. essentially different "objects" from integers.

Of course, it's clear that this is different problem altogether, but is it clear why it's easier?
• Jul 13th 2011, 12:10 PM
FernandoRevilla
Re: Fermat's Last Theorem For Polynomials
Quote:

Originally Posted by Drexel28
Of course, it's clear that this is different problem altogether, but is it clear why it's easier?

For me simply occurs, things are as they are. :)
• Jul 13th 2011, 07:35 PM
Drexel28
Re: Fermat's Last Theorem For Polynomials
Quote:

Originally Posted by FernandoRevilla
For me simply occurs, things are as they are. :)

Haha, I can see that. The only reason I would think it may be easier is that there is definitely a certain 'obvious' angle to attack the problem which, as I pointed out, turns out to be the correct way--the degree argument.
• Jul 13th 2011, 09:56 PM
NonCommAlg
Re: Fermat's Last Theorem For Polynomials
ok, both the ring of integers and polynomial rings are Euclidean domains, which i think is the reason you're comparing them.
but the proof of FLT for polynomials heavily depends on analytic properties of polynomial rings, something that the ring of integers does not have.
so it would be interesting to see a pure algebraic proof of FLT for polynomial rings! i don't think that can be done easily!
• Jul 13th 2011, 10:03 PM
Drexel28
Re: Fermat's Last Theorem For Polynomials
Quote:

Originally Posted by NonCommAlg
ok, both the ring of integers and polynomial rings are Euclidean domains, which i think is the reason you're comparing them.

Precisely

Quote:

but the proof of FLT for polynomials heavily depends on analytic properties of polynomial rings, something that the ring of integers does not have.
so it would be interesting to see a pure algebraic proof of FLT for polynomial rings! i don't think that can be done easily!
I'm not sure I totally agree. Unless I missed something in the proof of the FLT for polynomials via Mason-Stothers, there isn't any analysis used that I've seen. Am I being stupid? That said, I think the Mason-Stothers proof is one of those problem-solver proofs that is kind of...a fluke. It doesn't really tell you anything meaningful. The meaningful proof of the FLT for polynomials is done using alg. geo. in which case I'd agree with what you've said.
• Jul 13th 2011, 10:21 PM
NonCommAlg
Re: Fermat's Last Theorem For Polynomials
Quote:

Originally Posted by Drexel28
Precisely

I'm not sure I totally agree. Unless I missed something in the proof of the FLT for polynomials via Mason-Stothers, there isn't any analysis used that I've seen. Am I being stupid? That said, I think the Mason-Stothers proof is one of those problem-solver proofs that is kind of...a fluke. It doesn't really tell you anything meaningful. The meaningful proof of the FLT for polynomials is done using alg. geo. in which case I'd agree with what you've said.

by analytic properties i meant differentiating and this fact that by differentiating a polynomial the degree drops one unit.
although derivation can be defined over a ring algebraically but over the ring of integers the only derivation which can be defined is the zero map. so it's completely useless. even if we define a non-trivial derivation over some Euclidean domain, the degree of the derivative of an element will not necessarily be less than the degree of the element. this is a very special case that happens in polynomial rings.
• Jul 13th 2011, 10:24 PM
Drexel28
Re: Fermat's Last Theorem For Polynomials
Quote:

Originally Posted by NonCommAlg
by analytic properties i meant differentiating and this fact that by differentiating a polynomial the degree drops one unit.
although derivation can be defined over a ring algebraically but over the ring of integers the only derivation which can be defined is the zero map. so it's completely useless. even if we define a non-trivial derivation over some Euclidean domain, the degree of the derivative of an element will not necessarily be less than the degree of the element. this is a very special case that happens in polynomial rings.

Ah, I see what you mean.
• Aug 16th 2012, 11:50 PM
Vlasev
Re: Fermat's Last Theorem For Polynomials
I know it's been more than a month since this discussion took place but I have something to add. There is another big difference between the integers and the polynomials in the complex numbers and that is, factorization can be done in polynomial time for the latter but we do not know of such fast algorithm for the integers. This is not really an answer because the differences in proving the FLT maybe stem from the same reasons as the difficulties in factoring.

Also, there is a way to define a "derivative" on the integers. It's called an Arithmetic derivative but I wonder if it's of any use.