Posted by: Alexandre Borovik | May 24, 2008
Fermat’s Last Theorem for Polynomials
An old post by Nick Halloway at sci.math:
Someone mentioned to that an analog of FLT is true in ![\mathbb{C}[x]](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BC%7D%5Bx%5D&bg=ffffff&fg=4b5d67&s=0 "\mathbb{C}[x]")
as well, i.e. there are no relatively prime [non-constant] polynomials 
, 
and 
in with
for 
.
This is true. You can show it by adapting the false proof of FLT.
It’s only necessary to show it for 
an odd prime and for 
. The proof is similar for to an odd prime, so I will just show it for odd primes.
Proof by induction on the sum of the degrees of and .
Suppose the sum of the degrees is 
. Then 
factors as
,
where 
is a primitive 
‘th root of unity. These factors are all relatively prime polynomials of degree 
. But 
has repeated factors. So it won’t work for this case.
Now assume the sum of the degrees is 
. Again factor as
These factors are relatively prime. But has factors with multiplicity divisible by . So each factor 
must be a ‘th power. So we have
,
,
So 
, and
So 
Since neither 
nor 
are 0, and 
and 
are relatively prime, this violates the induction hypothesis.
This also shows it for polynomials in ![\mathbb{Z}[x]](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BZ%7D%5Bx%5D&bg=ffffff&fg=4b5d67&s=0 "\mathbb{Z}[x]")
, since polynomials that are relatively prime in are also relatively prime in .
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The remark at the end is false. For example, consider the constant polynomials f=2 and g=3. These are relatively prime in Z[x], but are multiples of each other in C[x].
By: Lucas on May 24, 2008
at 4:27 pm
Two elements f, g in a UFD are relatively prime if there is no irreducible or prime element which divides them both. Since 2 and 3 are units in C[x], neither has a prime factor, so they are relatively prime in this ring.
By: Todd Trimble on May 25, 2008
at 12:52 pm
Yes, you are right (I wasn’t thinking clearly), but then the result is false. We have that 2, 3, and (2^3+3^3)^(1/3) are all units in the ring, and clearly these yield a solution of a^n+b^n=c^n, with n=3.
In point of fact, now that I look at it, the proof assumes that the polynomials are nonconstant, but that isn’t stated in the statement of the theorem.
By: Lucas on May 25, 2008
at 3:14 pm