An old post by Nick Halloway at sci.math:
Someone mentioned to that an analog of FLT is true in
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
, since polynomials that are relatively prime in
are also relatively prime in
.
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