# [Abstract Algebra] GCD and Relatively Prime Proof

1. Feb 20, 2016

### RJLiberator

1. The problem statement, all variables and given/known data
If $gcd(f(x),g(x)) = 1$ and m,n ∈ ℕ, show that $gcd(f(x)^m, g(x)^n) = 1$.

2. Relevant equations

3. The attempt at a solution

So I had previously proved this for non-polynomials:

gcd(a,b)=1
then gcd(a^n,b^n)=1

Proof: a = p1*p2*...*pn
b = p1*p2*...*pm

then
a^n = p1^n*p2^n*...*pn^n
b^n = p1^n*p2^n*...*pm^n

Since a and a^n have the same prime factors and b and b^n have the same prime factors and a and b are relatively prime then a^n and b^n are relatively prime.

I guess, I am looking at this question from the same point of view.

Is there a difference here with polynomials? Why would there be?

2. Feb 20, 2016

### Staff: Mentor

As long as the underlying ring or field is of characteristic zero there shouldn't be a difference. (Not quite sure about other characteristics.)
However, your proof with the numbers uses the fact that you have a unique integer factorization into primes. You need to assure this for your polynomial ring if you want to adapt the same proof. As far as I can remember this is true for an integral domain. Another proof could be with contradiction: assume there is an irreducible $p=p(x)$ with $p|f^n$ and $p|g^m$.

3. Feb 20, 2016

### RJLiberator

How is this proof:

Since f and g are relatively prime, by Bezout's Identity there exists polynomials p and q such that fp+gq = 1.
Take the (2n-1)th power of both sides.
Any term of the binomial expansion of (fp+gq)^(2n-1) is divisible by one of the f^n or g^n.
Therefore:
1 = (fp+gq)^(2n-1) = f^nP+g^nQ
for some polynomials P and Q. It follows that f^n and g^n are relatively prime.

But does this hold for f^n and g^m ? (different degrees) hm...

I don't think so, right? Since we use (2n-1) as the exponential.

4. Feb 20, 2016

### RJLiberator

Perhaps this is cleaner / better:

Suppose p is a prime that divides gcd(f^n,g^m)
then p divides f^n and hence divides f. Similarly for g^m. Since gcd(f,g)=1 no such p can exist.

This is proven by:
If p divides f^n, than p divides f. for n>1
Proof using induction: true for n = 1
Assume it is true for n.
Then for n+1 p divides f^(n+1) = f*f^n implies that p divides f or it divides f^n. By hypothesis, p divides f.

However, not sure how this incorporates polynomials or if i even need to (the question doesn't necessarily make any statement about polynomials just is the question stated)

5. Feb 20, 2016

### Staff: Mentor

Your first proof is wrong because not every term of the binomial expansion is divided by $f^n$.
Let's say $p$ is irreducible instead.
$n \geq 1$
What is the underlying ring or field for the polynomials? $ℤ$?
This here is the crucial point: $p|f \cdot f^n ⇒ p|f ∨ p|f^n$?
Here I think you need the polynomial ring be an integral domain.
E.g. if $p(x) = x+1$ and $f(x)=2x$ both in $ℤ_4[x]$ then $p \nmid f$ but $p|f^2=4x^2=0$.
However, I suppose $ℤ[x]$ or a polynomial ring over field of characteristic zero $(ℚ[x],ℝ[x],ℂ[x])$ is meant.
Then everything looks ok.

6. Feb 21, 2016

### RJLiberator

Indeed. Thank you for analysis here. (as always)
fresh_42, I bow to you!