Can the GCD of Polynomials in a Field Always Be Reduced to 1?

[alexfresno]

Homework Statement

We know that the gcd of two polynomials can be written as

gcd(p(x),q(x))=p(x)m(x) + q(x)n(x) for some n(x) and m(x) in F[x] F a fieldI want to show gcd(n(x),m(x))=1 for a fixed gcd(p(x),q(x))

The Attempt at a Solution

Well, what I tried was to let D(x)=gcd(p(x),q(x))
then D(x)|q(x) and D(x)|p(x). Which implies that q(x)=D(x)k(x) and p(x)=D(x)s(x) for some k(x),s(x) in F[x].

Then i got D(x)=D(x)k(x)m(x)+D(x)s(x)n(x) by substitution

Can I divide both sides by D(x) since we are in a Field? If i can...
then won't 1=k(x)m(x)+s(x)n(x) imply gcd(m(x),n(x))=1?? or there is a stronger proof?

 

[micromass]

I think your proof is correct!