Thread: let q(x) and f(x) be relatively prime and elements of F[x] where F is a field

let q(x) and f(x) be relatively prime and elements of F[x] where F is a field. Let q(x) and f(x) be relatively prime. Let q(x)|f(x)g(x). Prove q(x)|g(x).

I dont know where to start guys. I know there gcd =1 obviously. Maybe it has something to do with their degrees?? I dont know! BIG HELP PLEASE!!

Since gcd(f,q)=1, there are polynomials a,b such that f(x)a(x)+q(x)b(x)=1. You also know something about f(x)g(x), so try to get that product involved somehow. It doesn't seem like I gave a big hint, but that's already most of the solution.

Wow! thank you very much. With the product, are you hinting to the degree(f(x)g(x))= m+n where m=deg(f(x)) and n= deg(g(x)). is that helpful at all.

You don't really need the degrees here. Try multiplying the equation f(x)a(x)+q(x)b(x)=1 through by g(x). Now q(x) divides both terms on the left side, so what can you conclude?