Let $f(x)$ and $g(x)$ be irreducible polynomials of different degrees in field $F[x]$. Prove that for all $p(x) \in F[x]$ that there exists some $\alpha(x), \beta(x)\in F[x]$ such that $p(x) = \alpha (x)f(x) + \beta (x)g(x).$
I think I have a legitimate proof for this, however I would like to check my thinking.
First, $f(x)$ and $g(x)$ are irreducible and have different degrees, then they are the polynomial equivalent to relatively prime numbers and therefore have a GCD of 1. I did use my definition about irreducible polynomials from the text, but just want to soldify thinking with something I'm already comfortable with.
Second, since the above GCD = 1, then we can say that $1 = a(x)f(x)+b(x)g(x)$.
Can I then multiply both sides of the equation by $p(x)$ to result in $p(x) = p(x)a(x)f(x) + p(x)b(x)g(x)$?
Does that then allow me to say that as long as we define $\alpha (x) = p(x)a(x)$ and $\beta(x) = p(x)g(x)$ and we know that $F$ is a field so is multiplicatively closed, then $\alpha(x), \beta(x) \in F$, that $p(x)=\alpha(x)f(x)+\beta(x)g(x)$?
Yes, if $p(x)$ and $q(x)$ are irreducible over the field $F$ and neither is a scalar multiple of the other their gcd is $1$, and then since $F[x]$ is a principal ideal domain you have Bézout's identity. In fact you just need $p(x)$ to be irreducible and not to divide $q(x)$ (or the same with $p$ and $q$ interchanged).