Explain why there are polynomials $p(x), q(x) \in \mathbb{R}[x]$ so that $f(x) = p(x)d(x), g(x) = q(x)d(x)$ and $\gcd[p(x), q(x)] = 1$.

Question

"Suppose that $f(x), g(x)$ are arbitrary polynomials in $\mathbb{R}[x]$, that are not both zero, and $d(x) = \gcd[f(x), g(x)]$. Explain why there are polynomials $p(x), q(x) \in \mathbb{R}[x]$ so that $f(x) = p(x)d(x), g(x) = q(x)d(x)$ and gcd$[p(x), q(x)] = 1$. "

My attempt:

"If $d(x) = \gcd[f(x), g(x)]$ then $d(x)\mid f(x)$ and $d(x)\mid g(x)$

Hence $f(x) = p(x)d(x)$ and $g(x) = q(x)d(x)$ for $p(x), q(x) \in \mathbb{R}[x]$

Suppose for a contradiction that $\gcd[p(x), q(x)] \not = 1$

Hence $\exists$ $a(x),s(x),t(x)$ with $\deg(a(x))\geq 1$ such that $p(x) = a(x)s(x)$ and $q(x)=a(x)t(x)$

Hence $f(x) = a(x)s(x)d(x)$ and $g(x) = a(x)t(x)d(x)$

Thus $\gcd[f(x), g(x)] = a(x) \not = 1 $"

I'm not sure if this is on the right track.

Answer 1

You are on the right track, and almost done. However, the last line isn't going anywhere.

"Hence $f(x) = a(x)s(x)d(x)$ and $g(x) = a(x)t(x)d(x)$
Thus $\gcd[f(x), g(x)] = a(x) \not = 1 $"

The conclusion should be that $a(x)d(x)|f(x)$ and $a(x)d(x)|g(x)$, thus by the definition of $\gcd$ for polynomials we have that $\deg(a(x)d(x))\leq \deg(d(x))$. This is a contradiction since $$\deg(a(x)d(x))=\deg(a(x))+\deg(d(x))=1+\deg(d(x))>\deg(d(x)).$$

It follows that $\gcd(p(x),q(x))=1$.

Answer 2

The technique is essentially what you see in my answer to this question except that polynomials rather than integers are used.