Show for univariate relatively prime polynomials over a perfect field, that $\gcd(fg,(fg)')=\gcd(f,f')\gcd(g,g')$.

Question

Let $K$ be a perfect field, and let $f,g\in K[x]$ be two univariate polynomials with $\gcd(f,g)=1$. I want to show that $\gcd(fg,(fg)')=\gcd(f,f')\gcd(g,g')$.

(I don't know if the assumption, that the field is perfect, is neccesary.)

What I have tried so far:

First we note that $(fg)'=f'g+fg'$. We see that $\gcd(f,f')\gcd(g,g')$ is clearly a divisor for $fg$, $f'g$ and $fg'$. It follows that it is a common divisor for $fg$ and $(fg)'$. Therefore, we have that there exists $h\in K[x]$ such that $\gcd(fg,(fg)')=\gcd(f,f')\gcd(g,g')h$. Now I want to show that $h$ is forced to be a unit, but I can't figure out how.

Any help would be appreciated.

Answer 1

I decided to revisit this question and found a solution.

We will make use of the following two lemmas:

1. $\text{gcd}(f,g)=1 \;\Rightarrow \; \text{gcd}(f,p)=\text{gcd}(f,pg)$ for any polynomial $p$.
2. $\text{gcd}(f,g)=1 \;\Rightarrow \; \text{gcd}(p,fg)=\text{gcd}(p,f)\text{gcd}(p,g)$ for any polynomial $p$.

We now do the computations: \begin{align*} \text{gcd}((fg)',fg)&\overset{2.}{=}\text{gcd}((fg)',f)\text{gcd}((fg)',g)=\text{gcd}(f'g+fg',f)\text{gcd}(f'g+fg',g) \\ &=\text{gcd}(f'g,f)\text{gcd}(fg',g)\overset{1.}{=}\text{gcd}(f',f)\text{gcd}(g',g). \end{align*}

proof of lemma 1: $$ \text{gcd}(f,p)=\text{gcd}(f,p\cdot\text{gcd}(f,g))=\text{gcd}(f,pf,pg)=\text{gcd}(f,pg). $$ proof of lemma 2: Recall we are working over an UFD. Since $f,g$ are relatively prime, we get that they have no common factors in their decomposition into irreducibles. The result then follows from looking at the decompositions on both sides of the equality.