Title: Relatively prime polynomials in $k[x]$
Tags: abstract-algebra
Question: suppose $k$ is a field and take two polynomials $p(x),g(x) \in k[x]$. If $(p,p&#39;)=1$ and $(g,g&#39;)=1$ then why is it true that $(pg,(pg)&#39;)=1$ where $&#39;$ denotes formal derivative and $(,)$ denotes gcd (i.e (pg)' denotes the derivative of pg). I tried writing the gcd as linear combination and using the chain rule but I don't see it. Can you please help me


Here is another answer: Note that $(p,p&#39;) = 1$ if and only if $p$ has no repeated roots (in the algebraic closure of $k$); so the conditions $(p,p&#39;)=1$ and $(g,g&#39;)=1$ mean that neither $p$ nor $g$ have repeated roots. However, it does not follow from this that $pg$ has no repeated roots, which is why the counterexample Mariano gives works: you could have a common factor to both $p$ and $g$ which will be a repeated factor of $p$ and $g$.
On the other hand, if $(p,p&#39;)=(g,g&#39;)=1$, then $(pg,(pg)&#39;)=1$ if and only if $(p,g)=1$. Because the only repeated factors in $pg$ must be common factors to both $p$ and $g$, and conversely, any common factor to $p$ and $g$ will be a repeated factor of $pg$. So $pg$ has no repeated factors if and only if $p$ and $g$ have no common factors (given the assumption that neither $p$ nor $g$ have repeated factors).

Here is another answer: Except for $\rm\:(pg)'=\: p'g+pg'\:,\:$ this has nothing to do with derivatives. Generally we have this
Theorem $\ $ Suppose $\rm\:p,p',g,g'$ are elements of a GCD domain with $\rm\:(p,p') = 1 = (g,g').\:$
Then $\rm\quad\ (p'g+pg',\ pg) = 1\ \iff\ (p,g) = 1\:.$
Proof $\rm\ \ (p'g+pg',\ pg)$
$\rm\quad\quad =\ \ (p'g+pg',\ (p,\:p'g+pg')\ (g,\:p'g+pg'))$
$\rm\quad\quad =\ \ (p'g+pg',\ (p,p'g)\ (g,pg'))$
$\rm\quad\quad =\ \ (p'g+pg',\ (p,g)\ (g,p))\ $ [$=1\ $ if $\rm\ (p,g) = 1$]
Conversely $\rm\:(p,g)\ |\ (p'g+pg',pg) = 1\ $ so $\rm\:(p,g)=1\ \ $ QED

Here is another answer: It is not true: consider the example where $p=g=x$.

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