Polynomial factors over a field with 0 characteritics

Question

Let $\mathbb K$ be a field, with $\mathrm{Char}(\mathbb K) = 0$, and $P\in \mathbb{K}[X]$ irreducible. Let $Q\ \in \mathbb{K}[X]$ such that $\gcd(P,Q) =1$. I'd like to show that $P$ does not divide $P'Q$ (or find counter examples if this is not correct).

It looks simple, so I guess I'm missing something obvious.

Thank you

Answer 1

If by the pretty weird notation $\;P\wedge Q=1\;$ you meant $\;P,Q\;$ are coprime, then you can argue as follows:

$$g.c.d.(P,Q)=1\implies \exists f(x),g(x)\in\Bbb K[x]\;\;s.t.\;\;f(x)P(x)+g(x)Q(x)=1$$

Now, suppose

$$P\mid P'Q\implies P'Q=m(x)P(x)\;,\;\;m(x)\in\Bbb K[x]\,,\;\;\text{but then}$$

$$P'=P'\cdot 1=fPP'+gP'Q=fPP'+gmP$$

and since $\;P\mid fPP'+gmP\;$ , then also $\;P\mid P'\;$ , which is absurd as we're in characteristic zero...

Answer 2

This is a simple exercise in the algebra of the $\gcd$ function. Assuming otherwise, we have:

$$ P = \gcd(P, P'Q) = \gcd(P, P') $$

because $\gcd(P,Q) = 1$, and so we would have $P | P'$, which can happen if and only if $P' = 0$.