Proving divisibility of a polynomial by a square of a polynomial.

Question

I need to prove that a polynomial $f \left( x \right) \in \mathbb{Q} \left[ X \right]$ is divisible by a square of a polynomial iff $f$ and $f'$ have a greatest common divisor of positive degree.

I have no idea where to start, except that I have to use the properties of Polynomial Rings as Euclidean Domain.

Answer 1

Hint: One direction is easy. For the other direction, note that $f$ and $f'$ have a greatest common divisor of positive degree iff $f$ and $f'$ have an irreducible common divisor.

Make sure to pinpoint where working over $\mathbb{Q}$ enters in your argument.

Answer 2

Let $q(x)$ be an irreducible common divisor of $f$ and $f'$. Division of $f$ by $q^2$ provides $$ f=kq^2+r, \quad \text{with}\,\,\,\deg r<\deg q^2 $$ and as $q$ divides $f$, then $q$ divides $r$, and thus $$ f=kq^2+r_1q, \quad \text{with}\,\,\,\deg r_1<\deg q. $$ Differentiating we obtain $$ f'=k'q^2+(2kq'+r_1')q+r_1q' $$ But since $q$ divides $f'$, then $q$ divides $r_1q'$, and since $q$ is irreducible, then $q$ divides $q'$or $r_1$. But, $\,deg q>\deg q'$, and hence $q$ does not divide $q'$, as $q'\ne 0$. Hence $q$ divides $r_1$, which implies that $r_1\equiv 0$, since $\deg r_1<\deg q$.