7
$\begingroup$

The problem is to show that a polynomial $f(x) \in F[x]$ (F is a field) has no repeated roots if and only if f(x) and f'(x) (the derivative of f(x)) are relatively prime. I've managed to prove one direction of this equivalence (if there are no repeated roots, f(x) and f'(x) are relatively prime), but the other one gives me headaches... can anyone help out?

$\endgroup$
  • 1
    $\begingroup$ You can look at the factorization of $f$ in some algebraic closure $\overline{F}$ of $F$. $\endgroup$ – Brandon Carter Apr 19 '11 at 21:42
  • 1
    $\begingroup$ Use Bezout's theorem. $\endgroup$ – Qiaochu Yuan Apr 19 '11 at 22:05
10
$\begingroup$

HINT $\rm\ \ g^2\ |\ f\ \Rightarrow\ g\ |\ gcd(f,f{\:'})\ $ since $\rm\ (g^2\:h)'\:=\ g\ (g\:h' + 2\:g'\:h)\:.$

$\endgroup$
1
$\begingroup$

So we are to prove that if f(x) and f'(x) are relatively prime, then there are no repeated roots. Let us consider the contrapositive: if there is a repeated root, then f(x) and f'(x) are not relatively prime. This, I think, is very simply to prove.

Denote the repeated root by r. Then $f(x) = (x-r)g(x)$ and $f'(x) = (x-r)h(x)$, for some $g(x)$ and some $h(x)$.

$\endgroup$
  • $\begingroup$ But here $x-r$ doesn't have to be in $F[x]$. $\endgroup$ – user9077 Apr 19 '11 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.