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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?

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

2 Answers 2

up vote 6 down vote accepted

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

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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 $h(x)$.

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

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