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?
When polynomials f(x) and f'(x) are relatively prime, f(x) has no repeated roots. Why?
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1Use Bezout's theorem. – 2011-04-19
2 Answers
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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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0But here $x-r$ doesn't have to be in $F[x]$. – 2011-04-19