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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1You can look at the factorization of $f$ in some algebraic closure $\overline{F}$ of $F$. – bzc Apr 19 '11 at 21:42
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1Use Bezout's theorem. – Qiaochu Yuan Apr 19 '11 at 22:05
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HINT $\rm\ \ g^2\ |\ f\ \Rightarrow\ g\ |\ gcd(f,f{\:'})\ $ since $\rm\ (g^2\:h)'\:=\ g\ (g\:h' + 2\:g'\:h)\:.$
Bill Dubuque
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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 simple 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 $h(x)$.
Afntu
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davidlowryduda
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