Given the definition; a polynomial f in the complex numbers is separable if it has only simple roots. My question is how do I prove f is separable if and only if f is relatively prime to its derivative.
Can someone help me go about doing this proof?
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This is not the correct statement! $f$ is separable if and only if the discriminant is not $0$. – diracdeltafunk Mar 11 '21 at 21:28
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2Anyway, the solution here depends quite a lot on your definition of "discriminant" -- the proof might be a single line or take lots of work. Please edit your post to include the definition you're working with, and what you've tried so far. – diracdeltafunk Mar 11 '21 at 21:29
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... you should not instead edit your post to make it a completely different question. This makes everything confusing to other readers in the future, since the current answers and comments make no sense. – diracdeltafunk Mar 11 '21 at 22:33
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One implication. – nowhere dense Mar 11 '21 at 22:40
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Hint: The discriminant of $f$ is (up to a constant) $\displaystyle \prod_{i \ne j} (\alpha_i - \alpha_j)$ where $\alpha_1, \cdots, \alpha_n$ are the roots of $f$.
Kenny Lau
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Recall that a polynomial is separable iff it doesn't have duplicate roots in the algebraic closure. Hence it is enough to prove that: $f\in \mathbb{C}[X]$ has only simple roots iff $\gcd_{\mathbb{C}[X]}(f,f')=1$.
Proof: $\gcd(f,f')\neq 1$ iff there is an $a\in K$ such that $x-a|f$ and $x-a|f'$. That is, iff $f(a)=f'(a)=0$. This happens when the first two coefficients of $$f(x)=f(a)+f'(a)(x-a)+\dots+\frac{f^{(k)}(a)}{k!}(x-a)^k$$ are $0$. Which is equivalent to $(x-a)^2|f(x)$.
As a remark, you can extend this proof to an algebraically closed field of positive characteristic by using the Hasse derivative in the expansion.
nowhere dense
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