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The question as written in the title:-

Irreducible polynomial over $\mathbb{Q}$ can not have repeated root in $\mathbb{C}$.

My attempt:- I know is $g=\gcd(f,f^{'})$ is not constant and hence separable. but how to go from here?

Jyrki Lahtonen
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henry
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2 Answers2

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If $f$ is irreducible and has a multiple root $\alpha \in \mathbb{C}$ then $\alpha$ is also a zero of $f'$. Hence $f$ divides $f'$. But $deg(f')<deg(f)$ thus $f'=0$. This implies that $f$ is constant.

math635
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    Why is this incorrect? f is irreducible and $\alpha$ is a root of $f$. So $f$ is the minimal polynomial(up to unit) of $\alpha$ and it divides any polynomial which has $\alpha$ as a root. See https://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory) – math635 Dec 26 '15 at 02:24
  • @PedroTamaroff this is certainly correct. If f is irreducible and alpha is a root of f, then f divides any polynomial g such that alpha is a root of g. The proof proceeds by using this fact to deduce that f | f' and hence f' must be zero. – hunter Dec 26 '15 at 02:50
  • I misunderstood where you were going with the proof. My apologies. I was thinking more in the lines of orangeskid's answer. – Pedro Dec 26 '15 at 02:56
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Almost there, if $f$ non-zero had a multiple root $a$ then $a$ would be a root of $f'$ too, so $(X-a)$ divides both $f$ and $f'$, so it divides $g = \gcd(f,f')$ that is not therefore a constant polynomial, and moreover, has degree $\le \deg f' < deg f$. So $g\in \mathbb{Q}[X]$ is a proper factor of $f$ and therefore, $f$ is reducible.

${\bf Added:}$ So if $f = c \prod (X-a_i)^{m_i}$ ( $ c \ne 0$ ) then $g = \prod (X-a_i)^{m_i-1}$ and $g \in \mathbb{Q}[X]$ ( the fact that $g \in \mathbb{Q}[X]$ is a consequence of Euclid's algorithm that takes place in $\mathbb{Q}[X]$ since both $f$ and $f'$ are there).

orangeskid
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