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I would like to prove that $P(x)=-x^5+10x^3-20x+12$ is irreducible over $\Bbb{Q}$

I tried using Eisenstein's criterion, reduction and factorizing but I didn't succeed, any idea ?

Adi Dani
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JeSuis
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  • What reductions have you tried? – Mark Bennet Oct 11 '16 at 08:49
  • @MarkBennet I tried modulo $2$ but I get $-x^5$ and modulo $3$ but I failed seeing why $-X^5+X^3-2X$ is irreducible, perhaps I am misleading somewhere. – JeSuis Oct 11 '16 at 08:54
  • Have you eliminated the possibility of a linear factor? – Mark Bennet Oct 11 '16 at 09:02
  • @MarkBennet I try it again. – JeSuis Oct 11 '16 at 09:03
  • @JeSuis $-X^5+X^3-2X$ is not irreducible in $\Bbb F_3[X]$ : it's clearly a multiple of $X$. If you want to prove irreducibility of a monic polynomial by reducing $\pmod p$, you have to pick a $p$ that does not divide the constant term. –  Oct 11 '16 at 09:19
  • @G.Sassatelli arf right... Thansk for the advise! – JeSuis Oct 11 '16 at 09:20
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    Further hint: $x^5+x^3-2x-1$ is irreducible over $\mathbb{F}_{11}$. – Jack D'Aurizio Oct 11 '16 at 10:19
  • @JackD'Aurizio thanks, can you please look at http://math.stackexchange.com/questions/1956386/explicit-expression-minimal-polynomial-which-is-equal-to-characteristic ? If you want only, thanks. – JeSuis Oct 11 '16 at 10:51

2 Answers2

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By the rational root theorem, $f(x)$ has no rational roots. For $f(p/q)=0$ would imply $p\mid 12$ and $q\mid 1$, and no such value works. By the Gauss Lemma it is enough to show that $f(x)$ is irreducible over $\mathbb{Z}$. Then $$ f(x)=(x^3+ax^2+bx+c)(-x^2+dx+e) $$ leads to a contradiction over $\mathbb{Z}$.

Dietrich Burde
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Eisenstein's criterion does apply to $f(-x+2)=x^5-10x^4+30x^3-20x^2-20x+20$ with respect to $p=5$.

Sil
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