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How can I prove that $x^5+6x^3+x^2+3x+2$ is irreducible in $\mathbb{Q}[x]$?

I tried with Eisenstein (also making the substitution $x\mapsto x-1$ and $x\mapsto x+1$ to see if I obtain an Eisenstein polynomial) but nothing. If we go modulo 2 we obtain a reducible polynomial, and modulo 3 we obtain $x^5+x^2+2$ and I don't know how to prove that it is irreducible.

Any help?

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There is always brute force. Modulo 3 there are only so many polynomials it could reduce to a product of, so you might check them all. –  Jon Beardsley Dec 30 '11 at 3:18
    
It's irreducible in $\mathbb F_{41}$. –  user16697 Dec 30 '11 at 3:18
    
@JBeardz - Is the brute force the only way to solve this problem? –  Victor Dec 30 '11 at 3:19
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mod 3, isn't $x^5 + x^2 + 2 \equiv 2(x^2+1)$? $x^2+1$ is irreducible over $\mathbb{Q}$... the equivalence is by Fermat –  mathmath8128 Dec 30 '11 at 3:21
    
@Victor No not at all. –  Jon Beardsley Dec 30 '11 at 3:27

1 Answer 1

up vote 10 down vote accepted

By the rational root theorem the only possible rational roots are $\pm 1, \pm 2$, and by inspection none of these are roots. If the polynomial is reducible, it therefore factors into the product of a quadratic and cubic factor (over $\mathbb{Z}$ by Gauss's lemma).

$\bmod 2$ the polynomial factors as $x(x^4 + x + 1)$. The latter factor has no root $\bmod 2$, so if it is reducible it is the product of two irreducible quadratics. But the only irreducible quadratic $\bmod 2$ is $x^2 + x + 1$, and $(x^2 + x + 1)^2 = x^4 + x^2 + 1$. Hence $x^4 + x + 1$ is irreducible $\bmod 2$.

But if the polynomial factored as the product of a qudaratic and cubic factor over $\mathbb{Z}$, it would only have at most cubic irreducible factors $\bmod 2$; contradiction. Hence the polynomial is irreducible.

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Scooped me by a minute...*precisely* the same answer! –  Bill Dubuque Dec 30 '11 at 3:28

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