$x^6+x^3+1$ is irreducible over $\mathbb{Q}$

Question

I have been trying to prove that $x^6+x^3+1$ is irreducible over $\mathbb{Q}$ (or $\mathbb{Z}$ since by Gauss' Lemma is the same), but I can't. Any idea of how to do so?

score 9 · Accepted Answer · answered Nov 01 '13 at 18:06

9

HINT: Let $y=x-1$, and apply Eisenstein's criterion for $p=3$.

answered Nov 01 '13 at 18:06

asatzhh

501

2

Thanks! How did you figure out that that substitution would work? – Francisco Martínez Nov 01 '13 at 18:22
HINT:$f(x)=g(x)h(x)\implies f(x+a)=g(x+a)h(x+a),\forall a\in \mathbb{Z}$. – asatzhh Nov 01 '13 at 18:26
@FranciscoMartínez The inspiration is to modify the proof that $x^p + x^{p-1} + \ldots + x +1$ is irreducible. If have you seen that one before, this seems a natural thing to do here. If you haven't, now you have a hint on how to do that problem :) – Ragib Zaman Nov 02 '13 at 12:13

score 5 · Answer 2 · answered Nov 01 '13 at 18:05

5

It's the $9$-th cyclotomic polynomial, and all cyclotomic polynomials are irreducible.

answered Nov 01 '13 at 18:05

Luca Bressan

6,845

score 3 · Answer 3 · answered Nov 01 '13 at 18:26

3

The generalised form of Cohn's criterion also works: all of the coefficients are non-negative and smaller than $2$, and $P(2) = 73$ is prime.

answered Nov 01 '13 at 18:26

Peter Taylor

13,425

$x^6+x^3+1$ is irreducible over $\mathbb{Q}$

3 Answers3

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