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I'm working through a (non-examined) question sheet and have this problem.

Is $x^6+x^3+1$ irreducible over the following fields; (I) $\mathbb{F}_2$, (II) $\mathbb{F}_3$, (III) $\mathbb{F}_{19}$ and (IV) $\mathbb{Q}$?

I'm not sure how this problem should be approached. Here are my thoughts so far:

First note that $(x^6+x^3+1)(x-1)(x^2+x+1)=(x^9-1)$

So the roots of this polynomial are also roots of $x^9-1$, i.e. roots of the form $e^{\frac{2 \pi n i}{9}}$ (for certain n). However, I think this only helps us with $\mathbb{Q}$.

Mathmo
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    For $\mathbb{F}_3$ it also helps you since $x^9-1=(x-1)^9$ in that field. Or you can get it directly from $x^6+x^3+1=(x^2+x+1)^3$. – OR. Nov 04 '13 at 19:28
  • For the last one, you can see here. – Cameron Buie Nov 04 '13 at 19:29
  • In $\mathbb{F}{3}$ and $\mathbb{F}{19}$ you should be able to find a root of the polynomial. In $\mathbb{F}_{2}$ you could list all irreducible polynomials of degree $\leq 3$ and see if any of them divide your polynomial. – Arthur Nov 04 '13 at 19:31
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    The non-examined life is not worth living. Socrates. — ὁ δὲ ἀνεξέταστος βίος οὐ βιωτὸς ἀνθρώπῳ – Will Jagy Nov 04 '13 at 20:04
  • @Arthur: Can you give a hint as to what the root will be in $\mathbb{F}_{19}$? – Mathmo Nov 04 '13 at 23:35
  • One of $0,1,2,\ldots,18$. Try them all out until you find one of the roots. Alternatively, note that $(x^6 + x^3 + 1)(x-1)(x^2 + x + 1)(x^9 + 1) = (x^{18} - 1)$ reduces into linear terms in $\mathbb{F}_{19}$. – Arthur Nov 05 '13 at 02:21
  • See this question for a method for finding out whether a cyclotomic polynomial is irreducible over a finite field. This will cover the fields of two and three elements. – Jyrki Lahtonen Nov 05 '13 at 22:31
  • The multiplicative group of $\Bbb{F}_{19}$ is cyclic of order 18, so it has the maximum number of primitive ninth roots of unity. Hint: all the non-zero squares in that field have order that is a factor of nine. – Jyrki Lahtonen Nov 05 '13 at 22:32

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