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Suppose that $p(x)=x^9+x^8+x^4+x^2+1 \in \mathbb{Z}_2[x]$. I have to show this polynomial is irreducible.

user26857
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armin
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    "A manual check shows that this polynomial is not divisible by any of the 256 polynomials of lower degree. Therefore, it is irreducible."

    What have you tried so far?

     – Eoin Apr 01 '15 at 06:57

1 Answers1

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If the polynomial is reducible, it must have an irreducible factor of degree at most $4$. It is clear by the Factor Theorem that $p$ has no factors of degree $1$. There is one irreducible quadratic in ${\Bbb Z}_2[x]$, there are two irreducible cubics and three irreducible quartics. Try them all.

Source for numbers: well known, or http://oeis.org/A001037.

Comment. As we are working in ${\Bbb Z}_2[x]$ with polynomials of degree less than $10$, you can do the calculations by counting on your fingers. This is not a joke, it really is possible - try it!!!

David
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  • why it must have an irreducible factor of degree at most 4? – armin Apr 01 '15 at 07:27
  • can you explain more? – armin Apr 01 '15 at 07:30
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    @armin If it has an irreducible factor of degree $>4$, the quotient has degree $\le 4$, so you'd already have found its irreducible factors. It's the same as when we look for prime factors of an integer $n$: if we reach above $\sqrt{n}$ without finding one, then we conclude that $n$ is prime. – egreg Apr 01 '15 at 07:32
  • can you explain more about there is one irreducible quadratic and there are two irreduvible cubics? – armin Apr 01 '15 at 08:00
  • There is a good discussion at http://math.stackexchange.com/questions/32197. – David Apr 01 '15 at 08:21
  • @ thank's a lot – armin Apr 01 '15 at 09:20