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Prove that the polynomial $t^8+2t^6+4t^4+t^2+1$ is reducible in $\Bbb F_p$, for all $p\in \Bbb P$. Here are some examples: $t^8+2t^6+4t^4+t^2+1=(1 + t + t^4)^2\pmod{2}$ $t^8+2t^6+4t^4+t^2+1=(1 + t) (2 + t) (1 + t^2) (2 + 2 t^2 + t^4)\pmod{3}$ $t^8+2t^6+4t^4+t^2+1=(2 + t^2) (3 + 4 t^2 + t^6)\pmod{5}$ $t^8+2t^6+4t^4+t^2+1=(10 + 9 t + 3 t^2 + 2 t^3 + t^4) (10 + 2 t + 3 t^2 + 9 t^3 + t^4)\pmod{11}$

Thanks in advance.

azimut
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lsr314
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    See http://math.stackexchange.com/questions/160847/polynomials-irreducible-over-mathbbq-but-reducible-over-mathbbf-p-for. – Potato Mar 23 '13 at 19:02
  • @Potato:Thank you very much!Could you show me how to compute the galois group of $t^8+2t^6+4t^4+t^2+1=0$? – lsr314 Mar 24 '13 at 04:14
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    By hand? I do not know. The effective Chebotarev density theorem implies that you only need to check primes up to a certain bound in order to show an $n$-cycle never occurs, but I don't know what that bound is. It's probably very large. If all you want is the answer (and not a proof), I recommend using a computer algebra system. – Potato Mar 24 '13 at 06:35

1 Answers1

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According to Polynomials irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$ you should calculate galois group of this polynomial over $\mathbb Q$. If you ask Magma:

P:=PolynomialAlgebra(Rationals()); f:=x^8+2*x^6+4*x^4+x^2+1; G:=GaloisGroup(f); print G;

then you have

Permutation group G acting on a set of cardinality 8
Order = 192 = 2^6 * 3
(1, 2, 3, 4)(5, 6, 7, 8)
(1, 2)(5, 6)
(1, 6)(2, 5)
(1, 6, 7, 8)(2, 3, 4, 5)

Now you should check that there are no 8-cycles in this group. Most simple way is to generate all 192 elements and check.

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Mikhail Ivanov
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