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I need to find cyclotomic cosets depending on $n=7$ and $q=4$ and find the factorization of $(x^7-1)$ into irreducible factors over $GF(4)$.

Thanks for any advice.

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Well, 1 is a root, so that's a start. –  Chris Eagle Oct 15 '12 at 15:39
    
So I have $(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$. –  James Oct 15 '12 at 15:46
    
Moreover, $x^7-1$ splits completely over $\mathbb{F}_8$, since its factorization in $\mathbb{F}_2$ is $(1+x)(1+x+x^3)(1+x^2+x^3)$. Note that there is no irreducible polynomial over $\mathbb{F}_2$ with degree $2$ that divides $x^7-1$. –  Jack D'Aurizio Oct 15 '12 at 15:49

1 Answer 1

As has been noted by Jack D'Aurizio in his comment, the polynomial $x^{7}-1$ splits into a product of $x-1$ and two different irreducible factors of degree $3$ over $F_{2}.$ This certainly gives the same factorization (but not a priori into irreducible factors) over $F_{4}.$ However $F_{4}$ and $F_{16}$ contain no element of multiplicative order $7,$ so contain no root of $x^{7}-1$ other than $1,$ so the two factors of degree $3$ remain irreducible in $F_{4}[x].$

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