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Math Help - Irreducible Polynomials f(x) = x^7 - x

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    Irreducible Polynomials f(x) = x^7 - x

    in Q[x], Z/2[x], and Z/7[x].

    in Q[x] we think its x(x-1)(x+1)(x^2-x+1)(x^2+x+1)

    in Z/2[x] we think its x(x-1)(x+1)

    and we have no idea about Z/7[x] we've tried using Little Fermats Theorem.

    THANKEEE YOU!!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Irreducible Polynomials f(x) = x^7 - x

    Quote Originally Posted by rpietrov View Post
    in Q[x], Z/2[x], and Z/7[x].

    in Q[x] we think its x(x-1)(x+1)(x^2-x+1)(x^2+x+1)

    in Z/2[x] we think its x(x-1)(x+1)

    and we have no idea about Z/7[x] we've tried using Little Fermats Theorem.

    THANKEEE YOU!!
    The polynomial P(x)= x^{7}-x contain in any case x. The other roots satisfy the equation...

    x^{6} \equiv 1\ , \text{mod}\ 7 (1)

    ... or, equivalently, one of the equations...

    x^{3} \equiv 1\ , \text{mod}\ 7 (2)

    x^{3} \equiv -1\ , \text{mod}\ 7 (3)

    The roots of (2) are 1, 2 and 4, the roots of (3) are -1, -2 and -4 so that is...

    x^{7}-x= x\ (x+1)\ (x-1)\ (x+2)\ (x-2)\ (x+4)\ (x-4) (4)

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Irreducible Polynomials f(x) = x^7 - x

    Quote Originally Posted by rpietrov View Post
    in Q[x] we think its x(x-1)(x+1)(x^2-x+1)(x^2+x+1)
    Right.


    in Z/2[x] we think its x(x-1)(x+1)
    Try again, it should be x^7-x=x(x+1)^2(x^2+x+1)^2 .
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