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    Hello
    my question is

    Show that \alpha=[x] is a primitive element of GF(8)= \mathbb{Z}_2[x]/<x^{3}+x^{2}+1>
    I have already worked out that the function in the < > is irreducible but I do not know where to go from this.

    please help me would be appreciated thanks
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    To be honest, I'm not totally sure what you're supposed to do here. The multiplicative group has 7 elements, which is prime, and so by Lagrange any element other than the identity will generate it.
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    how do you know it has seven elements? sorry, im not very good at this sort of stuff!! i have a vague understanding of lagrange, but do not really know the steps to showing this.

    thanks for the reply too
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    The multiplicative group is every element other than 0. Since there are 8 elements total, removing 0 leaves 7 non-zero elements.
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    The elements of GF(8) are the cosets of the possible remainders on division by x^{3}+x^{2}+1, right?

    How do you know what the possible remainders are though? is there a technique to this??
    thanks again
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    The remainders are simply the set of polys of degree < 3.
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    Would they be:

    0
    1
    x^{2}
    x^{2}+1

    but i cant think of anymore elements?
    would i use just x's?

    like:
    x+1
    x
    x^{2}+x
    x^{2}+x+1


    thanks for your help its much appreciated
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    The remainders are the 8 polys listed above.
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    Thanks !!
 
 
 
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