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Math Help - Proving a primitive element

  1. #1
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    Proving a primitive element

    Hi, I need to show that \alpha+1=[x] is a primitive element of GF(9)= \mathbb{Z}_3[x]/<x^{2}+x+2>
    I have already worked out that the function in the < > is irreducible but I do not know where to go from this.

    there are 8 elements in the multiplicative group, what would they be? I guess it would be: 0, 2, x+2, what else? Im very unsure how to do this.

    please help me would be appreciated thanks
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  2. #2
    MHF Contributor Swlabr's Avatar
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    The elements of your multiplicative group are,

    1, 2, x, x+1, x+2, 2x, 2x+1 and 2x+2. Not zero (as then your multiplicative group would not be a group...). Do you see where I got these elements from? (Strictly speaking, I should have brackets around these, like [1], [2], etc, but that doesn't really matter.)

    Your identity is 1.

    So, what is the order of x+1 (I presume that is the element you are trying to show is primitive)? That is, is (x+1)^2=1? (x+1)^4=1? What can you conclude?
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  3. #3
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    right, in Z3[x]/(x^2 + x + 2) = Z3[x]/I, we really ought to write x + I, or α instead of x, but it is a common abuse of notation.

    let's start with (α+1)^2 = α^2 + 2x + 1 = (-α - 2) + 2α + 1 = 2α + 1 + 2α + 1 = α + 2.

    make life easier on yourself. GF(9) is a field with 9 elements. thus the multiplicative group of GF(9) is cyclic of order 8.

    so what are the possible orders of α+1? if there are n possibilities, can you see it suffices to just check n-1 of them?

    basically we identify GF(9) with Z3[α] where α is a root of x^2 + x + 2. since Z3[α] is an extension field of Z3, it is a vector space

    of dimension deg(x^2 + x + 1) = 2 over Z3, with basis {1, α}. we can explictly list the elements of Z3 x Z3:

    (0,0) <--> 0 + 0α = 0
    (1,0) <--> 1 + 0α = 1
    (2,0) <--> 2 + 0α = 2
    (0,1) <--> 0 + 1α = α
    (1,1) <--> 1 + 1α= 1 + α
    (2,1) <--> 2 + 1α = 2 + α
    (0,2) <--> 0 +2α = 2α
    (1,2) <--> 1 + 2α
    (2,2) <--> 2 + 2α (since vector addition is abelian, we can re-write c+dα as dα + c, if we wish).

    the multiplicative group elements are all the non-zero ones.
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