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FInd non-zero elements are primitive in a field

  1. Jun 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Construct $\mathbb{F}_{16}$ as a quotient of $\mathbb{Z}_2[X]$. How many non-zero elements are primitive in this field? Calculate $|GL2_(\mathbb{F}_16)|$.

    2. Relevant equations

    Primitive Theorem

    3. The attempt at a solution

    For the first question, I don't know how to construct $\mathbb{F}_{16}$ as a quotient of $\mathbb{Z}_2[X]$. My guess is $\mathbb{Z}_2[X]/p(x)$ where $p(x)$ is not in $\mathbb{Z}_2[X]$.

    And the second part which should be 16^4-16^3
     
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  3. Jun 3, 2015 #2

    Fredrik

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  4. Jun 3, 2015 #3

    micromass

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    Right, you will have to construct ##\mathbb{F}_{16}## as ##\mathbb{Z}_2[X]/(p(x))## for some polynomial. Now there are two essential facts:

    1) The quotient will have to be a field
    2) The quotient will have to have 16 elements.

    Which conditions on ##p(x)## will guarantee this?
     
  5. Jun 3, 2015 #4

    micromass

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    If ##p(x)## is not in ##\mathbb{Z}_2[X]##, then the quotient makes no sense. So that is not correct.
     
  6. Jun 3, 2015 #5
    P (x) would be a maximal ideal and irreducible, I think x^3-x-1
     
  7. Jun 3, 2015 #6

    micromass

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    ##p(x)## is just a polynomial, and a polynomial being a maximal ideal makes no sense. You are correct that ##p(x)## will have to be irreducible. Indeed, if ##p(x)## is irreducible, then

    [tex]\mathbb{Z}_2[X]/(p(x))[/tex]

    will be a field, which is what you want. But you also want the field to have ##16## elements, under what conditions on ##p(x)## will that happen?

    That is not correct.
     
  8. Jun 3, 2015 #7
    x^2-2,
    If I didn't make mistakes, p (x)=x^2 - 3
     
  9. Jun 3, 2015 #8

    micromass

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    Can you stop guessing and actually think about it? In either case, if you put up a proposal, at least try to motivate it and say why it is true.
     
  10. Jun 3, 2015 #9
    P (x)=x^4+1, we have p (0) mod 2 =1 and p (1)=2 which show p (x) is irreducible in Z_2, hence F_16 isomorphic with Z_2 [x]/p (x)
     
  11. Jun 3, 2015 #10

    micromass

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    First of all, ##p(1) = 0## since ##2=0## in ##\mathbb{Z}_2##.
    Second of all, saying that the polynomials ##p(x)## has no roots in ##\mathbb{Z}_2## is not at all the same as saying that it is irreducible.
    Third of all, even if ##p(x)## were irreducible, that would only show that ##\mathbb{Z}_2[X]/(p(X))## is a field, you still need to find an argument for why it has ##16## elements.
    Fourth of all, ##\mathbb{Z}_2[X]/p(X)## makes no sense. You cannot quotient out an element of ##\mathbb{Z}_2[X]##, you can only quotient out an ideal.
     
  12. Jun 3, 2015 #11

    SammyS

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    Sorry, but I was having difficulty reading the OP, so here it is with readable Latex:
    Construct ##\ \mathbb{F}_{16}\ ## as a quotient of ##\ \mathbb{Z}_2[X]\ .\ ## How many non-zero elements are primitive in this field? Calculate ##\ |GL2_{({{\mathbb{F}}_{16}})}|\ ## .

    2. Relevant equations
    Primitive Theorem

    3. The attempt at a solution
    For the first question, I don't know how to construct ##\ \mathbb{F}_{16}\ ## as a quotient of ##\ \mathbb{Z}_2[X]\ ##. My guess is ##\ \mathbb{Z}_2[X]/p(x)\ ## where ##\ p(x)\ ## is not in ##\ \mathbb{Z}_2[X]\ ##.

    ... and there it is.
     
    Last edited: Jun 3, 2015
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