1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Counting elements of a field and more

  1. Nov 5, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the field K=Z_2[X]/<x^3+x+1>. How many elements does this field have?

    Then let f(x) = x^3+x+1 and b be a root of f(x) in Z_2[X], Find all other roots of f(x) in K and describe the Galois group

    2. Relevant equations


    3. The attempt at a solution
    So i'm trying to understand what K looks like, perhaps something like this:
    K={ax^3 + bx^2 + cx + d | a,b,c,d are elements of Z_2 and x^3+x+1 = 0}

    Then is it possible that since x^3 = x + 1 mod 2 that I could make this substitution in the description of the field? i.e.:
    K={a(x+1) + bx^2 + cx + d | a,b,c,d are elements of Z_2 and x^3+x+1 = 0}
    ={ax+a + bx^2 + cx + d | a,b,c,d are elements of Z_2 and x^3+x+1 = 0}
    = {bx^2 + (c+a)x + (d+a) | a,b,c,d are elements of Z_2 and x^3+x+1 = 0}
    ={bx^2 + ex + f | a,b,c,d are elements of Z_2 and x^3+x+1 = 0}

    So there would be 8 elements maybe? Is this logic correct?
     
  2. jcsd
  3. Nov 6, 2016 #2

    fresh_42

    Staff: Mentor

    I think this gives you an upper bound for the number of elements in ##\mathbb{Z}_2[x] / (x^3+x+1)## as the reasoning:
    "Since ##x^3## can be expressed by lower powers of ##x##, we will at most get polynomials of degree ##2## and thus ##8## possible coefficient arrays."
    would do. But it's not quite obvious (to me), whether all possibilities will actually occur, i.e. no double counting took place. (My guess is it is so, but I don't see it yet.)

    Have you tried to decompose ##x^3+x+1\,##?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted