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Constructing a splitting field for polynomial over F_5

  1. Nov 28, 2016 #1
    1. The problem statement, all variables and given/known data
    f(x) = x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x where the coefficients are elements of F_5. Show that this polynomial is divisible by x^5-x and construct a splitting field L for f over F_5 and computer [L:F_5]

    2. Relevant equations


    3. The attempt at a solution
    So the first thing I did was turn all the negative coefficients positive so that f(x) = x^7 + 3x^6 + 3x^5 + 4x^3 + 2x^2 + 2x and we want to divide this polynomial by x^5+4x. Upon doing the long division, I get a quotient x^2 + 3x + 3 with remainder 4x^3... I checked my steps and I got the same answer with the same remainder, can anyone check my work here?
     
  2. jcsd
  3. Nov 28, 2016 #2

    fresh_42

    Staff: Mentor

    ##x^2+3x+3## is correct, without remainder. You don't need to convert the coefficients. The equations keep being true, even outside the standard representation of ##\mathbb{F}_5.## I only used it to simplify the roots of this last polynomial of degree ##2,## e.g. ##\frac{1}{2}=3.##
    After that one can write down all ##7## roots which makes the splitting field quite obvious.
     
  4. Nov 28, 2016 #3
    Yes, excellent. And yet I am still having trouble finding all the roots. So the polynomial reduces to ##(x^2+3x+3)(x^5+4x)=(x^2+3x+3)(x^4+4)x. So the first three roots are the solutions to the quadratic and zero, are the other 4 roots c*i(4)^1/4 where c is a primitive fourth root of unity?
     
  5. Nov 28, 2016 #4

    fresh_42

    Staff: Mentor

    You can factor ##x^4+4##. It would be more obvious if you had kept the sign: ##x^4+4=x^4-1##. Now use ##x^4=(x^2)^2##.
    And ##x^2+3x+3=0## can be done by the usual formula. The only critical point in the formula is ##\frac{1}{2}## but this is simply ##3##.
     
  6. Nov 29, 2016 #5
    are the solutions to ##x^4-1=0## the powers of a primitive fourth root of unity? i.e. the powers of i? even in ##F_5##?
     
  7. Nov 29, 2016 #6

    fresh_42

    Staff: Mentor

    Not so complicated. Just remember the formula ##(x-a)(x+a)=x^2-a^2##. This gives you ##(x^2-1)(x^2+1)## in our case. The first factor splits again according to the same formula. To split the second, now write ##x^2+1=x^2-4## and consider ##2\cdot 2=4## even in ##\mathbb{F}_5##.

    Edit: The fourth roots of unity will work as well, but with the way above, it's easier to see, that ##i=2## in ##\mathbb{F}_5##.
     
    Last edited: Nov 29, 2016
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