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Finding a splitting field over Z_13

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Describe the splitting field of x^9-1 over Z_13.

    2. Relevant equations


    3. The attempt at a solution
    Well, seeing if any other of {1,......,12} would be some tricky arithmetic. Is my best bet here moving forward to just divide x^9-1 by x-1 since I know 1 is a root and then go from there?

    So basically i'm looking for elements in Z*_13 with orders of 1,3 or 9. But there will be no element of order 9 because 9 does not divide 12 ( the order of Z*_13) so i'm looking for all elements of order 3. This makes checking to see if {1,.....,12} work much easier.

    Also, there since 3 is prime there is a subgroup of order 3, so there are at least 2 elements such that x^3=1 (mod 13). I have found these elements, 3 and 9, therefore I have 3 out of the 9 roots. What do I do from here? Divide x^9-1 by (x-1)(x-3)(x-9) and then divide the quotient by (x-r)? Is there a simpler way?
     
    Last edited: Oct 30, 2016
  2. jcsd
  3. Oct 30, 2016 #2

    fresh_42

    Staff: Mentor

    Since ##\mathbb{Z}_{13}## is a field, do you expect any differences as if it was ##\mathbb{Q}##?
    ##1## is a root, as you've said, and you know what ##(x^9-1):(x-1)## is. What happens if you take a primitive 9-th root of unity, say ##\xi## and consider the powers ##\xi^i##?
     
  4. Oct 30, 2016 #3
    Yeah, char(Q) = 0 so there will be no repeated roots, here there will be repeated roots. I'm not exactly sure how to reduce a primitive 9-th root of unity mod 13. I'm pretty sure the only roots of x^9-1 in Z_13 are 1,3 and 9 though?
     
  5. Oct 30, 2016 #4

    fresh_42

    Staff: Mentor

    Then you should compute ##q(x)=(x-1)(x-3)(x-9)## and ##(x^9-1):q(x)## and see what is left.
     
  6. Oct 31, 2016 #5
    q(x) mod 13 is equivalent to x^3-1. (x^9-1)/(x^3-1) = x^6+x^3+1. In hindsight, it really should have been obvious that q(x) would reduce to x^3-1. Anyway, now i'm looking for the roots of f(x)=x^6+x^3+1, and since i've already found all the roots that are in Z*_13, the roots of f(x) must be in a some splitting field. I will now divide f(x) by x-r.

    f(x)/(x-r) = x^5+(r)x^4+(r^2)x^3+(r^3)x^2+(r^4)x+r^5. I feel this did not help me very much. Perhaps I should go back and try another strategy at finding the roots of x^6+x^3+1, my gut is telling me something to do with some kind of primitive root of unity. Any insights?
     
  7. Oct 31, 2016 #6

    fresh_42

    Staff: Mentor

    You will need only one element ##\xi## to adjoint. To split ##x^6+x^3+1## you could temporarily substitute ##y=x^3## and solve the quadratic equation. Since ##2## is a unit, you'll have to find ##2^{-1}## and use the known formula. Also the tables of ##a^2\, , \,a^3## for ##a \in \mathbb{Z}_{13}## are helpful.

    If you'll have the final splitting of ##x^9-1## it would be fine to see it so I can compare it with my solution.
     
    Last edited: Oct 31, 2016
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