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Finding splitting field and Galois group questions

  1. Dec 5, 2016 #1
    1. The problem statement, all variables and given/known data
    2 questions here:

    1)
    Let g(x) = x^6 - 10 be a polynomial in Q(c) where c is a primitive 6th root of unity. Find a splitting field for this polynomial and determine it's Galois group

    2)
    let f(x) = x^3 + x^2 + 2 with coefficients in ##F_3##. Find a splitting field K for this polynomial and determine it's Galois group

    2. Relevant equations


    3. The attempt at a solution
    Question 1)
    If we adjoin b = the 6th real root of 10 to Q(c) then we have a splitting field. Furthermore, the roots of f will be {b, bc, bc^2, bc^3, bc^4, bc^5}. The Galois group will permute the roots whilst fixing every element of Q(c). So I believe this galois group will be isomorphic to ##(Z_6,+)## where the generator will take b-->bc and fix all elements of Q(c).

    Question 2)
    I'm having trouble constructing a splitting field for this polynomial. If I assume r is a root and then divide f(x) by x-r I don't seem to arrive at anything helpful. Perhaps I should consider that the multiplicative group in F_7(r) will be cyclic of order 26 (since the degree of r is 3 and 3*9 = 27) so r^27=r, and then use this knowledge to see if any powers of r are also roots. Does this make sense? Does anyone else have any other insights to make this simpler?
     
  2. jcsd
  3. Dec 7, 2016 #2

    fresh_42

    Staff: Mentor

    I think this is correct. Only one remark: you can say ##g(x)## is a polynomial over ##\mathbb{Q}(c)## or in ##\mathbb{Q}(c)[x]##.
    I don't understand why you consider ##\mathbb{F}_7## here or even ##\mathbb{Z}_{26}##.
    If you divided ##f(x)=x^3+x^2+2## by ##(x-r)## with a root ##r## such that ##r^3+r^2+2=0## then you can also do the rest and split ##f(x)## completely. If I made no mistake, then you will need ##\sqrt{r+1}## in the splitting field. Together with ##r^2## this looks like a four-dimensional extension of ##\mathbb{F}_3## but I didn't check linear dependencies.
     
  4. Dec 8, 2016 #3
    I divided by x-r and then solved the resulting quadratic and calculated that it has a discriminant -3r^2 - 2r + 1. it appears our splitting fields will be a bit different.
     
  5. Dec 8, 2016 #4

    fresh_42

    Staff: Mentor

    I don't think so. ##-3=0## and ##-2=1## in ##\mathbb{F}_3##. :smile:
     
  6. Dec 8, 2016 #5
    Ahhh we are such genius's! lol
     
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