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Compute the degree of K over Q

  1. Feb 7, 2012 #1
    6f3xhi.png

    I'll skip out some tedious bits that I'm confident I have correct and get to the part I'm stuck on

    Let L = Q(31/3)

    [L:Q] = 3

    Clearly [K:L] >1

    w is a root of X3-1

    So

    [K:L] is less than or equal to 3, in particular it is either 2 or 3

    Here is where i'm stuck

    I tried to show that X3-1 is the min poly of K over L by supposing there exists an a,b in Q such that

    (a+b(31/3)) = 1

    with a hope of finding a contradiction, but when a=1 this is clearly a root and 1 is an element of L, so this canot be the minimal polynomial of K over L.

    Can anyone help me find the minimal polynomial of K over L?

    Thanks

    Matt
     
  2. jcsd
  3. Feb 7, 2012 #2

    micromass

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    Indeed, [itex]X^3-1[/itex] is not irreducible. Can you factor this polynomial over L?? I.e. can you factor out x-1? Maybe what's left WILL be the minimal polynomial?
     
  4. Feb 7, 2012 #3
    Ah ok so X2+X+1

    By the same method as before I showed thre is no element a,b satisfying

    (a+b(31/3))2 + (a+b(31/3)) = -1

    so it is irreducible over Q(31/3)

    By the tower law [K:Q]=6

    Are my basis elements

    {1, (31/3), (91/3), w, (31/3)w, (91/3)w}?

    Thanks
     
  5. Feb 7, 2012 #4

    micromass

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    That seems correct!!
     
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