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Splitting field with Galios group A4

  1. Nov 7, 2012 #1
    the polynomial x^4+8x+12=0 has the Galois group A4. I have all its roots, but can't figure out its splitting field. The roots are
    [tex]\alpha_1=\sqrt{2}(\sqrt{\cos{(\pi/9)}}+i\sqrt{\cos{(2\pi/9)}}+i\sqrt{\cos{(4\pi/9)}})[/tex]

    [tex]\alpha_2=\sqrt{2}(\sqrt{\cos{(\pi/9)}} - i\sqrt{\cos{(2\pi/9)}}-i\sqrt{\cos{(4\pi/9)}})[/tex]

    [tex]\alpha_3=\sqrt{2}(-\sqrt{\cos{(\pi/9)}} + i\sqrt{\cos{(2\pi/9)}}-i\sqrt{\cos{(4\pi/9)}})[/tex]

    [tex]\alpha_4=\sqrt{2}(-\sqrt{\cos{(\pi/9)}} - i\sqrt{\cos{(2\pi/9)}}+i\sqrt{\cos{(4\pi/9)}})[/tex]
     
    Last edited: Nov 7, 2012
  2. jcsd
  3. Nov 8, 2012 #2

    mathwonk

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    what do you mean? its the field generated by those roots. do you want a smaller set of generators? (presumably two of those would do.)
     
  4. Nov 10, 2012 #3

    mathwonk

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    hellooo?
     
  5. Nov 12, 2012 #4
    Sorry I didn't respond earlier, but I don't think it is as simple as just attaching two roots since the degree of the field in that case wouldn't be twelve.
     
  6. Nov 12, 2012 #5

    Hurkyl

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    Why not?

    What can you say about, e.g., the minimal polynomial of a2 in the field Q(a1)?

    It can't be x4 + 8x + 12, because this polynomial is reducible over Q(a1)! (it has a root)
     
    Last edited: Nov 12, 2012
  7. Nov 12, 2012 #6
    I think I get what you're saying let me mess with this idea for a bit and get back to you.
     
  8. Nov 12, 2012 #7
    Here is where I am at.

    [tex]x^4+8x+12=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)=(x^3+\alpha_1x^2+{\alpha_1}^2x+({\alpha_1}^3+8))(x-\alpha_1)=0[/tex]

    So I define [tex]F \equiv \mathbb{Q}(\alpha_1)[/tex] So it is obvious that the coefficients of that cubic are in the field [tex]F \equiv \mathbb{Q}(\alpha_1)[/tex] and that the roots of that cubic are [tex] \alpha_2, \alpha_3, \alpha_4[/tex] but it is not obvious to me that [tex]\alpha_3[/tex] can be written in terms of [tex] \alpha_2[/tex] and the field [tex]\mathbb{Q}(\alpha_1)[/tex] making the the Galois group of the cubic [tex]\mathbb{Z_3}[/tex]
     
  9. Nov 12, 2012 #8
    I guess I just have to assume that's the case since if it wasn't the Galois group would be S4 not A4. In that case I get a splitting field:
    [tex]\mathbb{Q}(\alpha_1,\alpha_2)= A +B\alpha_1 + C{\alpha_1}^2 + D{\alpha_1}^3 + E\alpha_2 + F\alpha_1\alpha_2 +G{\alpha_1}^2\alpha_2 +H{\alpha_1}^3\alpha_2 + I{\alpha_2}^2 + J\alpha_1{\alpha_2}^2 + K{\alpha_1}^2{\alpha_2}^2 +L{\alpha_1}^3{\alpha_2}^2[/tex] Is this correct?
     
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