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Understanding abelian Galois groups

  1. Aug 3, 2007 #1
    Hi, a quick question:

    If f is a degree n irreducible polynomial in Q[x] and the Galois group G of f is abelian, then
    1. How do we know that G has exactly n elements?
    2. Is the Galois group necessary cyclic?



    I think that since f is irreducible, the Galois group must contain an automorphism of order n. So n [itex]\leq |Gal(f)| [/itex]. But what about the other inequality?

    As for the answer to my second question, I thought it would be yes but now as I think about it, maybe not because G is a finitely generated abelian group. So by the Fundamental Theorem of Finitely Generated Abelian Groups, if [itex]n=(p_1)^{a_1} (p_2)^{a_2} ... (p_k)^{a_k} [/itex], then [itex]Gal(f) \cong \left[\frac{\mathbb{Z}}{p_1^{a_1}\mathbb{Z}} \times ... \times \frac{\mathbb{Z}}{p_k^{a_k} \mathbb{Z}} \right][/itex]????


    Please help...
     
  2. jcsd
  3. Aug 3, 2007 #2
    Primitive Element Theorem, show that finite extensions over [tex]\mathbb{Q}[/tex] are seperable and the are equal to the order of the Galois group since the field charachteristic of [tex]\mathbb{Q}[/tex] is [tex]0[/tex].

    No. Cyclotomic Extensions are a special case when Galois groups are cyclic. But consider for example [tex]E=\mathbb{Q}(\sqrt{2},\sqrt{3})[/tex] then [tex]\mbox{Gal}(E/\mathbb{Q})\simeq \mathbb{Z}_2\times \mathbb{Z}_2[/tex] which is not cyclic.
     
  4. Aug 3, 2007 #3

    Hurkyl

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    I don't think that does it; while you are guaranteed that the splitting field of f has primitive elements, a priori you have no guarantee that the roots of f are among them.
     
  5. Aug 3, 2007 #4

    mathwonk

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    the theorem of kronecker /weber? says that every abelian extension is a subfield of a cytclotomic extension. that pretty much describes them fully i guess.
     
    Last edited: Aug 3, 2007
  6. Aug 3, 2007 #5
    @Hurky, yes I appologize I made a mistake I was thinking about something else.

    Is Algebraic Number Theory your area of research? Because that is something I am (hopefully) going into. Though, there are many other beautiful areas as well, i.e. Harmonic Analysis. I am asking because I would like to see a proof of this nice theorem.
     
  7. Aug 5, 2007 #6

    mathwonk

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    i can find a reference for you but it is not my area and i have not read the proof.
     
  8. Aug 5, 2007 #7

    mathwonk

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    maybe silverman, advanced topics in the arithmetic of algebraic curves?
     
  9. Aug 5, 2007 #8

    mathwonk

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  10. Aug 5, 2007 #9

    mathwonk

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  11. Aug 5, 2007 #10

    mathwonk

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  12. Aug 6, 2007 #11
    Thank you MathWonk. I once sent an e-mail to Lord Peter Swinnerton-Dyer. Does anybody know if Bryan Birch is still alive?
     
  13. Aug 6, 2007 #12

    mathwonk

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  14. Mar 3, 2010 #13
    This does do it. We were considering polynomials which are irreducible in Q, but this is the Galois group of the polynomial f = ((x^2)-2)((x^2)-3), which is clearly not irreducible. However, you are correct.
    I am looking for a polynomial (preferably quintic) to research that is irreducible and has a nonabelian galois group. I also require it to not have the form (x^n) - a, where a is in Q. Does anyone have any suggestions of such a polynomial?
     
  15. Mar 3, 2010 #14
    The extension [tex]\mathbb{Q}(\sqrt{2},\sqrt{3})[/tex] is also generated by the roots of the polynomial: [tex]x^4-10x^2+1[/tex], which is irreducible.

    As for a quintic polynomials with non-abelian galois group...the vast majority of irreducible quintic polynomials have galois group [tex]S_5[/tex]. I suggest you pick one at random!
     
  16. Mar 4, 2010 #15
    Thanks. I'm just trying to work out how you found this out. [tex]x^4-10x^2+1[/tex] is factorized as [tex](x^2 + a)(x^2 + b)[/tex], where a = -5 +/- 4\sqrt{6} and
    b = -5 -/+\sqrt{6}. The four roots (a1,a2,a3,a4) are then the +/- roots of a. And q = a1+a2
    w=a1 + a3, e = a1 +a4 make up the fixed field Q(q^2,w^2,e^2) and the galois group is Q(q,w,e). so q,w or e equal to \sqrt{3}, \sqrt{2} and the other is 0? I am pretty sure I am doing something very wrong. any help would be greatly appreciated. Many thanks,
     
  17. Mar 4, 2010 #16
    You have made a couple of mistakes in you calculations. [tex]x^4-10x^2+1[/tex] actually factorizes as:[tex](x^2 - a)(x^2 - b)[/tex], where [tex]a=5+2\sqrt{6}[/tex] and [tex]b=5-2\sqrt{6}[/tex].

    So the roots are: [tex]\pm \sqrt{5+2\sqrt{6}}[/tex] and [tex]\pm \sqrt{5-2\sqrt{6}}[/tex].

    This is the same as: [tex]\sqrt{2}\pm \sqrt{3}[/tex] and [tex]-\sqrt{2}\pm \sqrt{3}[/tex]. (note that [tex]\sqrt{2}+\sqrt{3}[/tex] is the square root of [tex]5+2\sqrt{6}[/tex]).

    But that isn't the way I found this out! This is a fairly commonly used example of a polynomial with galois group the Klein 4-group, and I had remembered seeing it on wikipedia. ;-)
     
    Last edited: Mar 4, 2010
  18. Mar 5, 2010 #17
    Thanks! I am looking for a polynomial [tex]x^4+ax^2+b[/tex], that has a Galois group that is the dihedral group of order 8. Thus, I imagine, it must have roots a1,a2,a3,a4, where a1+a2 = sqrt{j}, a1 + a3 = sqrt{k} and a1 + a4 = sqrt{l}, with sqrt{j}, sqrt{k}, sqrt{l} irrational. pretty stuck. any help would be very much appreciated.
     
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