Understanding abelian Galois groups

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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...
 

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  • #2
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1. How do we know that G has exactly n elements?
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].

2. Is the Galois group necessary cyclic?
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.
 
  • #3
Hurkyl
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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].
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.
 
  • #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.
 
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@Hurky, yes I appologize I made a mistake I was thinking about something else.

the theorem of kronecker /weber? .
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.
 
  • #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.
 
  • #7
mathwonk
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maybe silverman, advanced topics in the arithmetic of algebraic curves?
 
  • #11
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Thank you MathWonk. I once sent an e-mail to Lord Peter Swinnerton-Dyer. Does anybody know if Bryan Birch is still alive?
 
  • #13
http://en.wikipedia.org/wiki/Primitive_element_theorem" [Broken], 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. http://en.wikipedia.org/wiki/Cyclotomic_field" [Broken] 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.
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?
 
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  • #14
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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?
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!
 
  • #15
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!
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,
 
  • #16
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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,
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 http://en.wikipedia.org/wiki/Galois_theory" [Broken] ;-)
 
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  • #17
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 http://en.wikipedia.org/wiki/Galois_theory" [Broken] ;-)
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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