# Understanding abelian Galois groups

1. Aug 3, 2007

### bham10246

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 $\leq |Gal(f)|$. 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 $n=(p_1)^{a_1} (p_2)^{a_2} ... (p_k)^{a_k}$, then $Gal(f) \cong \left[\frac{\mathbb{Z}}{p_1^{a_1}\mathbb{Z}} \times ... \times \frac{\mathbb{Z}}{p_k^{a_k} \mathbb{Z}} \right]$????

2. Aug 3, 2007

### Kummer

Primitive Element Theorem, show that finite extensions over $$\mathbb{Q}$$ are seperable and the are equal to the order of the Galois group since the field charachteristic of $$\mathbb{Q}$$ is $$0$$.

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

3. Aug 3, 2007

### Hurkyl

Staff Emeritus
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. Aug 3, 2007

### mathwonk

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
5. Aug 3, 2007

### Kummer

@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.

6. Aug 5, 2007

### mathwonk

i can find a reference for you but it is not my area and i have not read the proof.

7. Aug 5, 2007

### mathwonk

maybe silverman, advanced topics in the arithmetic of algebraic curves?

8. Aug 5, 2007

9. Aug 5, 2007

### mathwonk

10. Aug 5, 2007

### mathwonk

11. Aug 6, 2007

### Kummer

Thank you MathWonk. I once sent an e-mail to Lord Peter Swinnerton-Dyer. Does anybody know if Bryan Birch is still alive?

12. Aug 6, 2007

### mathwonk

13. Mar 3, 2010

### dave_hopkins

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?

Last edited by a moderator: May 4, 2017
14. Mar 3, 2010

### mrbohn1

The extension $$\mathbb{Q}(\sqrt{2},\sqrt{3})$$ is also generated by the roots of the polynomial: $$x^4-10x^2+1$$, which is irreducible.

As for a quintic polynomials with non-abelian galois group...the vast majority of irreducible quintic polynomials have galois group $$S_5$$. I suggest you pick one at random!

15. Mar 4, 2010

### dave_hopkins

Thanks. I'm just trying to work out how you found this out. $$x^4-10x^2+1$$ is factorized as $$(x^2 + a)(x^2 + b)$$, 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. Mar 4, 2010

### mrbohn1

You have made a couple of mistakes in you calculations. $$x^4-10x^2+1$$ actually factorizes as:$$(x^2 - a)(x^2 - b)$$, where $$a=5+2\sqrt{6}$$ and $$b=5-2\sqrt{6}$$.

So the roots are: $$\pm \sqrt{5+2\sqrt{6}}$$ and $$\pm \sqrt{5-2\sqrt{6}}$$.

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

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] ;-)

Last edited by a moderator: May 4, 2017
17. Mar 5, 2010

### dave_hopkins

Thanks! I am looking for a polynomial $$x^4+ax^2+b$$, 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.

Last edited by a moderator: May 4, 2017