Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What does it mean for an algebraic integer to have an abelian galois group ?

I thought all algebraic integers had an abelian galois group ?

Beware, I'm new to galois theory. Maybe examples and counterexamples will help me.

share|improve this question
    
What is the Galois group of an algebraic integer (or number)? Do you mean the Galois group if its minimal polynomial over the rationals (which is the same as over the integers)? –  DonAntonio Feb 10 '13 at 20:43
    
It occured here : en.wikipedia.org/wiki/Kronecker%E2%80%93Weber_theorem –  mick Feb 10 '13 at 20:48
    
Please read more carefully. The article doesn't talk about algebraic integers with abelian Galois groups. –  Martin Brandenburg Feb 10 '13 at 22:11
1  
@MartinBrandenburg What is the sentence "If an algebraic integer has abelian Galois group then it is expressible as a finite sum of roots of unity" talking about then? –  JSchlather Feb 10 '13 at 22:15
    
+1 for your comment @JacobSchlather :) –  mick Feb 11 '13 at 21:25
add comment

1 Answer

To say that an algebraic integer $\alpha\in\mathbb{C}$ has an abelian Galois group just means that the extension $\mathbb{Q}(\alpha)/\mathbb{Q}$ is a Galois extension whose Galois group is abelian. The Kronecker-Weber theorem can be viewed as a statement that all such algebraic integers are sums of roots of unity.

There certainly are algebraic integers for which this is not the case; take any finite non-abelian Galois extension $K/\mathbb{Q}$, and let $\alpha$ be a primitive element for $K$ (which exists by the primitive element theorem). You may then have to multiply $\alpha$ by an integer to ensure that it is an algebraic integer, and not just an algebraic number, but it will of course generate the same extension $K$.


An example is rather hard to give explicitly because any non-abelian extension is necessarily of degree $\geq 6$, so there is not necessarily going to be a nice expression in radicals.

However, Mathematica informs me that one (and hence, any) of the roots of $$x^6 - 3 x^5 + 6 x^4 - 11 x^3 + 12 x^2 + 3 x + 1$$ is a primitive element of the field $K=\mathbb{Q}(\sqrt[3]{2},\zeta_3)$, which has Galois group $\mathrm{Gal}(K/\mathbb{Q})\cong S_3$. Because the polynomial above is monic and has integer coefficients, its roots are algebraic integers.

enter image description here

share|improve this answer
    
See my other comment for the context. –  mick Feb 10 '13 at 20:49
    
@Zev: Shouldn't it rather be the normal closure of $\mathbb{Q}(a)$ which should be abelian and Galois? It is quite rare that $\mathbb{Q}(a)$ is already normal. –  Martin Brandenburg Feb 10 '13 at 22:12
    
It's worth noting that an element $\alpha \in \bar{\mathbb Q}$ has Abelian Galois group if and only if its minimum polynomial has an Abelian Group. Since any polynomial with Abelian Galois group splits completely in any extension containing one of its roots. This is a simple consequence of the fundamental theorem of Galois theory. –  JSchlather Feb 10 '13 at 22:13
    
@MartinBrandenburg If the minimum polynomial has abelian Galois group then the extensions $\mathbb Q(\alpha)$ are all conjugate in an abelian group, hence equal. So the polynomial splits over any extension containing a root. –  JSchlather Feb 10 '13 at 22:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.