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

