# Adjoining all roots of unity to an arbitrary field $F$, is an abelian extension?

I want to build an abelian extension of an arbitrary field $F$, such that any polynomial $x^n-1$ for all $n\in \mathbb{N}$ has an answer in it. So I want to adjoin all roots of unity to $F$. But I don't know how I can prove that this extension is abelian.

In my question I mean all roots of unity not just $n$th root for special number $n$.

furthermore, I know the theorem below is true but I want a reference for it:

If $N|K$ and $F|K$ are field extensions, then $Gal(N/K)\simeq‎ Gal(NF/F)$.

• See Kummer Theory. – Dietrich Burde Aug 11 '15 at 12:31
• possible duplicate of Kummer extensions – Dietrich Burde Aug 11 '15 at 12:33
• Since the compositum of any collection of abelian extensions of a field $F$ is abelian, there’s nothing much for you to prove, except perhaps for the claim I just made. – Lubin Aug 11 '15 at 14:30
• Ok, can you introduce a reference for the part 2 of my question? – MH.Fakharan Aug 11 '15 at 14:51
• This is, I think, called the Theorem on Natural Irrationalities. It says that if $N\supset K$ is Galois and $F\supset K$ is an extension of any kind, then $NF\supset K$ is Galois, and the Galois group is isomorphic to the group of $N\supset(N\cap K)$. (Hope I’ve quoted it correctly.) It’s rather easy to prove (if true). – Lubin Aug 12 '15 at 12:39

To make life easy, let's assume we're in characteristic zero. So you're given a field $F$, and you want to adjoin all the roots of all the polynomials $f_n(X) =X^n - 1$ for $n = 2, 3, ...$.
If $\zeta_n$ is a primitive $n$th root of unity, then $F(\zeta_n)$ is the field obtained by adjoining all the roots of $X^n - 1$ to $F$. So automatically $F(\zeta_n)/F$ is Galois. Since any $F$-automorphism of $F(\zeta_n)$ is completely determined by its effect on $\zeta_n$, we have an injection $$Gal(F(\zeta_n)/F) \rightarrow (\mathbb{Z}/m\mathbb{Z})^{\ast}$$ given by the formula $\sigma \mapsto a$, where $a$ is an integer satisfying $\sigma \zeta = \zeta^a$. Thus $F(\zeta_n)/F$ is abelian, since its Galois group is isomorphic to a subgroup of an abelian group. The injection is 'natural', in the sense that if $n \mid m$, the restriction map $Gal(\zeta_m) \rightarrow Gal(\zeta_n)$ commutes with the restriction $\mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/n \mathbb{Z}$ (immediate verification).
Now the field you are looking for (the splitting field of all $X^n - 1$ over $F$) is the union $$M := \bigcup\limits_{n \in \mathbb{N}} F(\zeta_n)$$ Now $M$ is the direct limit of the fields $F(\zeta_n)$ is the category of fields, the morphisms being inclusion. It satisfies a special universal property. And $Gal(-/F)$ is a contravariant functor from the category of fields containing $F$ to the category of groups, where an inclusion of fields $F(\zeta_n) \subseteq F(\zeta_m)$ is sent to a homomorphism of groups $Gal(F(\zeta_m)/F) \rightarrow Gal(F(\zeta_n)/F)$ (the restriction homomorphism).
You can check that this functor transforms direct limits into inverse limits. It follows that $$Gal(M/F) = Gal(\lim\limits_{\rightarrow} F(\zeta_n)/F) = \lim\limits_{\leftarrow} Gal(F(\zeta_n)/F)$$ Finally, there is a popular way to describe a group which is isomorphic to a given inverse limit of groups, namely the inverse limit is isomorphic to a special subgroup of the product $$\prod\limits_{n \in \mathbb{N}} Gal(F(\zeta_n)/F)$$ (the exact description I don't have room to explain here, although it's not too difficult). Since the Galois group of $M/F$ is isomorphic to a subgroup of $$\prod\limits_n (\mathbb{Z}/n\mathbb{Z})^{\ast}$$ it follows that $Gal(M/F)$ is abelian, which is what you wanted.
• Notice the inverse limit construction generalizes the case of a finite compositum of fields. If $L_1, ... , L_t$ are Galois extensions of $F$, then the Galois group of $L_1 \cdots L_t$ over $F$ is isomorphic to a certain subgroup of $$Gal(L_1/F) \times \cdots \times Gal(L_t/F)$$ – D_S Aug 16 '15 at 4:16