# 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

Not sure what your background is, so some of this you'll understand and some of it you may not. Feel free to ask for more details.

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