Galois group of successive quadratic extensions I'm trying to answer the following question:
"Suppose that $F=K(x_1,\dots,x_n)$ where $0\neq x_i^2=a_i\in K$. Show that $F/K$ is a Galois extension, with Galois group isomorphic to $(\mathbb{Z}/2\mathbb{Z})^m$ for some $m\leq n$."
My first thought was show that $[F:K]=2^m$ and then find $2^m$ K-homs (and hence automorphisms) from $F$ to $F$ and thus the Galois group.
I'm wondering if this is the "right" approach, since it seems we need to define maps sending $x_i$ to $\pm x_i$, and then prove that these are homs by considering $F$ as a $K$-vector space, which, at first sight, appears messy and unenlightening.
Is there a conceptually clearer way of doing things here?
Thanks.
 A: First we need to assume that $K$ has characteristic $\neq 2$.
Under this hypothesis, $F/K$ is separable, and to prove that it is Galois we need to show that it is normal.
As a matter of fact, we put all the roots of the polynomials $X^2-a_i$ (the $\pm x_i$) in $F$.
Finally, each element of the Galois group maps $x_i$ to $\pm x_i$, so we get a morphism $\mathrm{Gal}(F/K) \rightarrow \left( \{ \pm 1 \} \right)^n \simeq \left(\mathbb{Z}/2\mathbb{Z} \right)^n$, $\sigma \mapsto (\sigma(x_i)/x_i)_{1 \leq i \leq n}$, and it is easily shown to be injective.
We are left with the following statement to prove: a subgroup of $\left(\mathbb{Z}/2\mathbb{Z} \right)^n$ is isomorphic to $\left(\mathbb{Z}/2\mathbb{Z} \right)^m$ with $m \leq n$, and this follows from the fact that a subgroup of $\left(\mathbb{Z}/2\mathbb{Z} \right)^n$ is a sub-$\mathbb{Z}/2\mathbb{Z}$-vector space.
A: Let $K_i = K(\sqrt{a_1},\sqrt{a_2},\cdots,\sqrt{a_i}) = K_{i-1}(\sqrt{a_i})$.
Each time you go from $K_i$ to $K_{i+1}$ you're either doing a quadratic extension or not doing anything (in the case where $\sqrt{a_{i+1}} \in K_i$).
The theorem that fits this together is proved here. So the chain of groups is $(\mathbb Z/2 \mathbb Z)^{m}$ for $m$ being how many quadratic fields added were independent.
