Galois group of irreducible polynomial in a field of characteristic zero in which every element is a perfect square This is one of the exercises during my reading of Ian Stewart's Galois Theory. Whether the following statement is true: 

If $K$ is a field of characteristic zero in which every element is a perfect square, then the Galois group of any irreducible $n$-th degree polynomial over $K$ is isomorphic to $A_n$. 

I think it might not be true but the examples of such fields in my knowledge are very few. Any thoughts? Thanks!
 A: I proffer the following counterexample. 
First we need to construct such a field $K$. Let $K_0=\Bbb{Q}$, and recursively define fields $K_n, n\in\Bbb{N}$ by the recipe that $K_{\ell+1}$ is gotten from $K_\ell$ by adjoining the square roots of all the elements of $K_{\ell}$ to it. Note that we can do this construction inside $\Bbb{C}$. Then we let 
$$
K=\bigcup_{n\in\Bbb{N}}K_n\subset \Bbb{C}.
$$
Then any element of $K$ belongs to some field $K_\ell$, and thus has a square root in $K_{\ell+1}$, hence also in $K$.
If we pick any element $z\in K$, I claim that $[\Bbb{Q}(z):\Bbb{Q}]$ is a power of two. Clearly $z\in K_\ell$ for some $\ell$. W.l.o.g. we can assume that $\ell$ is the smallest natural number with that property. Then there exists a finite set of elements $z_1,z_2,\ldots,z_m\in K_{\ell-1}$ such that
$z\in K_{\ell-1}(\sqrt{z_1},\sqrt{z_2},\ldots,\sqrt{z_m})$. 
This means that we can write $z$ in terms of finitely many element of $K_{\ell-1}$ and their square roots. By induction hypothesis all those
elements are algebraic of degree a power of two over $\Bbb{Q}$. Thus so is their compositum. Adjoining the square roots of $z_i, i=1,\ldots,m,$ won't change that fact so we see that $\Bbb{Q}(z)$ is contained in a field of degree a power of two settling the claim.
Let's then consider the eleventh cyclotomic polynomial
$$
\phi_{11}(x)=x^{10}+x^9+x^8+\cdots+x^2+x+1.
$$
By the known theory of cyclotomic fields we know that $\phi_{11}(x)$ factors into a product of two quintic factors over $F=\Bbb{Q}(\sqrt{-11})$, the only quadratic
subfield of the eleventh cyclotomic field. Let $f(x)$ be one of those quintic factors. 
By the above observation $f(x)$ remains irreducible over $K$. If $L=K(e^{2\pi i/11})$ is the splitting field of $f(x)$, we easily see that $Gal(L/K)$ is isomorphic to $Gal(\Bbb{Q}(e^{2\pi i/11})/\Bbb{Q}(\sqrt{-11})\cong C_5$.
But $C_5$ is a proper subgroup of $A_5$.
