Properties of the extension $\mathbb Q(\sqrt{a\sqrt{D}})$ Let $a$ be a nonzero rational number, and let $D$ be a squarefree integer not equal to $1$. I want to show that, firstly, the extension $F=\mathbb Q(\sqrt{a\sqrt{D}})$ is not cyclic of degree $4$, and secondly, that if it's Galois, then $D=-1$.
So since that element is a root of $x^4-a^2D=(x^2+a\sqrt{D})(x^2-a\sqrt{D})$, this polynomial is the minimal polynomial (assuming the degree is $4$; otherwise the first statement is trivially true). I'm thinking that any automorphism of $F$ must permute the roots of the two quadratics which would prevent there from being an order $4$ automorphism, but I'm not sure why that has to be the case, since it's apparently possible for the extension to be Galois, and such an automorphism group is not transitive on the roots.
Hmm, apparently an extension can only be cyclic if it's Galois, so if I do the 2nd part, then I'm done, since I previously showed that $\mathbb Q(\sqrt{ai})$ is not cyclic.
As for the second part, if the extension is Galois, it must contain the $4$th roots of unity, including $i$. Then, since $F\neq\mathbb Q(i)$, $F$ is of degree $4$ over $\mathbb Q$. By the first part (assuming I figure it out), this extension has the Klein-$4$ Galois Group, so there are exactly $3$ proper nontrivial subfields. I can immediately see $3$: $\mathbb Q(i=\sqrt{-1})$,  $\mathbb Q(\sqrt{D})$, and $\mathbb Q(\sqrt{-D})$. Apparently, I want to in fact show that the first two are the same, and thus the third is trivial. 
Now, since $F$ contains imaginary numbers, $D<0$. Also, if $D\neq -1$, then $\sqrt{D}\notin \mathbb Q(i)$, so $F=\mathbb Q(i, \sqrt{D})$. Is there a way for me to get a contradiction from here? I tried writing $\sqrt{a\sqrt{D}}$ as a linear combination of $1$, $\sqrt{D}$. $i$, and $i\sqrt{D}$, but the resulting equations are messy. Is there a cleaner way?
 A: You can argue as follows. Let $b=-a^2D$. Then $F$ is a rupture
field for the irreducible polynomial $P=X^4+b$. Suppose that $F/{\mathbb Q}$ is 
Galois. Then $F$ contains all the fourth roots of $-b$. As $i$ can be written as a 
quotient of two such roots, we see that $i\in F$. It follows that $F$ contains
all the fourth roots of $b$, and hence also all the fourth roots of $|b|$.
If $X^4-|b|$ were irreducible over ${\mathbb Q}$, then considering the 
unique real root $\beta=\sqrt[4]{|b|}$ of this polynomial we would deduce 
$F={\mathbb Q}(\beta) \subseteq {\mathbb R}$, contradicting $i\in F$. So
$X^4-|b|$ is reducible, whence $X^4-|b|\neq X^4+b=P$, so that $b>0$.
So in ${\mathbb Q}[X]$, $X^4-|b|=X^4-b$ has either a rational root or
a factorization into degree-two polynomials.
In the first case, $b=c^4$ for some $c\in{\mathbb Q}$ hence
$c^4=-a^2D$ ; as $D$ is square-free we deduce $D=-1$.
In the second case, we have a factorization $X^4-b=(X^2+pX+q)(X^2+rX+s)$
with $p,q,r,s$ rational. Looking at the coefficient in $X^3$, we see that
$r=-p$. If $p=0$, we have a factorization $X^4-b=(X^2+q)(X^2-q)$, $b=q^2$
whence $D=-1$ again, by the same argument as above. If $p\neq 0$, 
looking at the coefficient in $X^1$ we see that $s=q$. So
$X^4-b=(X^2+pX+q)(X^2-pX+q)=(x^2+q)^2-(pX)^2$, and 
looking at the coefficient in $X^1$ we see that $q=\frac{p^2}{2}$. In the
end, $X^4-b=(X^2+pX+\frac{p^2}{2})(X^2-pX+\frac{p^2}{2})=X^4+\frac{p^4}{4}$,
so that $b=-\frac{p^4}{4}$ contradicting $b<0$. This concludes the proof.
