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Let $D\in\mathbb{Z}$ be a squarefree integer and let $a\in\mathbb{Q}$ be a nonzero rational number. Please show that $\mathbb{Q}(\sqrt{a\sqrt{D}})$ cannot be a cyclic extension of degree 4 over $\mathbb{Q}$. Also show that if $\mathbb{Q}(\sqrt{a\sqrt{D}})$ is Galois over $\mathbb{Q}$ then $D=-1$.

I have the case $D=-1,|a|\neq2b^2$ for some $b\in\mathbb{Q}$ that the Galois group is $\mathbb{Z_2}^2$, the Klein four group (which is not cyclic). How to show in general that a group is not cyclic?

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Some ideas for you to mull and complete:

$$x=\sqrt{a\sqrt D}\implies x^2=a\sqrt D\implies x^4=a^2D$$

The above almost proves the minimal polynomial (over the rationals) of $\,\sqrt{a\sqrt D}\,$ is $\,p(x)=x^4-a^2D\in\Bbb Q[x]\,$ (can you complete this?)

We get that a basis for the linear space $\,\Bbb Q(\sqrt{a\sqrt D})_{\Bbb Q}\,$ can be

$$\{1\,,\,w\,,\,w^2\,,\,w^3\}\;,\;\;w:=\sqrt{a\sqrt D}$$

But we have that

$$x^4-w^4=(x^2-w^2)(x^2+w^2)\in\Bbb Q(w^2)[x]$$

and this means that no embedding of $\,\Bbb Q(w)\,$ in some algebraic closed field containing it (take $\,\Bbb C\,$ , for example) can map a root of $\,x^2-w^2\,$ to a root of $\,x^2+w^2\,$ and neither the other way around, which already shows the extension cannot be cyclic.

That the extension is Galois means it is normal, and this in turn means $\,Q(w)\,$ contains all the roots of $\,x^4-w^4\,$ , among which there are the ones of $\,x^2+w^2\,$ :

$$\pm w\,i=\pm\sqrt{a\sqrt D}\,i=\pm\sqrt{a\sqrt D(-1)}\;\ldots$$

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Too much good work to go un-upvoted for an hour! +1 – amWhy Apr 4 '13 at 4:00
I was able to show $D<0$, but how do you show $|D|=1$? Of course assuming the degree extension is 4. Thanks. – user70520 Apr 4 '13 at 5:11

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