A problem related with Galois group Does there exist a finite Galois extension $L/Q$ which contains $Q(\sqrt3)$, such that $Gal(L/Q)\approx Z/4Z$?
 A: No such extension is possible. Let $L=\mathbb{Q}(x)$ where $x=\sqrt{a+b\sqrt{3}}$, and $a,b \in \mathbb{Q}$. Note that if $b=0$ then the Galois group is $\mathbb{Z}_2\oplus \mathbb{Z}_2$.
I shall use the fact that if $\sqrt{\beta}\in K(\sqrt{\alpha})-K$ then $\sqrt{\alpha\beta}\in K$.
Now if $L$ is Galois, then $$\sqrt{a-b\sqrt{3}}\in \mathbb{Q}(\sqrt{3})(\sqrt{a+b\sqrt{3}})$$
It follows that $$\sqrt{a^2-3b^2}\in \mathbb{Q}(\sqrt{3})$$. And then that
$$\sqrt{3a^2-9b^2}\in\mathbb{Q}$$
Now the equation $$3a^2-9b^2=k^2$$ is impossible, by a descent argument.
I was hoping to find a more conceptual Galois theoretic argument, but I have not succeeded and must go to bed now. 
EDIT: Note conversely that if $n=a^2+b^2$ then $\mathbb{Q}(\sqrt{n+b\sqrt{n}})$ has Galois group $\mathbb{Z}_4$.
A: Let $L/\Bbb Q$ be a cyclic quartic extension, and let $K=\Bbb Q(\sqrt d)$
be its quadratic subfield. Then $d$ is a sum of two squares of rationals.
That excludes $d=3$.
To see this, note that $L=K(\alpha)$ where $\alpha^2=a+b\sqrt d$
and moreover $\beta\in L$ where $\beta^2=a-b\sqrt d$. A generator $\sigma$
of the Galois group of $L/\Bbb Q$ takes $\alpha$ to $\beta$, and so $\beta$ to $-\alpha$. Then $\sigma(\alpha\beta)=-\alpha\beta$, and as
$(\alpha\beta)^2=a^2-bd^2\in\Bbb Q^\times$, then $\alpha\beta=c\sqrt d$
with $c\in\Bbb Q^\times$. So $a^2-b^2d=c^2d$. Re-arranging
$$d=\frac{a^2}{b^2+c^2}=\left(\frac{ab}{b^2+c^2}\right)^2
+\left(\frac{ac}{b^2+c^2}\right)^2.$$
A: There is a general criterion for embedding a cyclic extension into a larger cyclic extension: " Let $p$ be a given prime, k any field containing a primitive $p^r$ - primitive root of unity $w$, K/k a cyclic extension of degree divisible by $p$ . Then there exists an extension L/K such that L/k is cyclic and L/K is cyclic of degree $p^r$ iff $w$ is a norm from K . The proof is essentially a good exercise in Galois cohomology. See e.g.
Conceptual reason why a quadratic field has $-1$ as a norm if and only if it is a subfield of a $\mathbb{Z}/4$ extension?. Applying it here for $p=2$, your problem has a solution iff $-1$ is a norm from  $\mathbf Q (\sqrt 3)$, or equivalently $3$ is a norm from $\mathbf Q (\sqrt {-1})$, i.e. $3=x^2+y^2$ in $\mathbf Q$, or equivalently $3z^2=x^2+y^2$ in $\mathbf Z$. Reduction modulo $3$ shows this is impossible. 
