Galois extension of $\mathbb{Q}$ with Galois group $\mathbb{Z}/4\mathbb{Z}$ that contains $\mathbb{Q}(\sqrt{3})$? Does there exist a normal extension $L ⊃ \mathbb{Q}(\sqrt3) ⊃ \mathbb{Q}$ with Galois group $\mathrm{Gal}(L/\mathbb{Q}) \cong \mathbb{Z}/4\mathbb{Z}$?
 A: There's an alternative proof, using Kronecker-Weber and ramification: if the cyclic extension $L$ exists, it must be contained in the cyclotomic extension $C$ of $N$-th roots of unity, for some $N$. But if you look at the possible Galois groups of such cyclotomic extensions, they are abelian, so they are products of cyclic groups, and these cyclic groups have order divisible by $4$ --which is required because the Galois group of $L/\mathbb{Q}$ is a degree $4$ quotient of the one of $C/\mathbb{Q}$ -- only for cyclic groups corresponding to powers of primes $p$ in $N$ that are congruent to $1$ mod $4$, and to powers of the prime $2$ (here I am using standard facts about Euler's $\phi$ function). Thus a degree $4$ cyclic extension of $\mathbb{Q}$ can only ramify at such primes, it can never ramify at the prime $3$, so it can not contain $\mathbb{Q}(\sqrt{3})$. 
In other words: ramification at $3$, corresponding to $3^m$-roots of unity for $3^m \mid N$, contributes a cyclic factor of size $2 \cdot 3^{m-1}$ to the Galois group of $C/\mathbb{Q}$, and if you want to obtain ANY degree $4$ cyclic quotient of such a group it is clear that you have to kill this cyclic factor. Therefore, the degree $4$ extension that you obtain can not ramify at $3$.
If I am not mistaken, I am proving the more general claim that any degree $4$ cyclic extension of $\mathbb{Q}$ can not ramify at any prime $p$ congruent to $3$ mod $4$, and in particular, its quadratic subfield will have a discriminant $D$ not divisible by any such prime.
