Does there exist a normal extension $L ⊃ \mathbb{Q}(\sqrt3) ⊃ \mathbb{Q}$ with Galois group $\mathrm{Gal}(L/\mathbb{Q}) \cong \mathbb{Z}/4\mathbb{Z}$?
We will prove that such extension doesn’t exist.
Lemma 1. Equation $X^2=3Y^2+3Z^2$ doesn’t have integer solution $(a,b,c)$ such that $a\neq 0$. (In fact, $(0,0,0)$ is the only solution.)
Proof. If $(a,b,c)$ is an integer solution of $X^2=3Y^2+3Z^2$, we claim that $(a/3,b/3,c/3)$ is also an integer solution of the same equation. That it is a solution is obvious. We prove that $a/3,b/3,c/3$ are integers. Since $a^2=3b^2+3c^2$, we see that $3\mid a^2$, hence $3\mid a$ and $a’=a/3$ is an integer. Now, $9a’^2= 3b^2+3c^2$, hence $3a’^2= b^2+c^2$. Since a square of an integer equals $0$ or $1$ modulo $3$, $b^2+c^2$ is divisible by $3$ iff $b$ and $c$ are, i.e. $b/3,c/3$ are integers.
Now, we can prove that $3^n\mid a$, for all $n\geq 1$, hence $a=0$. $\square$
Lemma 2. Equation $X^2=3Y^2+3Z^2$ doesn’t have rational solution $(a,b,c)$ such that $a\neq 0$.
Proof. Assume that $(a,b,c)$ is a rational solution. Write $a=a_1/a_2,\,b=b_1/b_2,\, c=c_1/c_2$, where $a_2,b_2,c_2\neq 0$. Then $a^2=3b^2+3c^2$, hence $a_1^2/a_2^2= 3b_1^2/b_2^2+3c_1^2/c_2^2$ and we get $(a_1b_2c_2)^2= 3(a_2b_1c_2)^2+3(a_2b_2c_1)^2$. Therefore $(a_1b_2c_2,a_2b_1c_2,a_2b_2c_1)$ is an integer solution, so by Lemma 1. $a_1b_2c_2=0$, hence $a_1=0$. $\square$
Lemma 3. Let $X^4+pX^2+q\in\mathbb Q[x]$ be an irreducible polynomial, with roots $\pm\alpha,\pm\beta$, and $L=\mathbb Q(\alpha,\beta)$ its splitting field. Then:
$\mathrm{Gal}(L/\mathbb Q)\cong\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ iff $\alpha\beta\in\mathbb Q$;
$\mathrm{Gal}(L/\mathbb Q)\cong\mathbb Z/4\mathbb Z$ iff $\alpha/\beta-\beta/\alpha\in\mathbb Q$;
$\mathrm{Gal}(L/\mathbb Q)\cong\mathbb D_4$ iff $\alpha\beta,\alpha/\beta-\beta/\alpha\notin\mathbb Q$.
Proof. This is well-known theorem. $\square$
Assume now that $L/\mathbb Q$ is a Galois extension with Galois group isomorphic to $\mathbb Z/4\mathbb Z$, and that $\mathbb Q(\sqrt 3)\leq L$. Then $|L:\mathbb Q(\sqrt 3)|=2$, hence $L=\mathbb Q(\sqrt 3,\alpha)$, where $\alpha^2\in\mathbb Q(\sqrt 3)$. Let $\alpha^2=a+b\sqrt 3$, for some $a,b\in\mathbb Q$.
$\mathbb Q(\alpha)$ is not of degree $2$ over $\mathbb Q$; otherwise $L$ have two subfields of degree $2$ over $\mathbb Q$, and then $L$ is not cyclic extension. Therefore, $L=\mathbb Q(\alpha)$ is of degree $4$ over $\mathbb Q$.
Note that $b\neq 0$, since otherwise $\alpha^2=a\in\mathbb Q$. Also $a\neq 0$, since otherwise $\alpha^4=3b^2$, and $X^4-3b^3$ is a minimal polynomial for $\alpha$ over $\mathbb Q$. Its roots are $\pm\sqrt b\sqrt[4]3,\,\pm i\sqrt b\sqrt[4]3$, hence $i\in L$, and again $L$ has two subfields of degree $2$ over $\mathbb Q$. Therefore, $a,b\neq 0$.
Since $\alpha^2-a=b\sqrt 3$, we get that $\alpha^4-2a\alpha^2+(a^2-3b^2)=0$, hence $X^4-2aX^2+(a^2-3b^2)$ is a minimal polynomial for $\alpha$ over $\mathbb Q$ (since its degree is $4$). Its roots are $\pm\alpha,\pm\beta$, where $\beta=\sqrt{a-b\sqrt 3}$, and $L$ is its splitting field.
By Lemma 3. $\alpha\beta= \sqrt{a+b\sqrt 3}\sqrt{a-b\sqrt 3}= \sqrt{a^2-3b^2}\notin\mathbb Q$, but its square is in $\mathbb Q$, hence $\alpha\beta\in\mathbb Q(\sqrt 3)$, since $\mathbb Q(\sqrt 3)$ is the only subfield of $L$ of degree $2$ over $\mathbb Q$. So, write $\sqrt{a^2-3b^2}= d+c\sqrt 3$, where $c\neq 0$. Then $a^2-3b^2= d^2+2dc\sqrt 3+3c^2$, hence by linear independence of $\{1,\sqrt 3\}$ over $\mathbb Q$ we get $d=0$ (since $c\neq 0$). Therefore, $a^2-3b^2=3c^2$, i.e. $(a,b,c)$ is a non-zero rational solution of $X^2=3Y^2+3Z^2$. A contradiction by Lemma 2.
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.