$\mathbb{Q}(i)$ has no unramified extensions

It is a classical result that every extension of $\mathbb{Q}$ is ramified. Put differently: there are no unramified extensions of $\mathbb{Q}$. The classical proof follows from the following two statements:

(a) The only number field having discriminant $1$ is $\mathbb{Q}$ itself.

(b) A prime number $p$ ramifies in a number field $K$ if and only if it is a divisor of the discriminant $\Delta_K$.

Is it possible to argue for something similar for $\mathbb{Q}(i)$? The goal of my question is to construct an example of a Hilbert Class Field without appealing (yet) to Artin Reciprocity.

Using Artin Reciprocity, one has that $Gal(L/\mathbb{Q}) \cong Cl(\mathbb{Q})$, where $L$ is the Hilbert Class Field of $K=\mathbb{Q}$. Now, because $\mathbb{Z}$ is a PID, the class group becomes trivial, and we have that $L=\mathbb{Q}$. Since $L$ is by definition the maximal unramified abelian extension of $K=\mathbb{Q}$, this implies that there are no unramified extensions of $\mathbb{Q}$. Repeating the argument (and observing that $\mathbb{Z}[i]$ is a PID), one deduces that the Hilbert Class Field of $K=\mathbb{Q}(i)$ is $L=\mathbb{Q}(i)$. But this just means that $\mathbb{Q}(i)$ has no unramified extensions.

However, I would like to have a proof from scratch which argues that $\mathbb{Q}(i)$ has no unramified extensions.

EDIT: My question seems related to this MO post but I am having trouble following the argument there.

Solution 1:

It is possible to do this for $\mathbb Q(i)$, and this precisely what is done in the linked MO question.

Note also that CFT describes abelian everywhere unramified extensions. The statement about having no everywhere unramified extensions is stronger.