Let $L=\mathbb{Q}(\sqrt{r},\sqrt{q})$ and $K=\mathbb{Q}(\sqrt{rq})$ with $r \equiv1 \pmod4$ and $q \equiv3 \pmod 4$ prime numbers.
Like in this thread, I would like to show that $L/K$ is an unramified extension, but this time, I would like to use another approach using inertia groups and Galois theory which does not involve the discriminant of $L$.
I would like to know whether what I say is mathematically sound because I am still not very familiar with all of these notions. Could somebody please check my reasoning? It's a bit long near the end, perhaps it can be simplified.
Let $p\in \mathbb{Z}$ some prime number and $P$ a prime ideal divisor of $p$ in $O_L$ which I assume to be ramified with index $e>1$. Since it cannot be ramified at the same time in $\mathbb{Q}(\sqrt{r})$ and in $\mathbb{Q}(\sqrt{q})$ (due to coprime discriminants), $e \neq 4$. Since it has to divide $4$, then necessarily $e=2$.
First, all the extensions involved are obviously Galois, and $G=\text{Gal}(L/\mathbb{Q})\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ has cardinality $4$
Consider $D_P, I_P$ the decomposition group and the inertia group of $P$ over $p$, respectively. They are subgroups of $G$ and therefore have cardinality $1,2$ or $4$. Furthermore, we have the short exact sequence $1 \rightarrow I_P \rightarrow D_P \rightarrow \text{Gal}(k(P)/k(p)) \rightarrow 1$ where $k$ denotes the associated residual field.
If $|I_P|=4$, then by the short exact sequence we get that $\text{Gal} (k(P)/k(p))$ is trivial, so the inertial degree $f$ of $P$ over $p$ is $1$. If $g$ is the number of distinct prime divisors of $p$, we have $4=efg=2g$, so $g=2$ and $pO_L=P_1^2P_2^2$. But due to the transitivity of the Galois action upon prime divisors, $|D_P|<4$ (there's at least one automorphism which does not fix $P$ since there are two distinct divisors) so the exact sequence yields a contradiction (because $I_P$ injects into $D_P$ and since $I_P$ has cardinality $4$ it is the whole Galois group $G$). Another way to see this is that $g=[G:D_P]$, so if $I_P$ has cardinality $4$ and injects into $D_P$, then $g=1$, hence the contradiction.
If $|I_P|=1$, then the corresponding fixed field $F_P$ is $L$ itself, and according to class, the prime lying in $F_P$ below $P$ is ramified of degree $e$, but in this case this makes no sense since $e=2$ and the prime lying below $P$ is $P$ itself. Obviously $P \neq P^2$ since $P$ is a non-zero prime ideal.
So it must be that $|I_P|=2$. But in this case, the fixed field has to be one of the fields $\mathbb{Q}(\sqrt{r})$,$\mathbb{Q}(\sqrt{q})$,$\mathbb{Q}(\sqrt{rq})$.
If it is $\mathbb{Q}(\sqrt{rq})$, then it means that $\mathbb{Q}(\sqrt{r})$,$\mathbb{Q}(\sqrt{q})$ are not fixed by the non-trivial automorphism of $I_P$. So if $D_P=I_P$, the prime lying below $P$ in $\mathbb{Q}(\sqrt{r})$ or $\mathbb{Q}(\sqrt{q})$ has to split into two distinct primes in $O_L$ (because the decomposition group $D'P$ seen as a subgroup of the Galois group of $L$ over these two quadratic extensions is trivial so its index is $2=g'$). But $e=2$ means $p$ has to ramify into both of the quadratic extensions, which is a contradiction due to the coprimality of the discriminants. Therefore, the only remaining case is $D_P=G$. In this case, $|\text{Gal}(k(P)/k(p))|=2=f$ and $g=1$. Since $p$ cannot ramify in both $\mathbb{Q}(\sqrt{r})$ and $\mathbb{Q}(\sqrt{q})$ and also cannot split (because $g=1$), it has to be inert in at least one of them. It is also inert in $\mathbb{Q}(\sqrt{rq})$. This means that $X^2-rq$ is irreducible mod $p$, i.e. $rq$ is not a square mod $p$. Also, either $r$ or $q$ is not a square. If both are not squares, this leads to a contradiction since it would mean $rq$ is a square, so either $r$ or $q$ is a square, and since it cannot split, it ramifies into this extension. The discriminants of the quadratic extensions are $r$, $4q$ and $4qr$ so if it ramifies, it has to ramify in two extensions (since $2,q,r$ divide two discriminants), which is a contradiction.
So the fixed field associated to $I_P$ has to be one of $\mathbb{Q}(\sqrt{r})$, $\mathbb{Q}(\sqrt{q})$. Say $\mathbb{Q}(\sqrt{q})$. In this case, once again if $|D_P|=2$ then the prime lying below $P$ in $\mathbb{Q}(\sqrt{rq})$ has to split into $O_L$, and if $D_P=G$, by the reasoning above the only possibility is that $p$ is inert in $\mathbb{Q}(\sqrt{q})$ and ramifies in $\mathbb{Q}(\sqrt{r})$ and $\mathbb{Q}(\sqrt{rq})$, so in particular a divisor of $pO_{\mathbb{Q}(\sqrt{rq})}$ is unramified in $O_L$
In conclusion, $\mathbb{Q}(\sqrt{rq})$ is always unramified in $L=\mathbb{Q}(\sqrt{r},\sqrt{q})$.