Examples of quadratic extensions K, L of $\mathbb{Q}$ such that KL has some properties. Let $p$ be a prime integer, I want to find $p$ and K, L  extensions of $\mathbb{Q}$ such that 


*

*K, L contain each a unique prime lying over $p$ but KL does not.


Another, different, triplet such that


*The residue field extension of $\mathbb{Z}_p$ is trivial for K and L but not for KL.


Is there a way to easily compute such examples?
I can give examples of other cases (e.g p totally ramified in K and L but not in KL or inert in K and L but not in KL) but I'm finding the two above a bit more difficoult.
 A: (Answering the edited version with the misunderstanding cleared. The answer to the misunderstood version has been deleted, but can be seen in the edit history.)
Observe that because $K/\Bbb{Q}$ and $L/\Bbb{Q}$ are Galois, cyclic of order two, the extension $KL/\Bbb{Q}$ is also Galois with Galois group $C_2\times C_2$. Therefore:


*

*for each prime $p$, the equation $efg=4$ holds, and

*it is impossible to have $f=4$ for any prime $p$, because then the Galois group should have a subquotient isomorphic to $C_4$.

*Consequently $KL$ will have a third intermediate quadratic field $F$ that we can use as a tool for studying the primes of the compositum.


The second point is a bit more advanced, and not needed to answer this. But it does play a role.


*

*Namely, if $f(K/\Bbb{Q},p)=2=f(L/\Bbb{Q},p)$ then the condition (1) is automatic. So we can simply pick two integers $m,n$, coprime to $p$, such that neither is a quadratic residue modulo $p$. Then $K=\Bbb{Q}(\sqrt n)$ and
$L=\Bbb{Q}(\sqrt m)$ will work. For example with $p=5$, $n=2$, $m=7$ we see that 
$p$ is inert in $K=\Bbb{Q}(\sqrt2)$ and in $L=\Bbb{Q}(\sqrt7)$. As a product of two quadratic non-residues $2\cdot7=14$ is a quadratic residue modulo $5$, and therefore $5$ splits in $F=\Bbb{Q}(\sqrt{14})\subset KL$. Consequently, $p=5$ is a product of two prime ideals of $\Bbb{Q}(\sqrt2,\sqrt7)$, both with inertia degree $f=2$.

*To find an example of (2) we need to use ramification. The prime $p=3$ is ramified in both $K=\Bbb{Q}(\sqrt3)$ and $L=\Bbb{Q}(\sqrt{15})$, so $f(K/\Bbb{Q},3)=f(L/\Bbb{Q},3)=1$. But $\sqrt{3}\cdot\sqrt{15}=3\sqrt{5}$, so we see that $F=\Bbb{Q}(\sqrt5)\subseteq KL$. As $5$ is not a quadratic residue modulo $3$, it follows that $3$ is inert in $F$. Therefore $f(KL/\Bbb{Q},3)=2$ as well. 


It may be worth noting that we cannot get (2) when $p$ splits in both $K$ and $L$. For in that case $p$ will split compeletely in $KL$ (the product of two quadratic residues is a quadratic residue). The combination of ramified in $K$ + split in $L$ obviously won't work either. Ramified in both $K$ and $L$ is the only combo that works here. Of course, we still need to be careful for the prime $p$ might still split in the third quadratic intermediate field $F$. 
