I am wondering is there a theory on maximal inert fields? Consider a Galois extension of number fields $K\subset L$ with the Galois group $G=G(L/K)$ and rings of integers $R_K\subset R_L$. Let $p\subset R_K$ be a prime ideal, $\mathfrak{P}\subset R_L$ a prime ideal over $p$. Then $G(\mathfrak{P})<G$ is the subgroup fixing $\mathfrak{P}$ as a set, $T(\mathfrak{P})<G(\mathfrak{P})$ the subgroup that maps identically on $R_L/\mathfrak{P}$.
We now have a chain of subgroups
$$G>G(\mathfrak{P})>T(\mathfrak{P})>1,$$
which corresponds to a tower of fields
$$K\subset F_1\subset F_2\subset L.$$
It can be shown that $F_1$ is the largest intermediate field such that $F_1\cap \mathfrak{P}$ totally splits over $K$, $F_2$ is the largest intermediate field such that $F_2\cap \mathfrak{P}$ is unramified over $K$.
I wonder if there is a similar theory for intermediate fields $F$ where $p$ is inert in $F$. I have two approaches in my mind. One is to proceed as above and I need to find a subgroup of $G$ with order $eg$, where $e$ is the ramification index of $p$ in $L$ and $g$ the number of prime ideals over $p$ in $L$. The other is to prove that composition of two intermediate fields where $p$ is inert still has $p$ being inert.
The composition of two intermediate fields in which $p$ is inert need not have $p$ be inert.
Example. For an odd prime $p$, let $q$ and $r$ be different odd primes such that $(\frac{q}{p}) = -1$ and $(\frac{r}{p}) = -1$. Then $p$ is inert in $\mathbf Q(\sqrt{q})$ and $\mathbf Q(\sqrt{r})$. If $p$ were inert in the composite field $\mathbf Q(\sqrt{q},\sqrt{r})$ then it would inert in every subfield, but it splits in the subfield $\mathbf Q(\sqrt{qr})$ since $(\frac{qr}{p}) = 1$.
The property of a prime being totally ramified, rather than inert, also does not behave well for composites of fields.
Example. An odd prime $p$ is totally ramified in $\mathbf Q(\sqrt{p})$ and $\mathbf Q(\sqrt{-p})$, but if $p$ were totally ramified in their composite field $\mathbf Q(\sqrt{p},\sqrt{-p}) = \mathbf Q(\sqrt{p},i)$ then $p$ would be totally ramified in every subfield, but it is unramified in the subfield $\mathbf Q(i)$.