I am aware of How to determine the density of the set of completely splitting primes for a finite extension? and similar posts but these do not answer my question.
Let $L:K$ be a Galois extension of number fields. Then the primes of $K$ which split in $L$ have Dirichlet density $\frac{1}{[L:K]}$. The proof is as follows.
Let $S$ denote the set of primes of $L$ which lie above a prime of $K$ which splits in $L$. For every prime $\mathfrak{p} \in \text{Spl}(L:K)$ there are precisely $[L:K]$ primes above it, which are all in $S$.
The Dirichlet density of $S$ is 1. Indeed, for every prime $\mathfrak{q}$ not in $S$, and $\mathfrak{p}$ the prime of $K$ below $\mathfrak{q}$ there are two possibilities for $\mathfrak{p}$: either it ramifies, or the residual degree of $\mathfrak{q}$ over $\mathfrak{p}$ at least 2. In both cases $\mathfrak{q}$ is in a set of Dirichlet density zero. Thus the complement of $S$ has Dirichlet density zero, which implies $d(S) = 1$. Combining this with the fact that the primes above some $\mathfrak{p} \in \text{Spl}(L:K)$ have the same norm, namely $N(\mathfrak{p})$, since $\mathfrak{p}$ is split, we have $$ \lim_{s \to 1^+} \frac{\sum_{\mathfrak{p} \in \text{Spl}(L:K)} N(\mathfrak{p})^{-s}}{\sum_{\mathfrak{p}} N(\mathfrak{p})^{-s}} = \lim_{s \to 1^+} \frac{\frac{1}{[L:K]} \sum_{\mathfrak{q} \in S} N(\mathfrak{q}^{-s})}{ \log \frac{1}{s-1}} = \frac{1}{[L:K]} d(S) = \frac{1}{[L:K]}. $$ Therefore the primes of $K$ which split in $L$ have Dirichlet density $\frac{1}{[L:K]}$.
Now my issue is that I do not see the necessity of the assumption that $L:K$ is Galois. Even if it is not Galois, and some prime $\mathfrak{q}$ is not in $S$, then the same case distinction should still apply.
