Let $K ⊂ L$ number fields. Prove that there are infinitely many primes of K that split completely in $L$.
My attempt:
Using this question: Infinitely many primes in the ring of integers , we know there are infinitely many primes $\mathfrak p$ in $K$ such that inertia degree is $f (\mathfrak p|p) = 1$.
Consider a normal extension $M/L$ (of finite degree over $L$).
Then there are infinitely many primes $\mathfrak q_i$ in $M$ such that inertia degree is $f (\mathfrak q_i|p_i) = 1$.
All the primes $p_i\in \Bbb Z$ lying under those primes $\mathfrak q_i$ such that $p_i> Disc(M)$ do not ramify in $M$ since $p\nmid Disc(M)$.
Let $p_i=\prod_{i=1}^r \mathfrak q_i$ be the splitting in $M$ of such a prime, hence: $\forall i\in \{1,...,r\} e(\mathfrak q_i|p_i)=e(\mathfrak q_1|p_1)=1$ since $M$ is normal (all ramification indices are equal).
Now taking the primes $\mathfrak p_i, i=1...s, s<r$ we have the splitting in $L$ of the $p_i$'s:
$p_i=\prod_{i=1}^r \mathfrak p_i^{e_i}$ with $ \forall i, \mathfrak p_i\in L$
By multiplicativity of ramification indices in field extensions, $\forall i, e_i|1 \Rightarrow e_i=1$ .
Hence taking all the primes in $K$ lying under the $\mathfrak p_i's$ there are infinitely many primes in K that split completely in $L$ since $[L:K]=efs=s$.
Thank you for any comments or help in this elementary proof.