Splitting of primes in the compositum of fields If $L_i/K$ are Galois extensions of number fields, $i=1,\ldots,n$, and $L=L_1\cdots L_n$ is the compositum. Then it's true that a prime $\mathfrak{p}$ of $K$ splits in $L$ if and only if it splits in all of $L_i$. Does this also hold if the $L_i$ are not Galois extensions of $K$?
The proof that I know regarding the compositum uses the fact that the $L_i/K$ are Galois, so is this true in the more general setting and how would one prove it?

This holds always. QiL posted a local solution, and explained that one direction is trivial. For the other direction I would argue as follows using basic facts about the decomposition groups. Let $E/K$ be a normal closure of $L/K$, and write $G=Gal(E/K)$, $G_i=Gal(E/L_i)\le G$ and $H=Gal(E/L)$. Let $\mathfrak{P}$ be any prime ideal of $E$ above $\mathfrak{p}$, and let 
$$D=D(\mathfrak{P}/\mathfrak{p})
=\{\sigma\in G\mid \sigma(\mathfrak{P})=\mathfrak{P}\}\le G$$ 
be the decomposition group. It is known that if $M$ is any intermediate field, and $\mathfrak{p}_M=\mathfrak{P}\cap M$, then
$e(\mathfrak{p}_M\mid\mathfrak{p})=f(\mathfrak{p}_M\mid\mathfrak{p})=1$ if and only if
$M$ is contained in the decomposition field $Inv(D)$, or equivalently $D\le Gal(E/M)$.
By our assumption this is true, whenever $M$ is one of the fields $L_i$. Therefore $D\le G_i$
for all $i$. Thus also $D\le H=\cap_i G_i$. Applying the above result in the opposite direction tells us then that for the prime ideal $\mathfrak{p}'=\mathfrak{P}\cap L$ we have $e(\mathfrak{p}'\mid\mathfrak{p})=f(\mathfrak{p}'\mid\mathfrak{p})=1$. All the prime
ideals $\mathfrak{p}'$ of $L$ above $\mathfrak{p}$ are gotten in this way, so the claim follows.
With basic facts about decomposition groups in place this is more or less a tautology, so I don't know, if this answer is useful :-(
