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?

Solutions Collecting From Web of "Splitting of primes in the compositum of fields"

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
=\{\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 🙁

It is true in general. A trivial direction is if $\mathfrak p$ splits in $L/K$, then it splits in any sub-extension because the spliting means unramified and no residue extensions. In particular $\mathfrak p$ splits in all $L_i$’s.

Suppose now that $\mathfrak p$ splits in all $L_i$. Without lose of generalities, we can suppose $n=2$. As the problem is local at $\mathfrak p$, we can localize and suppose $K$ is a discrete valuation field (you can stick to number fields if you prefer). Denote by $O_K$ the valuation ring of $K$ and by $O_{L_i}$ the integral closure of $O_K$ in $L_i$. Let $\pi$ be a uniformizing element of $O_K$. By hypothesis $O_{L_i}/(\pi)$ is a direct sum of copies of $k$, the residue field of $K$.

Consider the tensor product $A=O_{L_1}\otimes O_{L_2}$ over $O_K$. Its generic fiber $A\otimes K$ is $L_1\otimes_K L_2$ and is reduced because $L_i/K$ is separable. And $A/\pi A=O_{L_1}/(\pi) \otimes_k O_{L_2}/(\pi)$ is a direct sum of copies of $k$.
I claim that $O_L$ is a quotient of $A$. This will imply that $O_L/(\pi)$ is a direct sum of copies of $k$ hence $\mathfrak p$ splits in $L$.

Proof of the claim. Consider the canonical map
$$f : A\otimes K=L_1\otimes L_2\to L, \quad x_1\otimes x_2\mapsto x_1x_2.$$
The image $f(A)$ is a subring of $L$, finite over $O_K$ because $A$ is finite over $O_K$. Hence $f(A)\subseteq O_L$. Moreover $f(A)/(\pi)$ is a quotient of $A/\pi A$, so it is a direct sum of copies of $k$. In particular $O_K$ is unramified in $f(A)$, so $f(A)$ is regular hence equal to $O_L$.

If I understand the question correctly the answer should be no.

Using magma I verified the following facts:

$(5)$ splits in the splitting field of $x^3-2$ over $\mathbb{Q}$ (it is the product of 3 distinct prime ideals) but it is inert in $\mathbb{Q}(\zeta_3)$. Since $split(x^3-2)=\mathbb{Q}(\sqrt[3]{2}) \mathbb{Q}(\zeta_3)$ where $\mathbb{Q}(\sqrt[3]{2})$ is not a Galois-extension this should yield a counterexample.

Edit: After thinking a bit about the question I realized you probably meant “splits completely”. So far I have not been able to construct a counterexample for this case

A simple way to see this is that if p splits completely in an extension then it splits completely in the normal closure of that extension. The result then follows easily.