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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?

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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 🙁

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.

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