Do integer primes split completely in intermediate fields? Is it true that if the ideal, generated by the integer prime $p,$ splits in $KL,$ then it splits completely in both $K$ and $L$? I only know that if $p$ has a first degree prime divisor in $KL,$ then it has a first degree prime divisor in $K$ and $L,$ but I don't know how to approach complete splitting.
 A: If we have a chain of extensions of number fields $K \subset L \subset E$
with rings of integers $\mathcal O_K,\mathcal O_L$ and $\mathcal O_E$ respectively, then for any maximal ideal $\mathfrak{p}$ of $\mathcal O_K$ with an ideal $\mathfrak q$ in $\mathcal O_L$ lying above it and an maximal ideal $\mathfrak r$ in $\mathcal O_E$ lying above $\mathfrak q$, we have following equalities which are easy to verify:
$$f(\mathfrak r / \mathfrak p)=f(\mathfrak r / \mathfrak q) f(\mathfrak q / \mathfrak p)$$ (this is just the tower rule for field extensions applied to the extensions of residue fields.) Where $f$ denotes the inertia degree, i.e. the degree of the residue field extension.
We also have the equality:
$$e(\mathfrak r / \mathfrak p)=e(\mathfrak r / \mathfrak q) e(\mathfrak q / \mathfrak p)$$ where $e(a/b)$ is the exponent of $a$ in the factorization of the extension of $b$. This follows immediately from unique factorization into prime ideals.
Now a prime ideal $\mathfrak{p}$ splits completely in $L$ iff for all $\mathfrak{q}$ that lie above $\mathfrak{p}$ we have $ e(\mathfrak q / \mathfrak p) =  f(\mathfrak q / \mathfrak p) = 1$.
If we take any prime ideal $\mathfrak{q}$ above $\mathfrak{p}$, then there is some prime ideal $\mathfrak{r}$ above $\mathfrak{q}$, then we get, as $\mathfrak{p}$ splits completely in $E$:
$$1 =f(\mathfrak r / \mathfrak p)=f(\mathfrak r / \mathfrak q) f(\mathfrak q / \mathfrak p) \Rightarrow f(\mathfrak q / \mathfrak p) = 1$$
$$1 =e(\mathfrak r / \mathfrak p)=e(\mathfrak r / \mathfrak q) e(\mathfrak q / \mathfrak p) \Rightarrow e(\mathfrak q / \mathfrak p) = 1$$
So $\mathfrak{p}$ splits completely in $L$.
In your situation you can take $E$ to be the compositum of $L$ with some other field and $K= \mathbb Q$.
A: Yes.  If $\Bbb Q\subseteq K\subseteq F$ are finite extensions, and $p$ splits
completely in $F$ then it splits completely in $K$. To see this, note that
as $p$ splits completely in $F$ then every embedding of $F$ into the 
algebraic closure of $\Bbb Q_p$ (the $p$-adic numbers) has image 
contained in $\Bbb Q_p$. Each embedding of $K$ into the 
algebraic closure of $\Bbb Q_p$ extends to an embedding of $F$ into the 
algebraic closure of $\Bbb Q_p$, so also has image inside $\Bbb Q_p$.
Therefore $p$ splits completely in $K$.
A: As @MathinBoulomenos said, local methods may be not necessary, but they can make things more "visible". For an extension $E/k$ and a prime ideal (or a place) $P$ of $k$, the complete splitting of $P$ in $E$ means that the completions $k_P$ and $E_Q$ coincide for any prime $Q$ above $P$. Thus in a tower $E/F/k$, if $P$ splits in $E$, it must split in $F$.
