compositum of an unramified and a totally ramified extension Suppose $L$ and $K$ are abelian extensions of $\mathbb Q$ such that $LK$ is also an abelian extension of $\mathbb Q$. Let $p$ be totally ramified in $L$ with ramification index $e$ and split into $r$ primes in $K$ each with residue field degree $f$ and ramification index 1. Then how does $p$ behave in $LK$? 
 A: First, note that the given conditions force $ LK/K $ to be totally ramified at every prime of $ \mathcal O_K $ lying over $ p $. Indeed, choosing a prime $ \mathfrak p $ lying over $ p $ in $ \mathcal O_K $, and letting $ \mathfrak q $ be a prime (we will see it is the prime) lying over $ \mathfrak p $ in $ LK $, we have
$$ [L : \mathbf Q] = e \leq e_{\mathfrak q | p} = e_{\mathfrak q | \mathfrak p} e_{\mathfrak p | p} = e_{\mathfrak q | \mathfrak p} \leq [LK : K] $$
However, we always have $ [L : \mathbf Q] \geq [LK : K] $, so it follows that we have equalities throughout, and $ e_{\mathfrak q | \mathfrak p} = [LK : K] = [L : \mathbf Q] $, i.e $ LK/K $ is totally ramified at any prime of $ K $ lying over $ p $. Now, let $ \mathfrak q $ be any prime of $ LK $ lying over $ p $, and let $ \mathfrak p $ be the prime downstairs in $ K $. Since $ LK/K $ is totally ramified at $ \mathfrak p $, the inertia degrees must be trivial, so we have
$$ f_{\mathfrak q | p} = f_{\mathfrak q | \mathfrak p} f_{\mathfrak p | p} = 1 \cdot f = f $$
Thus, the shared inertia degree is $ f $ and the shared ramification index is $ e $. Finally, we have
$$ [LK : \mathbf Q] = [LK : K][K : \mathbf Q] = [L : \mathbf Q][K : \mathbf Q] = efr $$
so the fundamental identity implies that $ p $ splits into $ r $ primes in $ LK $. Note that the assumption that the extensions are abelian is redundant, the result is the same for arbitrary Galois extensions.
