Splitting of a prime in compositum Let $K$ be a number field. Let $L_1$ and $L_2$ be two extensions of $K$. Let $P\subseteq O_K$ be a prime ideal. We know that if $P$ is unramified in $L_1$ and $L_2$ it remains unramified in the compositum $L_1L_2$.
Do we have a similar theorem when $P$ splits completely in both $L_1$ and $L_2$, namely: 
If $P$ splits completeely in both $L_1$ and $L_2$, can we say that $P$ splits completely in the compositum $L_1L_2$?
 A: @Gal Porat. It is perhaps more "visible" to use the completion of the number field $K$ at $P$. The prime $P$ splits completely in $L_i$ if and only if the local field $K_P$ coincides with $(L_i)_Q$ for all primes $Q$ of $L_i$ above $P$. Hence if $P$ splits completely in both $L_1$ and $L_2$, it will split completely in their compositum.
A: The answer is yes, essentially by the same proof. 
In general, if $E/K$ is a Galois extension of number fields, and $Q$ is a prime of $O_E$ lying over a prime $P$ of $O_K$, we can consider the decomposition group $D=D(Q|P)$. Then the fixed field $E^D$ is called the decomposition field. Suppose we have an intermediate field $K\subset L\subset E$, with $I$ being the prime ideal lying between $Q$ and $P$. It can be shown that $L\subset E^D$ if and only if $e(I|P)=f(I|P)=1$.
In regards to your question, embed $L_1L_2$ in some Galois extension $E/K$.
Let $Q$ be any prime of $E$ over $P$, and let $D=D(Q|P)\leq Gal(E/K)$ be the decomposition group as before. Let $I_1,I_2$ be the primes in $L_1,L_2$ respectively lying between $Q$ and $P$. Since by the assumption $I_1,I_2$ have $e(I_i|P)=f(I_i|P)=1$, it follows that $L_1,L_2\subset E^D$. Hence also $L_1L_2\subset E^D$. 
Therefore any prime of $L_1L_2$ that lies over $P$ has inertia degree and ramification equal to $1$ over $P$, i.e. $P$ splits completely in $L_1L_2$. 
The proof for ramification is the same when we replace the decomposition group with the inertia group.
