Let $E,F/Q$ be number field extensions where $Q$ is rational number. Suppose $p$ is a prime number s.t. $p$ does not ramify in $E$ and $F$.
Note the following reasoning does not use $E/Q$ is galois or $F/Q$ is galois. (I feel this is suspicious.)
$\textbf{Q:}$ Is this reasoning correct? WLOG, assume $E,F$ linearly disjoint over $K=E\cap F$. Since ramification index is 1 for $E/K$ and $F/K$, it suffices to replace $Q$ by $K$ and assumes $E,F$ linearly disjoint over $K$. Let $S$ be the prime in $K$ corresponding to $p$. By assumption, $S$ does not ramify at $E$ and $F$. Complete $K$ by $S-$adic valuation. Ramification index is invariant under completion. Hence it suffices to assume $E/K,F/K$ are extensions of linearly disjoint local fields by flatness of completion. Since $E/K,F/K$ are not ramify at $S$, discriminant ideal of $O_E,O_F$ are coprime to $S$. Hence both discriminant ideals are $O_K$. By coprimeness of discriminant ideals, I deduce that $EF$'s discriminant ideal is $O_K$. Hence $p$ does not ramify in $EF$.
$\textbf{Q:}$ Is it true that $EF$ is non-ramified at $p$?