Suppose $K,L$ are number fields with coprime discriminants. Let $N/\mathbb{Q}$ be the galois closure of $K/\mathbb{Q}$. Is it true that $$[NL:KL] = [N:K] $$ I feel that this would be true, with some argument having to do with the discriminants being coprime but I cant quite get it.
Edit: I know that $\gcd(D_N, D_L) = 1$.
By basic field theory we have $[NL:KL]$ = $[N:N\cap KL]$, so we need to show that $K = N\cap KL$. $N\cap KL\supseteq K$ is trivial, so it suffices to show that $N\cap L\subseteq K$. In fact we can show that under your hypotheses, $N\cap L =\mathbb{Q}$.
Indeed, a prime ramifies in $N$ iff it ramifies in $K$ iff it divides $D_K$, and similarly a prime ramifies in $D_L$ iff it divides $D_L$. Since $(D_L,D_K)=1$ by assumption, the sets of primes ramifying in these fields are disjoint. Now suppose some prime $p$ ramifies in $N \cap L$. Then $p$ must ramify in both $N$ and $L$, but this is impossible by what we've shown.
Thus $K\cap L$ is an unramified extension of $\mathbb{Q}$, which must be trivial since the (narrow) class group of $\mathbb{Q}$ is $1$. Alternatively for a proof not using class field theory, as in Mindlack's comment you can use Minkowski's theorem to show that any extension of $\mathbb{Q}$ has non-trivial discriminant and hence some prime ramifies, giving the necessary contradiction.