Elementary proof that $\mathfrak{p}$ unramified in $L,L'$ implies unramified in $LL'$?

Let $K$ be a number field, let $\mathfrak{p}$ be a prime of $K$, and let $L,L'$ be extensions of $K$. Suppose $\mathfrak{p}$ doesn't ramify in either $L$ or $L'$. Is there a simple proof that $\mathfrak{p}$ is unramified in the composite field $LL'$?

Google searching, I found a proof in Narkiewicz, Elementary and Analytic Theory of Algebraic Numbers, p. 159, but it uses the different, which I am not comfortable with yet. Can this be proven without the different?

If it would simplify things, I would be mostly satisfied with the case that $K=\mathbb{Q}$.

Thanks in advance.

Solution 1:

The result you are looking for is Theorem 31 of Marcus' Number Fields. I am appending the proof here in case you don't have a copy of it.

Suppose $\mathfrak{p}$ is unramified in $L$ and $L'$; let $P$ be a prime of $LL'$ lying over $\mathfrak{p}$. We wish to show that $e(P|\mathfrak{p}) = 1.$ Let $M$ be the normal closure of $LL'$ and let $Q$ be a prime in $M$ lying over $P$. Thus we have inclusions $$\mathfrak{p} \subseteq P \subseteq Q.$$

Let $E$ denote the inertia group $E(Q|\mathfrak{p})$. Now the primes $Q \cap \mathcal{O}_L$ and $Q \cap \mathcal{O}_{L'}$ are unramified in $L$ and $L'$ respectively because $\mathfrak{p}$ is unramified in those places. By virtue of the inertia field $M^E$ being the largest extension in which $\mathfrak{p}$ is unramified, we have $L \subseteq M^E$ and $L' \subseteq M^E$. However this means that the composite

$$LL' \subseteq M^E$$

from which it follows that $\mathfrak{p}$ is unramified in $LL'$.