I'm using the following (nonstandard?) definition of an etale ring homomorphism: $f: A \rightarrow B$ is etale if $B$ is finite, projective, and separable as an $A$-module. Here separable is defined as saying that the trace map $B \rightarrow A$ is nondegenerate, as in these notes that I'm reading. The picture below is from the same set of notes (pg. 23).
$B$ is a finitely generated module over $A$ (maybe we need $L/K$ separable for this, not sure), and since $A_{\mathfrak p}$ is a PID, $B$ is projective as an $A$-module. I'm trying to see why $B/A$ being unramified implies nondegeneracy of the trace, and moreover how a prime of $A$ being ramified in $B$ would prevent this from happening.
Let $\mathfrak d$ be the different of $L/K$ (assumed finite and separable), which controls ramification.
The trace pairing $\mathfrak d^{-1} \times B \rightarrow A$ induces $\mathfrak d^{-1} \cong B^{\vee} = \text{Hom}_A(B,A)$, and no fractional ideal $I \subsetneq \mathfrak d^{-1}$ satisfies $I \cong B^{\vee}$ via the trace pairing.