Let $R$ be a Dedekind domain with quotient field $K$, let $L$ be a finite separable extension of $K$, and let $S$ be the integral closure of $R$ in $L$. If $\mathfrak{p}$ is a nonzero prime ideal of $R$ that is contained in the union of the prime ideals of $R$ that split completely in $S$, does it follow that $\mathfrak{p}$ splits completely in $S$? It follows easily if $\mathfrak{p}$ is principal or (by prime avoidance) if only finitely many prime ideals of $R$ split completely in $S$. If necessary, assume that $R$ is the ring of integers in a number field and/or $L/K$ is Galois.



The union of the totally split primes
If the class group is finite, then writing $\mathfrak{p}^h = (\alpha)$, it follows that
$\alpha$ is contained in a prime ideal $\mathfrak{q}$ which splits completely in $S$, and thus $\mathfrak{p}^h \subseteq \mathfrak{q} \Rightarrow \mathfrak{p} = \mathfrak{q}$ (because $R$ has dimension one). It sounds like that suffices for your purposes.


