Let $R\subset S$ be integral domains, with $R$ integrally closed in its field of fractions, and $S$ integral over $R$. Suppose that the fraction field of $S$ is a finite Galois extension of the fraction field of $R$.
(1) If $\mathfrak{a}$ is an ideal of $R$, is it possible for $\mathfrak{a}S$ to be principal in $S$ without $\mathfrak{a}$ being principal in $R$?
(2) If "yes", what is an example? If "no", does this change if I drop the stipulation that the fraction field of $S$ is a finite Galois extension of the fraction field of $R$?
Motivation: I ask because in the examples that are coming to my mind of a ring containment $R\subset S$ with $\mathfrak{a}\triangleleft R$ not principal but $\mathfrak{a}S$ principal, the reason this happens is because $S$ contains some inverses of nonunit elements of $R$, but in the case that led me to wonder about all this, $R$ and $S$ were the rings of integers of algebraic number fields, whereupon $S$'s integrality over integrally-closed $R$ ruled out that particular way for it to happen.