# If $S$ is integral over integrally closed $R$, can $\mathfrak{a}S$ be principal without $\mathfrak{a}\triangleleft R$ being principal?

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.

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Just to add to Prof Garrett's helpful remarks, if you think of the usual first example $\mathbb Q(\sqrt{-5})$ of a quadratic field of class number $> 1$ and its Hilbert class field $\mathbb Q(i, \sqrt{5})$, and extend the non-principal ideal $(2, 1 + \sqrt{-5})$, then this should give an example. –  Dylan Moreland Sep 1 '12 at 2:26

As in Hilbert's examples, one can directly see that in, for example, complex quadratic extensions of $\mathbb Q$ with class number 2, there is an obvious unramified quadratic extension with obvious interactions with the ideals in the lower field, etc.