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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
up vote 2 down vote accepted

This idea was Kummer's (and maybe Kronecker's) way to overcome non-PID-ness of rings of algebraic integers: anticipating the "Haupt-Ideal-Satz" of Furtwangler (illustrated in the quadratic case by Hilbert in his 1896 write-up on alg no th), in the "Hilbert class field" (Furtwangler prove that the maximal unramified abelian extension has this property) of an alg no field, every ideal become principal.

Golod-Shafarevich showed in 1964 that there are infinite "Hilbert classfield towers", so that in general it is futile to try to make rings of alg integers PIDs by repeatedly moving to Hilbert classfields.

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.

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