Proof. Let $\mathfrak{a}$ be an arbitrary ideal of a Dedekind domain $R$ . Let $\mathfrak{b}$ be such an ideal of $R$ that $\mathfrak{ab}$ is a principal ideal $(\beta)$ . The lemma to which this entry is attached gives also an element $\gamma$ and an ideal $\mathfrak{c}$ of $R$ such that $\mathfrak{ac} = (\gamma)$ and $\mathfrak{b+c} = R$ . Then we have $$\mathfrak{a} = \gcd(\mathfrak{ab},\,\mathfrak{ac}) = \gcd((\beta),\,(\gamma)) = (\beta,\,\gamma)$$ because $\gcd(\mathfrak{b},\,\mathfrak{c}) = \mathfrak{b+c} = R = (1)$ .

The Dedekind domains are trivially Prüfer domains, but the two-generator property can not be generalized to the invertible ideals of all Prüfer domains (and Prüfer rings): Schülting has constructed an invertible ideal of a Prüfer domain that can not be generated by less than three generators. The example of Schülting is the fractional ideal $(1,\,X,\,Y)$ of the Prüfer domain $\bigcap_j B_j$ where the $B_j$ 's run all valuation rings of the rational function field $\mathbb{R}(X,\,Y)$ which have the residue fields formally real.

Bibliography

1

EBEN MATLIS: ``The two-generator problem for ideals''. - The Michigan Mathematical Journal 17 $\mbox{N}\sp\circ$ 3 (1970).

2

HEINZ-WERNER SCHÜLTING: ``Über die Erzeugendenanzahl invertierbarer Ideale in Prüferringen''. - Communications in Algebra 7 $\mbox{N}\sp\circ$ 13 (1979). [Zentralblatt 432.13010]