mathbb { Z } ) } $ $ have the stacked bases property ?

It is known that a Prüfer domain either with dimension 1 or with finite character has the stacked bases property. Following Brewer and Klinger, some rings of integer-valued polynomials provide, for every n ≥ 2, examples of n-dimensional Prüfer domains without finite character which have the stacked bases property. But, the following question is still open: does the two-dimensional Prüfer domain Int(Z) = { f ∈ Q[X ] | f (Z) ⊆ Z} have the stacked bases property? By means of the UCS-property, we reduce the question to the search for some 2 × 2 matrices with coefficients in Int(Z). Mathematics Subject Classification 13F20 · 13C10 · 13F05 1 The stacked bases property If you cannot solve the proposed problem, try to solve first some related problem. Could you imagine a more accessible related problem? ... G. Pólya [8] The following result is well known. It is itself a generalization of the structure theorem for the finitely generated abelian groups. Proposition 1.1 Let D be a principal ideal domain. If M is a free D-module with finite rank m, then every submodule N of M is also a free D-module with rank n ≤ m. Moreover, there exists a basis (e1, . . . , em) of M and elements a1, . . . , an of D such that (a1e1, . . . , anen) is a basis of N . We still may ask that a j divides a j+1 for 1 ≤ j ≤ n − 1. J.-L. Chabert (B) Départment de Mathématiques, LAMFA CNRS-UMR 6140, Université de Picardie, 80039 Amiens, France E-mail: [email protected]