Does Int $${(\mathbb {Z})}$$ have the stacked bases property?

Integer-valued polynomials, Prüfer domains and the stacked bases property

“…We already studied the case of Int(Z) in [12] where we proved that in the previous proposition, under some hypotheses, we may assume that m = l = 2 thanks to the following technical lemma: (1) the ring S −1 R has the BCU property, (2) for every a ∈ S, the ring R/aR has the BCU property. Then, for every n, every finitely generated submodule of R n with unit content contains a submodule with unit content which may be generated by two elements.…”

“…The last assertion about the contents is not in [12,Proposition 4.3] but is in its proof. Now, let us recall the definition of the BCU property which is a strong form of the UCS property: Definition 5.3.…”

“…In order to prove this result, using Frisch's results [16], we give first a complete description of the spectrum of Int(E, V ) when Int(E, V ) is Prüfer (Theorem 3.5). In a second part (section 5), we study the 'global case' with Int(O K ) where O K denotes the ring of integers of a number field K and we reduce the question to know whether Int(O K ) has the stacked bases property to questions concerning 2 × 2-matrices (Theorem 5.6) by continuing the work undertaken in [12].…”

“…[12, Proposition 4.3] Let R be a ring and S be a multiplicative subset of R without zero divisors such that…”

“…[12, Corollary 4.4] Let R be a ring and S be a multiplicative subset of R without zero divisors such that (1) the ring S −1 R has the BCU property,(2) for every a ∈ S, the ring R/aR has the BCU property. Then, R has the UCS property if and only if ( * ) ∀B ∈ M n×2 (R) such that cont(B) = R, ∃C ∈ M 2×2 (R) such that cont(BC) = R and every 2 × 2-minor of BC is zero Moreover, we may consider only matrices B = (B 1 B 2 ) such that cont(B 1 ) ∩ S = ∅ and cont(B 2 ) + aR = R where a ∈ cont(B 1 ) ∩ S.…”