AN APPLICATION OF SB–RINGS

Amir M. Rahimi

Abstract. All rings are commutative rings with identity and J(R) denotes the Jacobson radical of a ring R. A ring R is called a SB–ring provided that for any sequence a_{1}, a_{2}, …, a_{s}, a_{s+1} of elements in R with s ≥ 2 and (a_{1}, a_{2}, …, a_{s−1}) ⊈ J(R), there exists b ∈ R such that (a_{1}, a_{2}, … a_{s}, a_{s+1}) = (a_{1}, a_{2}, …, a_{s} + ba_{s+1}). By applying some of the properties of SB–rings, it is shown that R[X] is not a Prüfer domain for any Noetherian domain R which is not a field.

Preliminaries and the Main Result. All rings are commutative rings with identity and J(R) denotes the Jacobson radical of a ring R. For any s ≥ 1, a sequence a_{1}, a_{2}, …, a_{s}, a_{s+1} of elements in a ring R is called a unimodular sequence provided that (a_{1}, a_{2}, …, a_{s}, a_{s+1}) = R. R is said to be a B–ring, if for any unimodular sequence a_{1}, a_{2}, …, a_{s}, a_{s+1} of elements in R with s ≥ 2 and (a_{1}, a_{2}, …, a_{s−1}) ⊈ J(R), there exists b ∈ R such that (a_{1}, a_{2}, …, a_{s} + ba_{s+1}) = R. R is said to be a strongly B–ring (SB–ring) provided that for any sequence a_{1}, a_{2}, …, a_{s}, a_{s+1} of elements in R with s ≥ 2 and (a_{1}, a_{2}, …, a_{s−1}) ⊈ J(R), there exists b ∈ R such that (a_{1}, a_{2}, …, a_{s}, a_{s+1}) = (a_{1}, a_{2}, …, a_{s} + ba_{s+1}). For a detailed study of B–rings and SB–rings, see [2]. Furthermore, for a more general case of B–type rings see the dissertation of the author [3].

A Prüfer domain is an integral domain in which every nonzero finitely generated ideal is invertible. A Dedekind domain is an integral domain in which every nonzero ideal is invertible.

\underline{Lemma 1}. If R is a Dedekind domain, then R is a SB–ring.

\underline{Proof}. See Theorem 3.2 in [2].

\underline{Lemma 2}. R[X] is a SB–ring if and only if R is a field.

\underline{Proof}. See Theorem 3.4 in [2].

\underline{Theorem}. If R is a Noetherian domain which is not a field, then R[X] cannot be a Prüfer domain.

\underline{Proof}. Suppose R[X] is a Prüfer domain. Since every ideal in a Noetherian domain is a finitely generated ideal, then R[X] must be a Dedekind domain. Now by applying Lemma 1 and Lemma 2 above, we can conclude that R is a field and this is a contradiction to the choice of R.

\underline{Remark}. From the above theorem, it is easy to see that Z[X] is not a Prüfer domain, where Z is the ring of rational integers. See also [1] for an argument that shows Z[X] is not a Prüfer domain.

\underline{Acknowledgment}. This work was made possible by the sponsorship of Professor Yosi Shibbery of Terre Haute, Indiana.

\underline{References}

1. D. Brizolis, “A Theorem on Ideals in Prüfer Rings of Integer–Valued Polynomials,” Commutative Algebra, 7 (1979), 1065–1077.

2. M. Moore and A. Steger, “Some Results on Completability in Commutative Rings,” Pacific Journal of Mathematics, 37 (1981), 453–460.

3. A. M. Rahimi, Some Results on Stable Range in Commutative Rings, Ph.D. dissertation, 1993, University of Texas at Arlington.

Amir M. Rahimi Insititute for Studies in Theoretical Physics and Mathematics Department of Mathematics Niavaran Building P.O. Box 19395-5746 Tehran, IRAN