Dedekind-Mertens lemma and content formulas in power series rings
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- Title
- Dedekind-Mertens lemma and content formulas in power series rings
- Authors
- Park, Mi Hee; Kang, Byung Gyun.; Phan Thanh Toan
- Date Issued
- 2018-08
- Publisher
- ELSEVIER SCIENCE BV
- Abstract
- Let R[X] be the power series ring over a commutative ring R with identity. For f is an element of R[X], let A(f) denote the content ideal of f, i.e., the ideal of R generated by the coefficients of f. We show that if R is a Priifer domain and if g is an element of R[X] such that A(g) is locally finitely generated (or equivalently locally principal), then a Dedekind-Mertens type formula holds for g, namely A(f)(2)A(g) = A(f)A(fg) for all f is an element of R[X]. More generally for a Prefer domain R, we prove the content formula (A(f)A(g))(2) = (A(f)A(g))A(fg) for all f, g is an element of R[X]. As a consequence it is shown that an integral domain R is completely integrally closed if and only if (A(f)A(g))(v) = (A(fg))(v) for all nonzero f,g is an element of R[X], which is a beautiful result corresponding to the well-known fact that an integral domain R is integrally closed if and only if (A(f)A(g))(v) = (A(fg))(v) for all nonzero f,g is an element of R[X], where R[X] is the polynomial ring over R. For a ring R and g is an element of R[X], if A(g) is not locally finitely generated, then there may be no positive integer k such that A(f)(k+1)A(g) = A(f)(k)A(fg) for all f is an element of R[X]. Assuming that the locally minimal number of generators of A(g) is k +1, Epstein and Shapiro posed a question about the validation of the formula A(f)(k+1)A(g) = A(f)(k)A(fg) for all f is an element of R[X]. We give a negative answer to this question and show that the finiteness of the locally minimal number of special generators of A(g) is in fact a more suitable assumption. More precisely we prove that if the locally minimal number of special generators of A(g) is k + 1, then A(f)(k+1)A(g) = A(f)(k)A(fg) for all f is an element of R[X]. As a consequence we show that if A(g) is finitely generated (in particular if g is an element of R[X]), then there exists a nonnegative integer k such that A(f)(k+1)A(g)= A(f)(k)A(fg) for all f is an element of R[X]. (C) 2017 Elsevier B.V. All rights reserved.
- URI
- https://oasis.postech.ac.kr/handle/2014.oak/95718
- DOI
- 10.1016/j.jpaa.2017.09.013
- ISSN
- 0022-4049
- Article Type
- Article
- Citation
- JOURNAL OF PURE AND APPLIED ALGEBRA, vol. 222, no. 8, page. 2299 - 2309, 2018-08
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