• 호남수학학술지
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• 제28권1호
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• pp.69-81
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• 2006

First, we give a complete description of the multiplication modules over local Dedekind domains. Second, if R is the pullback ring of two local Dedekind domains over a common factor field then we give a complete description of separated multiplication modules over R.

• 호남수학학술지
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• 제22권1호
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• pp.1-7
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• 2000

• Kyungpook Mathematical Journal
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• 제51권1호
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• pp.1-10
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• 2011

Let (R, T) be a normal pair of commutative rings (i.e., R ${\subseteq}$ T is a unita extension of commutative rings, not necessarily integral domains, such that S is integrally closed in T for each ring S such that R ${\subseteq}$ S ${\subseteq}$ T) such that the total quotient ring of R is a von Neumann regular ring. Let P be one of the following ring-theoretic properties: going-down ring, extensionally going-down (EGD) ring, locally divided ring. Then R has P if and only if T has P. An example shows that the "if" part of the assertion fails if P is taken to be the "divided domain" property.

• Kyungpook Mathematical Journal
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• 제55권4호
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• pp.817-826
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• 2015

Let R be a commutative ring with total quotient ring K. Each monomorphic R-module endomorphism of a cyclic R-module is an isomorphism if and only if R has Krull dimension 0. Each monomorphic R-module endomorphism of R is an isomorphism if and only if R = K. We say that R has property (${\star}$) if for each nonzero element $a{\in}R$, each monomorphic R-module endomorphism of R/Ra is an isomorphism. If R has property (${\star}$), then each nonzero principal prime ideal of R is a maximal ideal, but the converse is false, even for integral domains of Krull dimension 2. An integral domain R has property (${\star}$) if and only if R has no R-sequence of length 2; the "if" assertion fails in general for non-domain rings R. Each treed domain has property (${\star}$), but the converse is false.

• Kyungpook Mathematical Journal
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• 제60권1호
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• pp.53-69
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• 2020

Let (A, M) ⊂ (B, N) be commutative quasi-local rings. We consider the property that there exists a ring D such that A ⊆ D ⊂ B and the extension D ⊂ B is inert. Examples show that the number of such D may be any non-negative integer or infinite. The existence of such D does not imply M ⊆ N. Suppose henceforth that M ⊆ N. If the field extension A/M ⊆ B/N is algebraic, the existence of such D does not imply that B is integral over A (except when B has Krull dimension 0). If A/M ⊆ B/N is a minimal field extension, there exists a unique such D, necessarily given by D = A + N (but it need not be the case that N = MB). The converse fails, even if M = N and B/M is a finite field.