منابع مشابه
Essentially Retractable Modules
We call a module essentially retractable if HomR for all essential submodules N of M. For a right FBN ring R, it is shown that: (i) A non-zero module is retractable (in the sense that HomR for all non-zero ) if and only if certain factor modules of M are essentially retractable nonsingular modules over R modulo their annihilators. (ii) A non-zero module is essentially retractable if and on...
متن کاملEssentially Reductive Hilbert Modules
Consider a Hilbert space obtained as the completion of the polynomials C[z ] in m-variables for which the monomials are orthogonal. If the commuting weighted shifts defined by the coordinate functions are essentially normal, then the same is true for their restrictions to invariant subspaces spanned by monomials. This generalizes the result of Arveson [4] in which the Hilbert space is the m-shi...
متن کاملEssentially Reductive Hilbert Modules II
Many Hilbert modules over the polynomial ring in m variables are essentially reductive, that is, have commutators which are compact. Arveson has raised the question of whether the closure of homogeneous ideals inherit this property and provided motivation to seek an affirmative answer. Positive results have been obtained by Arveson, Guo, Wang and the author. More recently, Guo and Wang extended...
متن کاملessentially retractable modules
we call a module essentially retractable if homr for all essential submodules n of m. for a right fbn ring r, it is shown that: (i) a non-zero module is retractable (in the sense that homr for all non-zero ) if and only if certain factor modules of m are essentially retractable nonsingular modules over r modulo their annihilators. (ii) a non-zero module is essentially retractable if and on...
متن کاملRings and modules
Proof. Consider the map g:A/I → C , a+I 7→ f (a). It is well defined: a+I = a′ +I implies a− a′ ∈ I implies f (a) = f (a′). The element a + I belongs to the kernel of g iff g(a + I) = f (a) = 0, i.e. a ∈ I , i.e. a + I = I is the zero element of A/I . Thus, ker(g) = 0. The image of g is g(A/I) = {f (a) : a ∈ A} = C . Thus, g is an isomorphism. The inverse morphism to g is given by f (a) 7→ a + I .
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ژورنال
عنوان ژورنال: Journal of Algebra
سال: 2006
ISSN: 0021-8693
DOI: 10.1016/j.jalgebra.2005.08.018