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arXiv · 2410.14750

Free modules with isomorphic duals

Theodoros Kyriopoulos
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Abstract

Let M, N be free modules over a Noetherian commutative ring R and let F be a field such that card(F) does not exceed the continuum. Then : (1) The assertion that [Any two F-vector spaces with isomorphic duals are isomorphic] is equivallent to the ICF (Injective continium function) hypothesis and it is a non-decidable statement in ZFC. (2) If the dual of M is a projective R-module and rank(M) is infinite then the ring R is Artinian. (3) If R is Artinian and card(R) does not exceed the continuum then the the dual of M is free. (4) Assume that R is a non-Artinian ring that is either Hilbert or countable. Then : (a) If M, N have isomorphic duals then they are themselves isomorphic (b) Any free direct summand of the dual of M is finitely generated, if Card(R) is not omega-measurable. (c) If R is connected and both Card(R), rank(M) are not omega-measurable then [Any direct summand of the dual of M that is not finitely generated is a dual of a free R-module]. (5) If R is a non-local domain then R is a half-slender ring. (6) If R is Artinian ring and it's cardinality card(R) does not exceed the continuum then the assertion that [any two free R-modules with isomorphic duals are isomorphic] is non-decidable in ZFC. (7) If R is a domain and rank(M) is infinite then the Goldie dimension of the dual of M is equal to it's cardinality . (8) If R is a complex affine algebra whose corresponding affine variety has no isolated points then [any two projective R-modules with isomorphic duals are themselves isomorphic]. (9) Let V be an F-vector space of infinite dimension . The dimension of it's dual is cardinality of the powerset of the the dimension of V.

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Theodoros Kyriopoulos. 2024-10-17. Free modules with isomorphic duals. https://arxiv.org/abs/2410.14750

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