# Torsion-free modules by Eben Matlis

By Eben Matlis

The topic of torsion-free modules over an arbitrary crucial area arises obviously as a generalization of torsion-free abelian teams. during this quantity, Eben Matlis brings jointly his examine on torsion-free modules that has seemed in a few mathematical journals. Professor Matlis has remodeled a few of the proofs in order that basically an straightforward wisdom of homological algebra and commutative ring conception is important for an realizing of the theory.

The first 8 chapters of the booklet are a basic advent to the idea of torsion-free modules. This a part of the booklet is acceptable for a self-contained easy path at the topic. extra really good difficulties of discovering all integrally closed D-rings are tested within the final seven chapters, the place fabric coated within the first 8 chapters is applied.

An quintessential area is expounded to be a D-ring if each torsion-free module of finite rank decomposes right into a direct sum of modules of rank 1. After a lot research, Professor Matlis came across that an integrally closed area is a D-ring if, and provided that, it's the intersection of at so much maximal valuation rings.

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From the reports of the 1st printing of this publication, released as quantity fifty eight of the Encyclopaedia of Mathematical Sciences:". .. This booklet may be very helpful as a reference and consultant to researchers and graduate scholars in algebra and and topology. " Acta Scientiarum Mathematicarum, Ungarn, 1994 ". .

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Since Mr C (Rb: Ra)= P, and thus M= P. This shows that every nonzero prime ideal of R is maximal; that is R has Krull dimension 1. 55 M- 1 If M is a maximal ideal of R, then by Theorem 35(3), can be generated by two elements. (3) =:> (4). Let I be a nonzero ideal of R, and choose a non- zero element a e I. sion one, Since R is a Noetherian domain of Krull dimen- I/Ra has finite length as an R-module. • , n-1. is a reflexive ideal of R 1 and ExtR{Ra, R) = 0. hypotheses of Theorem 39 are satisfied.

We have a derived exact sequence obtained by applying the functor HomR(·, R) to exact sequence (a): (b) 0 ~I'~ A'~ Now Ext~(I, R) 2 to ExtR(R/I, R). + rank R' = 2. R' --~ Ext~(I, R) is a torsion R-module, since it is isomorphic Hence it follows from (b) that rank A'= rank 11 But then rank A = rank A', and so A is torsionless by Theorem 27. Let J = Im a' in exact sequence (b). J We have just shown that is not zero, and thus it is isomorphic to a non-zero ideal of R. We now have a commutative diagram with exact rows: 0->R~A~>I~O (c) AAl 0-> AI!

THEOREM 28. If A is a torsion-free R-module of finite rank, then A' is a reflexive R-module. Proof. A 1 :A' ..... A"' bethecanonical R-homomorphisms. A induces a homomorphism µ:A"'_,. A'. A, is the identity on A'. A, is a monomorphism. But then A' is a torsionless R-module, and hence by rank A' = rank A'". A' DEFINITIONS. can not be a is an isomorphism. An R-module will be called a universal injective R-module if it is injective and contains a copy of every simple R-module. J } the injective envelope of J.