What is the (natural) bijection between the set of all sub modules upto isomorphism and set of all isomorphic quotient modules upto isomorphism of a finitely generated torsion module over a PID. Is there any inclusion relation between these classes?
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What is «the isomorphic class of sub modules» – Mariano Suárez-Álvarez Apr 17 '13 at 05:28
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collection of all sub modules upto isomorphism of modules – GA316 Apr 17 '13 at 05:29
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So, the «set of isomorphism classes of submodules». – Mariano Suárez-Álvarez Apr 17 '13 at 05:30
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3This (or half of this) was essentially asked for abelian groups here and for PIDs the reasoning is just the same. The conclusion there is that every quotient of an finite abelian group is isomorphic to a subgroup. – Mariano Suárez-Álvarez Apr 17 '13 at 05:31
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yes. this is what I mean exactly. – GA316 Apr 17 '13 at 05:31
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Well, then edit the question so that it is clearly expressed :-) – Mariano Suárez-Álvarez Apr 17 '13 at 05:31
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1On the other hand, if there is a sensible bijection between the sets of isomorphism classes of quotients and of submodules, there iss exactly one such bijection, so naturality does not mean much in this context! – Mariano Suárez-Álvarez Apr 17 '13 at 05:34
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ya. but I am looking for a bijection which is natural. – GA316 Apr 17 '13 at 05:35
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I dont understand your statement clearly. please explain me – GA316 Apr 17 '13 at 05:36
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Natural in what sense? – Mariano Suárez-Álvarez Apr 17 '13 at 05:36
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I think it is better for you to read the related question I linked to, think a bit about it, and then try to solve your problem. – Mariano Suárez-Álvarez Apr 17 '13 at 05:37
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For abilion groups (Z-modules) I have understand the answer. but how to prove for general free torsion module over a PID? thanks. – GA316 Apr 17 '13 at 05:37
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natural in the sense given a sub module we can able to guess the corresponding quotient module. there should be some natural relation between them. I cant able to tell clearly. sorry. – GA316 Apr 17 '13 at 05:39
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Well, you apparently can do this for abelian groups: what is the correspondence in that case? – Mariano Suárez-Álvarez Apr 17 '13 at 05:40
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If $M$ is a f.g. torsion module over a PID $R$, one can show that every submodule of $M$ is isomorphic to a quotient of module of $M$, and vice versa, by looking at a decomposition of $M$ into direct summands of cyclic modules. However, this correspondence is not natural.
However, the following works: Let $K$ be the field of fractions of $R$. Then $\hom(-,K/R)$ ("Pontrjagin dual") is a contravariant functor from $R$-modules to $R$-modules. For every $R$-module $M$ there is a canonical homomorphism $M \to \hom(\hom(M,K/R),K/R)$. If $M$ is finite cyclic, this is easily seen to be an isomorphism. Hence, it is an isomorphism for all f.g. torsion modules. It follows that $\hom(-,K/R)$ is an anti-equivalence of categories from f.g. torsion modules to itsself.
Now it follows by abstract nonsense that submodules (quotients) of $M$ correspond naturally to quotients (submodules) of $\hom(M,K/R)$.
Besides, for every f.g. torsion module $M$ there is some isomorphism $M \cong \hom(M,K/R)$ (which is not natural). But once this is fixed, we get a correspondence between submodules of $M$ and quotients of $M$, where these modules are abstractly isomorphic.
Martin Brandenburg
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