Can Direct Sum Isomorphism Imply Module Equality in PID?

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SUMMARY

The discussion centers on the assertion that if M and N are finitely generated modules over a Principal Ideal Domain (PID) and M ⊕ M ≅ N ⊕ N, then M must be isomorphic to N. Participants clarify that this statement is not universally true without the finitely generated condition, as demonstrated by counterexamples involving free abelian groups of infinite rank. The structure theorem for finitely generated modules over PIDs is highlighted as a crucial tool in understanding the implications of the direct sum isomorphism.

PREREQUISITES
  • Understanding of finitely generated modules over Principal Ideal Domains (PIDs)
  • Familiarity with the structure theorem for modules
  • Knowledge of direct sum and isomorphism concepts
  • Basic concepts of abelian groups and their ranks
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  • Study the structure theorem for finitely generated modules over PIDs in detail
  • Explore counterexamples involving free abelian groups of infinite rank
  • Learn about the properties of direct sums in module theory
  • Investigate the implications of isomorphism in algebraic structures
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Mathematicians, algebraists, and students studying module theory, particularly those focusing on finitely generated modules over PIDs and their properties.

bruno321
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Hi. I'm trying to prove this "little fact": let M, N be finitely generated modules over a PID. Then if M+M=N+N (where = means isomorphism and + means direct sum) then M=N.

I'm sure it can be done with the structure theorem (it is obvious from the hypotheses); it looks like it should be trivially proven, but alas, I don't think it can be.

What do you think?

Cheers,
 
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EDIT: this isn't true the way I originally thought it was...thinking harder!

Well it's certainly not true without the finitely generated over a PID hypothesis (just think about free abelian groups of infinite rank), so I'm guessing it's some particular property of finitely generated modules over PIDs. The structure theorem immediately comes to mind.
 
M + m = n + n.
2m = 2n
m = n
 

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