Let [tex]R[/tex] be a ring with [tex]1_R[/tex]. If [tex]M[/tex] is an

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SUMMARY

The discussion centers on the properties of R-modules in the context of ring theory, specifically addressing the scenario where M is an R-module that is not unitary. It is established that for some element m in M, the product Rm equals zero, denoted as Rm = 0. The definition of Rm is clarified as the set of all products of the form {r · m | r ∈ R}. The distinction between unitary and non-unitary modules is emphasized, particularly the implication that 1_R · x ≠ x for some x in M.

PREREQUISITES
  • Understanding of ring theory, specifically the definition of rings and R-modules.
  • Familiarity with the concept of unitary modules and their properties.
  • Knowledge of the notation and operations involving elements of rings and modules.
  • Basic comprehension of proofs in abstract algebra.
NEXT STEPS
  • Study the properties of unitary versus non-unitary R-modules in depth.
  • Explore examples of rings and their corresponding modules to solidify understanding.
  • Learn about the implications of Rm = 0 in various algebraic structures.
  • Investigate the role of the identity element in ring theory and its impact on module behavior.
USEFUL FOR

Students and researchers in abstract algebra, particularly those focusing on ring theory and module theory, will benefit from this discussion.

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Let [tex]R[/tex] be a ring with [tex]1_R[/tex]. If [tex]M[/tex] is an R-module that is NOT unitary then for some [tex]m \in M[/tex], [tex]Rm = 0[/tex].

I'm pretty sure [tex]Rm = \{ r \cdot m \mid r \in R \}[/tex]. While M being not unitary means that [tex]1_R \cdot x \neq x[/tex] for some [tex]x \in M[/tex]. I'm thinking this problem should be an obvious and direct proof but I can't see it.
 
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If m ≠ 1m, then consider m - 1m ≠ 0.
 


:blushing:

that's embarassing.
 

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