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].(adsbygoogle = window.adsbygoogle || []).push({});

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

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