Let $R$ be a ring with unity. Take for granted that an $R$-module $M$ is simple if and only if it is cyclic, and that if $m \in M$ then $Rm$ is a submodule of $M$.

Let $M$ be an $R$-module and suppose $N=\{m_1, \ldots ,m_k \}$ generates $M$. Then each $Rm_i$ is a simple module, and since $span(N)=M$, it follows that $$Rm_1 \oplus \ldots \oplus Rm_k=M ,$$ and the result is proven.

Is the above true?


Take for granted that an $R$-module $M$ is simple if and only if it is cyclic,

At this point, you have implicitly assumed $R$ is a division ring, so every $R$ module is a sum of simple modules (copies of $R$).

since $span(N)=M$, it follows that $$Rm_1 \oplus \ldots \oplus R_k=m ,$$

This assumes that your generating set is “a basis” which is true when $R$ is a Division ring if you additionally assume it is a minimal generating set.

So the result is true, but you haven’t established it.

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  • $\begingroup$ Why do I need to assume that $R$ is a division ring if I don't use that in the proof? And is assuming that $M$ is torsion-free enough $Rm_1 \oplus \ldots \oplus Rm_k = M$ to hold? $\endgroup$ – Manj Mar 6 at 0:16
  • $\begingroup$ @MonzurRahman You do use it to conclude such a thing as a basis exists. That is not possible in all commutative rings, even. $\endgroup$ – rschwieb Mar 6 at 0:29
  • $\begingroup$ Sorry, I meant in the proof of 'A module is simple if and only if it's cyclic.' Surely we don't need $R$ to be a division ring for that to hold? And if I add the assumption that $M$ is torsion-free, is that enough? $\endgroup$ – Manj Mar 6 at 0:32
  • $\begingroup$ @MonzurRahman With your hypothesis: $R$ itself is cyclic, therefore simple. A ring which is a simple right module over itself is necessarily a division ring. Your assumption is very restrictive. $\endgroup$ – rschwieb Mar 6 at 0:43

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