# Can every finitely generated module be written as a sum of simple modules?

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.

• 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? – Manj Mar 6 at 0:16
• @MonzurRahman You do use it to conclude such a thing as a basis exists. That is not possible in all commutative rings, even. – rschwieb Mar 6 at 0:29
• 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? – Manj Mar 6 at 0:32
• @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. – rschwieb Mar 6 at 0:43