Direct sum of modules and length.

Question

How do you prove this: let be $_RM$ and $(_RM_i)_{i\in I}$ such that $M=\sum_{i\in I}M_i,$ suppose that $l(M)$ is finite and $l(M)=\sum_{i\in I}l(M_i)$ then $M=\bigoplus_{i\in I}M_i.$?

Answer 1

Take $I=\{1,2\}$. There is a property of length: $$l(M_1+M_2)+l(M_1 \cap M_2)=l(M_1)+l(M_2).$$ So if $M=M_1+M_2$ and $l(M)=l(M_1)+l(M_2)$, then $M_1 \cap M_2 = 0$. Then $M= M_1 \oplus M_2$. If $I$ is a finite set, then we can get the result similarly: since $M=\Sigma_{i\in I}M_i$, then $l(M)+l(M_j \cap(\Sigma_{i \in I\setminus j}M_i))=l(M_j)+l((\Sigma_{i \in I\setminus jM_i})) \le \Sigma_{i \in I} l(M_i)$, while $l(M)=\Sigma_{i \in I} l(M_i)$. Then $M_j \cap(\Sigma_{i \in I\setminus j}M_i)=0$ for any $j\in I$, so we can get that $M=\oplus_{i \in I} M_i$.