Let $R$ a ring and $M$ a $R-$module. I want to prove that if $M\cong \bigoplus_{i=1}^n M_i$ where $M_i$ are simple $R-$ submodule $M$, then $M=\sum_{i=1}^n M_i$ (I know that he converse is wrong a priori). The fact that $\sum_{i=1}^n M_i\subset M$ is obvious. For the converge, let $m\in M$. Let $$\varphi: \bigoplus_{i=1}^n M_i\longrightarrow M$$ the bijection. Then, there is $(k_1,...,k_n)\in \bigoplus_{i=1}^n M_i$ s.t. $$m=\varphi(k_1,...,k_n).$$ I would like in fact take $\varphi(k_1,...,k_n)=k_1+...+k_n$ but this is unfortunately not an isomorphism. Any idea ?
If $M\cong \bigoplus_{i=1}^n M_i$, then $M=\sum_{i=1}^n M_i$
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abstract-algebra
modules
1 Answers
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You don't need that the $M_i$ are simple. Now that is the meaning of $M= \sum M_i$? It just says you have submodules $M_i \subset M$ and you can write every element of $M$ as a sum of elements of the $M_i$.
If you are given an isomorphism $$ \varphi \colon \bigoplus M_i \to M $$ you can identify $\varphi(M_i)$ with a submodule $M_i$. If you now want a decomposition of $x\in M$ you can take $\varphi^{-1}(x)$ and in the direct sum you have the decomposition. Now you can map the summands separately back to $M$ and you get the result.