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We define a ring has unity and semisimple module is a direct sum of simple submodules.

There is a theorem states that

A direct sum of semisimple left $R$-modules is semisimple.

Proof: Suppose $M=\bigoplus_{i\in I} M_i$, where $M_i$ is semisimple module. Since $M_i$ is semisimple, we write $M_i=\bigoplus_{j\in J_i}M_{ij}$, where $M_{ij}$ is simple module. Therefore, we have $$M=\bigoplus_{i\in I}\left(\bigoplus_{j\in J_i}M_{ij}\right).$$

My Question: How do I proof that $M$ is still a direct sum?

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Perhaps you are hesitating because you are worried about checking condition for internal direct sums?

Think of it this way: you can view $M_i$ as an external direct sum of simple modules. Then you are forming an external direct sum of the $M_i$.

Then writing $$M=\bigoplus_{i\in I}\left(\bigoplus_{j\in J_i}M_{ij}\right)$$

is really no different than forming $J=\{(i,j)\mid i\in I, j\in J_i\}$

$$M=\bigoplus_{(i,j)\in J}M_{ij}$$

and every $M_{ij}$ is a simple $R$ module, so...

You could also argue with isomorphisms that the two things are the same, too, if you want to chase some arrows.

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  • $\begingroup$ Thanks. That was just my question. But I used the Theorem 2.4 in Lam's A First Course in Noncommutative Rings to avoid checking the condition. That is, it is sufficient to show that $M=\sum_{i\in I}\left(\sum_{j\in J_i}M_{ij}\right)$. $\endgroup$ – bfhaha Nov 8 '17 at 6:44

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