R-module is semisimple iff every cyclic R-submodule is semisimple

456 Views Asked by At

I am asked to prove the following statement:

An $R$-module $M$ is semisimple iff every cyclic $R$-submodule of $M$ is semisimple

The $\implies$ direction is trivial since every submodule of a semisimple module is semisimple

The other direction is more difficult since I do not see why for example $M$ cannot have the form $M=\prod_{i=1}^{\infty}Rm_i$ where $Rm_i$ are all cyclic and simple as $R$-modules. What am I missing here?

1

There are 1 best solutions below

8
On BEST ANSWER

The other direction is still pretty straightforward, as long as you know a module is semisimple iff it is a sum of simple submodules.

A module is always the sum of its cyclic submodules.

If each of those submodules is semisimple, then the module is a sum of sums of simple submodules, hence it is a sum of simple submodules, i.e. semisimple.


Note: you wrote $M=\prod_{i=1}^\infty Rm_i$ when talking about semisimple modules. This is not correct. You should be aware that the product $\prod$ is a completely different animal from $\oplus$ when the index set is infinite. Semisimplicity of modules has to be proven with sums or direct sums but not products.