For a module $M$ one has $rad(M)=0$ if and only if $M$ is isomorphic to a submodule of a direct product of simple modules

modulessemi-simple-rings

Definition: For a module $M$, the intersection of all the maximal submodules and module $M$ is called as radical of the module and denoted by $rad(M)$.

For a module $M$ one has $rad(M)=0$ if and only if $M$ is isomorphic to a submodule of a direct product of simple modules.

Above statement is written in a review report of a well reputed journal. I have doubt about the validity of above statement. Please clarify it to me.

My thought: submodule of a direct product of simple modules is semisimple and so $M$ will be semisimple if and only if $rad(M)=0$ (according to above statement). Which is not true in general. For example, $\mathbb{Z}$ has zero radical but not semisimple.

Best Answer

If $\{N_\alpha\mid \alpha\in\kappa\}$ is an indexing of the maximal submodules of $M$, then the canonical map $M\to \prod_{\alpha\in\kappa} M/N_\alpha$ has kernel $rad(M)$.

If $rad(M)$ is zero this is injective so that $M$ is a submodule of the product of simple modules.

It is important not to confuse sums of modules with products. When the index set is infinite, they can be quite different.

Actually in the scenario above, since the projections on each coordinate are onto $M/N_\alpha$, there is another common terminology: one says that $M$ is the subdirect product of the $M/N_\alpha$s.