If $M$ isn't semisimple, then the radical is not trivial.

Question

I read in some papers that if a module is semisimple then the radical is trivial. How can I prove that if a module isn't semisimple then the radical isn't trivial or which book can I read to find the proof of this?

Answer 1

You can't prove this statement because it isn't true.

As egreg mentions, a simple counterexample is $\mathbb Z$.

I had a more complicated example since I had these rings in mind: if you take an infinite dimensional $F$ vector space $V$, then $End(V_F)$ is a ring that isn't a semisimple module over itself, but its Jacobson radical is trivial.

The problem is that from the statement $A\implies B$ you appear to want to say that $\neg A\implies \neg B$ must also be true, but as the example shows, this is not a valid deduction.

It would be valid to ask about the contrapositive ($\neg B\implies \neg A$) though:

How can I demonstrate that if the Jacobson radical is nontrivial, then the module is not semisimple directly?

Suppose $M$ is semisimple and $J(M)\neq\{0\}$. Then there exists a simple submodule $S$ of $J(M)$ and a submodule $N$ such that $S\oplus N=M$. But now you can easily show $N$ is a maximal submodule of $M$, and moreover it cannot contain $J(M)$. But this contradicts the definition of $J(M)$. So, it is not possible for $M$ to be semisimple if $J(M)\neq\{0\}$.

This, however, is logically equivalent to the other statement, and if you've already seen the other proof, nothing is new here really.

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There are some elementary conditions that, along with $J(M)=\{0\}$ imply $M$ is semisimple. If $M$ is Artinian, or more generally finitely cogenerated, then $M$ must be semisimple (but it'll also be finitely generated.)