Determining which rings have a module which is not semisimple

Question

I have a homework question as follows:

Let $M$ be an $R$-module with submodule $K$. A complement to $K$ in $M$ is a submodule $L$ of $M$ such that $M = K \bigoplus L$. An $R$-module $M$ is semisimple if every submodule of $M$ has a complement.

For each of the following rings $\mathbb{Z}, \mathbb{C}[t]$, and $\mathbb{C}[\mathbb{Z}]$ find a module which is not semisimple.

For the integers, I think I found one: $K = 2\mathbb{Z}$, which is a $\mathbb{Z}$-module, but the odd integers are not a module, so $K$ does not have a complement. Is this a correct example?

For the second one, I thought perhaps $\mathbb{R}[x]$ would work, but I'm not sure that this is an $R$-module for the ring $\mathbb{C}[t]$, and for the third one, I am unsure of where to start.

Any hints/examples would be appreciated.

Answer 1

All three are integral domains, and no proper ideal of an integral domain can be a summand of the ring. This is because $\{0\}\neq IJ\subseteq I\cap J$ for every pair of nontrivial ideals $I$, $J$.

As for your title question, the answer is easy: a ring has a non-semisimple module precisely when it is not a semisimple ring.

Answer 2

Let $R$ be a noetherian ring with a noninvertible nonzero divisor. Then $R$ is not semisimple.

Suppose we have a direct sum decomposition $R=(a) \oplus M$ and consider the quotient $\pi: (a) \oplus M \to (a)$. If $(a)$ is a nonzero divisor then $R \cong (a)$, so this gives us a surjective endomorphism $R \to R$. By the fact that $R$ is noetherian, any such map is necessarily an isomorphism, which implies $M=ker(\pi)=0$. Hence $(a)=(1)$ and $a$ is therefore invertible, a contradiction.

EDIT: By the discussion below, to show any ring is not semisimple, it suffices to show it has a noninvertible nonzero divisor.