# Determining which rings have a module which is not semisimple

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.

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.

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.

• Any ring with a noninvertible nonzero divisor is not semisimple. It follows right from the fact that in an Artinian ring, all elements are either zero divisors or units... – rschwieb Nov 11 '19 at 23:06
• @rschwieb Interesting. How exactly do you use your second claim to show your first? – leibnewtz Nov 11 '19 at 23:09
• every semisimple ring is Artinian... So the first sentence implies the ring is not Artinian, hence not semisimple. – rschwieb Nov 11 '19 at 23:11
• Ah right makes sense. Thanks – leibnewtz Nov 11 '19 at 23:12
• @rschwieb Maybe it's worth pointing out that a semisimple ring is Noetherian as well, so the above argument also shows that a semisimple ring can never have a noninvertible nonzero divisor – leibnewtz Nov 11 '19 at 23:21