Semisimple-ish rings!

Question

Let S be the class of all rings R which have 1 and satisfy this condition:

for every "non-zero" right ideal I of R there exists a "proper" right ideal J of R such that I + J = R. (The + here is not necessarily direct.)

All semisimple rings are in S and (commutative) local rings which are not fields are not in S. The ring of integers Z is also in S and so S properly contains the class of semisimple rings.

My questions:

Will this condition by itself force an element of S to have any (known, interesting) structure?

A more important question:

What about simple rings which are in S? For example, do they have to be semisimple? (Unlikely!)

Answer 1

By Zorn's lemma, each right ideal is contained in a maximal right ideal, therefore if $I+J = R$ then $I+M = R$ where $M$ is a maximal right ideal. If $I+M\ne R$ for all maximal right ideals $M$ then $I\subseteq M$ for all maximal ideals $M$. Thus $I\subseteq J(R)$, the Jacobson radical of $R$ which is the intersection of all maximal right ideals of $R$. Hence condition $S$ is equivalent to $J(R)=0$.

A ring with vanishing Jacobson ideal is called semiprimitive. As $J(R)$ is also the intersection of the maximal left ideals of $R$ then the property of semiprimitivity is left-right symmetric. There are plenty of examples of semiprimitive rings which are not semisimple. For instance every simple ring is semiprimitive and every subdirect product of semiprimitive rings is semiprimitive ($\mathbb{Z}$ is a subdirect product of finite fields). As a reference see Section 10.4 of P. M. Cohn Algebra (2nd ed. vol 3) Wiley 1991.