Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!)

share|improve this question

1 Answer 1

up vote 13 down vote accepted

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.

share|improve this answer
That's great! Thank you very much. –  carlos Jun 28 '10 at 7:06
The type of right ideals which do not have such a complement are exactly the superfluous (or small) right ideals. As the excellent answer above shows, rings with $J(R)=0$ are the rings without small right or left ideals. Going one step further, semisimple rings (right Artinian +$J(R)=0$) are the rings without essential (or large) right ideals. It's interesting that "no essential right ideals" implies it's dual relative "no superfluous right ideals". It's someone akin to right Artinian implying right Noetherian in rings. –  rschwieb Dec 17 '11 at 12:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.