1. Is there a name for a ring $A$ (commutative with $1 \neq 0$) all of whose ideals are radical?
  2. Is there a name for a ring $A$ (commutative with $1 \neq 0$) such that for each $x \in A$, there is $n \geq 2$ such that $x^n = x$?

I am currently calling $A$ in 1 a radical ring and $A$ in 2 a semi-Boolean ring.

  • $\begingroup$ "Radical ring" apparently already means a (necessarily non-unital) ring which is equal to its own Jacobson radical. Anyway, that first condition seems quite strong. In the second condition, is $n$ fixed or is it allowed to depend on $x$? $\endgroup$ – Qiaochu Yuan May 24 '13 at 6:28
  • $\begingroup$ Maybe I shouldn't use that word then. A Boolean ring $A$ satisfies the first condition since for an ideal $\mathfrak{a} \subseteq A$, we have $\sqrt{\mathfrak{a}} = \{x \in A : x^{n} \in \mathfrak{a} \text{ for some } n \geq 2\} = \{x \in A : x \in \mathfrak{a}\} = \mathfrak{a}$. For the second condition, the integer $n \geq 2$ depends on $x$. $\endgroup$ – Gil May 24 '13 at 6:32
  • $\begingroup$ The first condition implies that $A$ is von Neumann regular (en.wikipedia.org/wiki/Von_Neumann_regular_ring) and I think the converse holds if $A$ is Noetherian. $\endgroup$ – Qiaochu Yuan May 24 '13 at 6:40
  • 1
    $\begingroup$ @QiaochuYuan It's equivalent even without the Noetherian condition. $\endgroup$ – rschwieb May 24 '13 at 13:13

(I'm pretty sure that "radical" here means that $\sqrt{I}=I$)

Under the conditions of #1, all ideals of the ring are semiprime. Taking any element $a$, $(aR)^2$ is a semiprime ideal, so it must contain $aR$, and we have proven that $aR=(aR)^2$. Since this says that $a=ara$ for some $r\in R$, we have also shown that $R$ is von Neumann regular. An observation that we'll need shortly is that $ar$ is idempotent, and $aR=araR\subseteq arR\subseteq aR$, so $aR=eR$ for an idempotent $e=ar$.

Conversely, if $R$ is von Neumann regular, we can show that all its ideals are semiprime, since they are all idempotent.

As observed before, for any element $a\in R$, there exists an idempotent $e$ such that $eR=aR$. Suppose $I\lhd R$ and $I^2\neq I$. Clearly $aR\subseteq I$, and $(aR)^2\subseteq I^2$. Pick $a\in I\setminus I^2$. Then $(aR)^2=(eR)^2=e^2R=eR=aR$, and that implies that $a\in (aR)^2\subseteq I^2$, a contradiction. Hence $I^2=I$ for all ideals, and $I$ is semiprime (=radical).

It turns out that for commutative rings, being VNR is exactly the same as saying that every ideal is an intersection of maximal ideals. As a result, every proper quotient ring of $R$ has zero Jacobson radical. Since Jacobson used "radical ring" to describe a ring in which the Jacobson radical is the entire ring, this is an unfortunate conflict with your choice of term. Alternatives appear in the list below.

In the noncommutative case, things separate a bit, and I can tell you some more about terminology.

  1. If all ideals are semiprime, the ring is sometimes called fully semiprime, and sometimes fully idempotent, since all ideals being semiprime is equivalent to having all ideals idempotent. VNR rings are fully idempotent (same proof as above) but clearly not all fully idempotent rings are VNR: there exist non-VNR simple rings which are trivially fully idempotent.

  2. If all ideals are the intersection of maximal right ideals, the ring is called a right V-ring (where V stands for Villemajor). There are non-VNR V-rings and non-V VNR rings. This is less relevant to your situation than the above point, but interesting enough to mention :)

I don't immediately recall a name for condition 2, but I'm sure there is one. Of course, condition 2 implies condition 1. You might be interested to know that you don't have to assume commutativity: a theorem by Jacobson proves that if for every $x$ there is $n>1$ such that $x=x^n$, then the ring is commutative.


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