# Nilpotent or non-Nilpotent Jacobson Radical

Let $R$ be a ring with identity element such that every ideal of which is idempotent or nilpotent. Is it true that the Jacobson radical $J(R)$ of $R$ is nilpotent?

If $R$ is Noetherian and $J(R)$ is idempotent the Nakayama lemma yields $J(R)=0$. So, for Noetherian rings whose Jacobson radicals are nonzero we have an affirmative answer to the raised question.

Thanks for any help or suggestion!

• The example is easy: just take a field (or the full matrix ring over a field, if you want a noncommutative one). – egreg Jul 3 '16 at 11:17
• For less trivial examples, you could take any von Neumann regular ring (all ideals are idempotent) or any local ring with a nilpotent maximal ideal (all proper ideals are nilpotent.) Haven't thought of any with nonnilpotent radical, though. – rschwieb Jul 4 '16 at 11:38

As noted in Rostami's answer, you do get that $J(R)$ is nil, hence is nilpotent if it is finitely generated. (Here is an alternative proof, just for fun. Since idempotents are locally 0 or 1, these assumptions imply $J(R) =$nil$(R)$ holds locally, hence globally.)

But here is the main point of my answer: an example of a quasilocal ring whose maximal ideal is idempotent and all other proper ideals are nilpotent. So the answer to your question is "no".

(Edited with parenthetical remarks justifying the claims, as requested.)

Let $K$ be a field, $D := K[\{X^s \mid s \in \mathbb{Q}^+\}]$, $M$ be the maximal ideal consisting of elements with zero constant term, and $\overline D_M := D_M/(X)_M$. Note that the nonzero elements of $D_M$ are each a unit multiple of a power of $X$. (Given $f \in D$, factor out the biggest power of $X$ that you can to get $f = X^sf_0$, where $f_0 \notin M$. Since elements of $D$ that are not in $M$ are units in $D_M$, any element of $D_M$ with a numerator of $f$ is a unit multiple of $X^s$.) Thus $M_M = M_M^2$, hence $\overline M_M = \overline M_M^2$. If $I_M$ is any other nonzero proper ideal of $D_M$, then there is a positive lower bound on the powers of $X$ it contains, hence $\overline I_M$ is nilpotent. (If $I_M$ is a proper ideal and there is no positive lower bound on the powers of $X$ that $I_M$ contains, then for each positive rational $s$, there is a $t < s$ with $X^t \in I_M$, hence $X^s = X^{s-t}X^t \in I_M$. So $I_M$ would contain every power of $X$, hence $I_M = M_M$.)

• This is a nifty example: the ideals are linearly ordered and (I believe) this is a dual ring. I had been doing something similar but I didn't have the localization trick to simplify things. Every nontrivial and nonmaximal ideal is contained in a principal (nilpotent) ideal. – rschwieb Jul 25 '16 at 20:58
• @karparvar It's clear that if $X^q\in I_M$, then $X^{q'}\in I_M$ for any $q'\geq q$. Further, everything in $I_M$ looks like $X^qu$ for some unit $u$. If there is a positive rational bound $r$ for these powers of $X$ in $I_M$, that means $(X^r)\supseteq I_M$ are both nilpotent, whether or not $X^r\in I_M$. If no such $r$ exists, then $I_M$ contains arbitrarily small $X^q$, and hence all $X^q$, thus $I_M=M_M$. Everything seems in order to me. – rschwieb Jul 29 '16 at 10:42
• @karparvar More to the point I guess: why would you expect $X^{s_0}\in I_M$? Nothing requires that... – rschwieb Jul 29 '16 at 10:46
• Rephrasing the proof another way, the claims are: (1) If I contains arbitrarily small positive powers of X, then I contains all positive powers of X, hence equals M. (2) If I does not contain arbitrarily small powers of X, then it is nilpotent. Both claims seem fairly clear to me. The parenthetical remark at the end is just a proof of claim (1). – Jason Juett Jul 29 '16 at 13:06
• And you are right about what I proved. I proved that either there is a lower bound, which you denote $s_0$, or there is no lower bound and $I = M$. It does not matter if $X^{s_0}$ is in $I$ or not. Sometimes it is, sometimes it is not. Either way, there is a positive integer $n$ with $ns_0 \ge 1$, and $\overline I_M^n = (0)$. – Jason Juett Jul 29 '16 at 13:11

Let $a$ be an element of $R$. If $\left<a\right>=\left<a\right>^2$, then it is clear that the exists $e^2=e\in \left<a\right>$ such that $\left<a\right>=\left<e\right>$. Now, if $0\not= a\in J (R)$, then $\left<a\right>$ is nilpotent or idempotent. If $\left<a\right>$ is idempotent, by above argument there exists $e^2=e\in\left<a\right>$ such that $\left<a\right>=\left<e\right>$. Since $e\in J (R)$ , $1-e$ is a unit idempotent and so $1-e=1$. Thus, $\left<a\right>=0$, a contradiction. Therefore, every element of $J (R)$ in nilpotent and so $J (R)$ is nil. If $J (R)$ is finitely generated $J (R)$ is nilpotent.

• @user26857: Is there any example where $J(R)$ is "not" finitely generated? Thanks, of course, for your answer. – karparvar Jul 5 '16 at 2:40
• Aside from $J(R)=N(R)$, we also can conclude $R/J(R)$ is von Neumann regular. But finding or disproving the existence of a ring with $J(R)$ idempotent and not nilpotent is the main trick. – rschwieb Jul 5 '16 at 18:43
• Clearly if $J (R)$ is finitely generated and idempotent, then $J (R)$ must be zero ideal, by Nakayama' Lemma. For example $\J (k [x _1, X2, ..., x_n,...]/<x_1^2, x_2^3, ...x_n^{n+1},...>)= <x_1, x_2, ...x_n^,...>/<x_1^2, x_2^3, ...x_n^{n+1},...>$, $k$ is a field, which is not finitely generated. – E.Rostami Jul 6 '16 at 0:02
• @E.Rostami: Now, why each ideal of the ring is idempotent or nilpotent? I think you refer to "math.stackexchange.com/questions/1711322/…"> , but the answer is incorrect in the infinitely generated case. – karparvar Jul 6 '16 at 6:43
• The hard part is to ensure all ideals in the Jacobson radical are idempotent or nilpotent. – rschwieb Jul 6 '16 at 18:28