I am confused about the notation $\operatorname{rad}^2 A$. It can be considered as $\operatorname{rad}(\operatorname{rad}(A))$ or as $(\operatorname{rad}(A))^2$. Are $\operatorname{rad}(\operatorname{rad}(A))$ and $(\operatorname{rad}(A))^2$ the same?

Here $\operatorname{rad}(\operatorname{rad}(A))$ is the radical of the Jacobson radical of $A$ and $(\operatorname{rad}(A))^2$ consists of all finite sums of the elements of the form $x_1 x_2$, where $x_1, x_2 \in \operatorname{rad}(A)$.

Since $(\operatorname{rad}(A))^2$ is a nilpotent ideal of $\operatorname{rad}(A)$ and $\operatorname{rad}(\operatorname{rad}(A))$ is the maximal ideal of $\operatorname{rad}(A)$, we know that $(\operatorname{rad}(A))^2 \subseteq \operatorname{rad}(\operatorname{rad}(A))$. But how to prove that $\operatorname{rad}(\operatorname{rad}(A)) \subseteq (\operatorname{rad}(A))^2$? I have another question: is $\operatorname{rad}(A)$ the unique maximal ideal of $A$? Thank you very much.

Edit: this question comes from the question.


2 Answers 2


I guess you are pretty much confused by the definition. Assume throughout $A$ is finite dimensional (I suppose it works for artinian rings in general).

The (Jacobson) radical $J(A)$ of an algebra $A$ has many equivalent definitions, which you can simply look up from, for example, wikipedia. For the radical rad$M$ of a (left) $A$-module $M$, you can use either (1) $\mathrm{rad}M=J(A)M$ or (2) $\mathrm{rad}M$=intersection of maximal submodule of $M$. Again (1) and (2) are equivalent, by exactly the same reason(s) as why different definitions for radical of $A$ are the same. If you want more detailed explanation, please consult Auslander-Reiten-Smalo's book "Representation theory of Artin algebras" Proposition 3.5.

So when you consider the radical of the radical of $A$, you are considering the radical of an $A$-module, namely, $M=\mathrm{rad}A$, which if you go by definition (1), will become $\mathrm{rad}A\mathrm{rad}A$.

For your second question, it is false in general, unless $A$ is local.

P.S. When I first started learning representation theory of algebras, I used both Assem-Skowronski-Simson and ARS at the same time, to complement each other contents; ARS arguments are usually the one that is preferred because it is very concise and the proofs are elegant; ASS has a very down-to-earth, example-rich approach which is very good to understand what is going on concretely, which naturally becomes a very good supplement to ARS. (and it has more interesting theories such as tilting modules and torsion theory)

EDIT: Sorry for being lazy on tex-typing; and missing authors' names. Added them back in (Assem-Skowronski-Simson wrote book that OP is constantly raising questions from; so I assumed he knows but I forgot about the general audience.)

  • $\begingroup$ If use \operatorname{rad} instead of rad outside the $$, then all looks better. $\endgroup$
    – user26857
    Apr 21, 2013 at 15:05
  • $\begingroup$ No one has mentioned ASS so it would be a good idea to spell out the names at least once so that everyone knows what book uou are referring to :-) $\endgroup$ Apr 21, 2013 at 22:08

If $B$ is an ideal of $A$ then $\operatorname{rad}(B)=\operatorname{rad}(A)\cap B$ [Herstein, Noncommutative Rings, Theorem 1.2.5]. So $\operatorname{rad}(\operatorname{rad}(A))=\operatorname{rad}(A)$.

  • $\begingroup$ thank you very much. But I think that maybe $\operatorname{rad}(\operatorname{rad} A) \neq \operatorname{rad} A$. This question comes from another question. I put in link of the related question above. $\endgroup$
    – LJR
    Apr 21, 2013 at 8:14
  • $\begingroup$ I don't understand. Do you mean that Theorem 1.2.5 from [Herstein] is not true? $\endgroup$ Apr 21, 2013 at 9:17
  • $\begingroup$ I'm sorry I don't have a copy of Herstein's book. May I ask is the ideal $B$ in the statement a two-sided ideal? $\endgroup$
    – Aaron
    Apr 21, 2013 at 14:39
  • $\begingroup$ @Aaron: Yes, it is. Moreover, this theorem is not true for one-sided ideals. $\endgroup$ Apr 21, 2013 at 14:45
  • $\begingroup$ What is Herstein's definition of the Jacobson radical of an ideal? I ask because I've seen the definition of the Jacobson radical of a ring -- which doesn't directly apply to ideals if you insist, as usual, that your ring have -- unit and also applied to modules. As in Aaron's answer, the latter is simply the intersection of the maximal submodules. Also as he says, if you apply the latter definition to an ideal $I$ you'll get $rad I = rad(R) I$ and thus $rad (rad R) \neq rad R$ in general. $\endgroup$ Apr 21, 2013 at 15:08

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