I should start by saying I found this post

Equivalent definitions of the Jacobson Radical

which is about the same two formulations of the Jacobson radical but it didn't really to answer my question. I'm trying to prove for myself that these two definitions are equivalent for a ring $R$ with $1$ (we'll call them $J$ and $J'$):

$$\newcommand{\Ann}{\operatorname{Ann}} J(R)=\cap\{\Ann(M)\mid\text{$M$ is a simple left $R$-module}\}$$

where $\Ann(M)$ is the annihilator of $M$, and

$$J'(R)=\cap\{I\mid\text{$I$ is a maximal left ideal of $R$}\}$$

To show $J(R)\subseteq J'(R)$: let $a\in J(R)$ and $I$ be a maximal left ideal. Then $R/I$ is a simple left $R$-module, so $a\bar x=0$ for any $\bar x\in R/I$. In particular, $\bar a=a\bar 1=0$, so $a\in I$, and hence $a\in J'(R)$.

I'm having trouble with the other direction. Of course, I need to let $a\in J'(R)$ and $M$ be a simple left $R$-module, but I see no immediate way to relate this to maximal ideals of $R$. For what it's worth I'd prefer a hint rather than a complete answer.

  • $\begingroup$ Hint: What is the annihilator of a nonzero element in M? $\endgroup$ – Cihan May 19 '16 at 5:57
  • $\begingroup$ @Cihan it's a left ideal of $R$. And I guess it's supposed to be maximal too, but I'm not sure how to show that. $\endgroup$ – Alex Mathers May 19 '16 at 6:22

For every x in M, there is a unique left R-module homomorphism from R to M that sends 1 to x. If x is nonzero, this homomorphism is surjective (why?). The kernel is precisely the annihilator of x. Now use the so-called correspondence theorem to see this is a maximal left ideal. Can you finish the argument from here?

  • $\begingroup$ I see now. If $K$ is the kernel then $R/K\cong M$ as $R$-modules, so $R/K$ is simple i.e. has no left ideals, and hence $K$ is a maximal left ideal. And hence $a\in K=\text{Ann}(x)=\text{Ann}(M)$. $\endgroup$ – Alex Mathers May 19 '16 at 6:56
  • $\begingroup$ Careful: The annihilator of x can be strictly larger than the annihilator of M. $\endgroup$ – Cihan May 19 '16 at 7:10
  • $\begingroup$ yeah I see now why that was wrong, so what am I missing? $\endgroup$ – Alex Mathers May 19 '16 at 17:43
  • $\begingroup$ Although the annihilator of M can be smaller, it is still equal to the intersection of the annihilators of the nonzero elements in M. That's enough for what we need. $\endgroup$ – Cihan May 19 '16 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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