# Proving equivalent formulations of Jacobson radical

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.

• Hint: What is the annihilator of a nonzero element in M? – Cihan May 19 '16 at 5:57
• @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. – Alex Mathers May 19 '16 at 6:22

• 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 $a\in K=\text{Ann}(x)=\text{Ann}(M)$. – Alex Mathers May 19 '16 at 6:56