Prove that the Jacobson radical of a ring $R$ contains no idempotents other than $0$.could anyone give me a hint please?

  • $\begingroup$ I edited your post to $\LaTeX$ify it. Also added the "ring-theory" tag. Cheers! $\endgroup$ Jan 8 '18 at 0:26

If $e$ is an idempotent and in the Jacobson radical, $e^2=e$ and $(1-e)$ is invertible, ($x$ is in the Jacobsoon radical if and only if $1+ax$ is invertible for every $a$) $(1-e)^2=1-2e+e^2=1-e$ implies that $(1-e)^{-1}(1-e)^2=1$ implies that $1-e=1$ and $e=0$.


Another way to see it, based on the idea that the Jacobson radical only contains "nongenerators."

Suppose you have a proper right ideal $T$: I claim that $T+J(R)\neq R$. This is true because $T$ must be contained in some maximal right ideal $M$, and then $T+J(R)\subseteq M+M=M\subsetneq R$.

Now, if you had a nonzero idempotent $e$, then $eR+(1-e)R=R$. If $eR\subseteq J(R)$, this would say that $J(R)+(1-e)R=R$. But as we just established, this is a contradiction since $(1-e)R\neq R$.


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.