Prove that the Jacobson radical of a ring $R$ contains no idempotents other than $0$.could anyone give me a hint please?
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 $(1-e)R\neq R$.