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

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

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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 $(1-e)R\neq R$.

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