zero Jacobson Radical

Question

I read in a paper that if every nonzero left (or right) ideal of R (non commutative) contains a nonzero idempotent so $J(R)=0$, but i don't know why? I want understand why happens this.

Thank you

Answer 1

One characterization of things in the Jacobson radical is that $x\in J(R)$ iff $1-xr$ is a unit for every $r\in R$.

If $J(R)\neq\{0\}$, and then contained a nonzero idempotent $e$, it would be impossible for $1-e$ to be a unit, because it is a zero divisor on both sides.

Answer 2

Suppose $e\in J(R)$ is idempotent, $e\ne0$. Then $1-e$ is idempotent as well and $1-e\ne1$, so $1-e$ is not invertible. Take $I$ a maximal right ideal such that $1-e\in I$: then $1=e+(1-e)\in I$: contradiction.

Thus the only idempotent in $J(R)$ is $0$.