The jacobson radical of a ring $R$ contains no idempotents other than $0$. Prove that the Jacobson radical of a ring $R$ contains no idempotents other than $0$. Could anyone give me a hint please?
 A: 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 $e=0$.
A: 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 $(1-e)R\neq R$.
A: This is true even if $R$ does not contain a multiplicative identity $1$. I shall use the following definitions in Hungerford's algebra book.

*

*$a \in R$ is called left quasi-regular if there exists $r \in R$ such that $a + r + ra = 0$

*A left ideal $I$ of $R$ is called left quasi-regular if every element of $I$ is left quasi-regular

Then the following is true. (See Hungerford Lemma 2.15 in chapter 9)

For $a \in R$, $a \in J(R)$ if and ony if $Ra$ is a left quasi-regular left ideal of $R$.

Now suppose $J(R)$ contains an idempotent $e$, i.e. $e^2 = e$. Then $Re$ should be a left quasi-regular left ideal. Since $-e = (-e)\cdot e \in Re$, $-e$ is left quasi-regular. Thus there exists $r \in R$ such that $-e + r - re = 0$. Hence $e = r- re$ and $e^2 = re - re^2 = 0$.
