# Ideals generated by idempotent elements

Let $R$ be a unitary associative ring and $J$ be an ideal of $M_{n}(R)$. We know that there is an ideal $I$ of $R$ such that $J=M_{n}(I)$. Now there is a question.

Question: If $J$ is generated by a subset of idempotent elements of $M_{n}(R)$ say $S$, is $I$ generated by a subset of idempotent elements of $R$, which related to $S$?

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Dear all, I think if $R$ is an integral domain, in the ring $M_2(R)$ your statement is trivially true. because in this case, we could characterize all of idempotents. all of idempotents have the form $0$, $I_2$ or a matrix whose entries are $a_{11}=a$, $a_{12}=b$ , $a_{21}=c$, $a_{22}=1-a$ and $a,b,c$ satisfy in the relation $bc=a-a^2$. so in this case the only nonzero ideal in $M_2(R)$ which contains an idempotent is the total ring. so in this case, your claim is true. – Ali Reza Nov 27 '12 at 7:34
What do you mean by "$J$ is generated by idempotents?" Wouldn't it contain one of such, and then contain a unit, thus being the whole $M_n(R)$? – Filippo Alberto Edoardo Nov 28 '12 at 1:56
Dear Filippo, $J$ is generated by a subset of idempotent elements as a two-sided ideal of $M_[n}(R)$. If one is unit, then is trivial. – Ali Nov 28 '12 at 4:44

(1) If the ideal $J=M_n(I)$ of $M_n(R)$ is generated by some idempotents, then $I$ is an idempotent ideal, i.e., $I^2=I$.
(2) If $J=M_n(I)$ is generated by finitely many idempotents, then $I$ is a finitely generated ideal.
Thus in the case where $R$ is commutative and $J$ is generated by finitely many idempotents, there exists an idempotent $e\in R$ such that $I=Re$.
I think this solves the problem. Choose a set of generators of $J$. For each generator, look at the ideal generated by just that, and apply this argument to find an idempotent. Then the ideal generated by all those idempotents is $I$. – Will Sawin Jan 18 '13 at 18:04