# can we find an isomorphism between $R$ and $eR$ ,where $R$ is von Neumann regular ring and $e$ is idempotent of $R$?

In essay i am reading, we have an finitely generated ideal, so it is generated by an idempotent element $e\in R$ (since finitely generated ideals generated by an idempotent element in von Neumann regular ring)

Then it is written that ''we can replace $eR$ by $R$''. I can't understand how and why. I am on the opinion that there should be isomorphism between $R$ and $eR$,but couldn't see yet. ($R$ is not domain)

Could anyone help me ?

• Please take a look at my edits that introduce TeX markup (and whitespacing to make things more readable) and try to do that in future posts... it will benefit your questions greatly! – rschwieb Apr 13 '17 at 13:08
• And apparently you are working with commutative VNR rings? – rschwieb Apr 13 '17 at 13:09
• Really, there must be some more context near the phrase "we can replace $eR$ by $R$"... could you provide it? We've already resolved the question you originally asked, but now maybe we can work to resolve what the author means. – rschwieb Apr 13 '17 at 13:10
• i am sorry about my writing mistakes.The ring that i study is not commutative,just VNR.Actually i have $n$ vectors from $R^{n}$,and sum of their contents equals $R$.So these sum equals eR, for some idempotent e.Then it says ''we can replace $eR$ by $R$,so sum of these content equals $R$''.There is nothing more.I quess it thinks homomorphism $eR$ to $R$ taking $e$r to $r$.it is onto and one-to-one. – idem Apr 15 '17 at 10:08
• I'm not sure what you mean by content actually... could you define it? – rschwieb Apr 15 '17 at 11:44

Look at $R=F_2\times F_2$ where $F_2$ is the field of two elements.
$R$ is von Neumann regular and is clearly not isomorphic to $(1,0)R\cong F_2$.
But on the other hand, if $S=\prod_{i\in \mathbb N} F_2$, there exists a nontrivial idempotent $e$ such that $eS\cong S$ (and there exist other idempotents which do not satisfy that.)