TITLE: The structure of a Noetherian ring in which every element is an idempotent.
QUESTION [7 upvotes]: Let $A$ be a ring which may not have a unity.
Suppose every element $a$ of $A$ is an idempotent. i.e. $a^2 = a$.
It is easily proved that $A$ is commutative.
Suppose every ideal of $A$ is finitely generated.
Can we determine the structure of $A$?

REPLY [9 votes]: It can be proven with simple steps which are lemmas that come in handy in other situations:

Noetherian Boolean rings have identity. Proof: let $\hat{R}$ be the unitization of $R$, so that $R\lhd \hat{R}$. $R$ is finitely generated in $\hat{R}$ and $R^2=R$, so by  Nakayama's Lemma, there exists an $x\in R$ such that $(1-x)R=0$. But this means that $x$ is an identity for $R$. 
Boolean rings are trivially von Neumann regular
a Noetherian von Neumann regular ring is semisimple
a commutative semisimple ring is a finite product of fields
the only Boolean field is $\Bbb F_2$, so all the fields are $\Bbb F_2$.

So the structure of all such rings is that of finite products of $\Bbb F_2$.