The structure of a Noetherian ring in which every element is an idempotent.

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 $A$?

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 $\Bbb F_2$.

A boolean ring is $0$-dimensional. Hence a noetherian boolean ring $A$ is artinian, therefore $\mathrm{Spec}(A)$ is finite and discrete. But $A \cong C(\mathrm{Spec}(A),\mathbb{F}_2)$ by Stone's Theorem. Hence, $A \cong \mathbb{F}_2^n$ for some $n$. If $A$ is not unital, consider it as an ideal of the unitalization $\tilde{A}$ as an $\mathbb{F}_2$-algebra, which is also boolean and noetherian. Hence $\tilde{A} \cong \mathbb{F}_2^n$, and we see $A \cong \mathbb{F}_2^{n-1}$.