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

Question

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 $A$?

Answer 1

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 $\Bbb F_2$.

Answer 2

Short answer: $A$ is a finite ring with unit.

As you may know, these are called Boolean rings, and the category of Boolean rings with unit is equivalent to that of Boolean algebras and dual to the category of totally disconnected compact Hausdorff spaces. This is a mouthful of adjectives which is usually shortened to "Stone space" or, when Stone is writing, to "Boolean space," though his need not be compact. A proof that in this case Noetherianness is equivalent to finiteness appears in these notes of Pete Clark.

In this paper that introduced the duality between Boolean rings and spaces, Stone himself does not restrict to the unital case. In the more general setting the category of Boolean rings is shown to be dual to the category of totally disconnected Hausdorff spaces (warning: confusingly, Stone calls Hausdorff spaces "H-spaces." Also, "bicompact" = "compact.") Most relevant is his

Theorem 8 The non-compact Boolean spaces (i.e. non-compact locally compact totally disconnected Hausdorff spaces) are exactly the non-closed open subsets of compact Boolean spaces. In particular the natural one-point compactification maps non-compact Boolean spaces to compact Boolean spaces. Accordingly, the Boolean rngs without unit are characterized algebraically as the non-principal ideals in Boolean rings, i.e. each of the former is functorially isomorphic to one of the latter.

I've tried to translate his language for the current century. So, this leads us to

Claim $A$ as in the question statement is a finite ring with unit.

To prove the claim, suppose $A$ has no unit, so by Stone's theorem above we may imbed $A$ into a Boolean ring $\hat{A}$ as a non-principal ideal $I$. The point is that $I$ is then actually infinitely generated: finitely generated ideals in Boolean rings are principal. For the finitely generated ideal $(a_1,...,a_n)$ is actually equal to $(a_1,...,a_{n-1}+a_n+a_{n-1}a_n),$ and by induction the result follows. We check the induction step: observe that certainly $a_{n-1}+a_n+a_{n-1}a_n\in(a_1,...,a_n)$. Conversely, $a_n(a_{n-1}+a_n+a_{n-1}a_n)=2a_na_{n-1}+a_n^2=a_n, a_{n-1}(a_{n-1}+a_n+a_{n-1}a_n)=a_{n-1}$, and indeed $a_{n-1},a_n\in (a_1,...,a_{n-1}+a_n+a_{n-1}a_n)$.

We've shown $A$ is isomorphic to an infinitely generated ideal of $\hat{A}$. Let $(a_i)_{i\in\Bbb{N}}$ be a reduced generating set for $A$, so that $(a_1,...,a_{n-1})\subsetneq (a_1,...,a_n)$ for every $n$. Each $(a_1,...,a_n)$ is an ideal of $\hat{A}$, a fortiori an ideal of $A$, and thus $A$ cannot be Noetherian. We've reached a contradiction, and see that in fact all Noetherian Boolean rings are finite Boolean rings with unit. Since finite rings are Noetherian, this gives a complete characterization.

Answer 3

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 $A \cong \mathbb{F}_2^{n-1}$.