# 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 $A$?

• Are you aware that $A$ is finite if it does have a unity? – Kevin Carlson Aug 9 '13 at 2:59
• If we interpret your ring as the Boolean algebra of subsets of $\Bbb{N}$, the ring and order ideals coincide-and shouldn't a maximal ideal dual to a free ultrafilter be infinitely generated? – Kevin Carlson Aug 9 '13 at 3:17
• Ah, here's a more elementary counterargument: can you find a finite generating set for the ideal of all elements with finitely many $1$s? – Kevin Carlson Aug 9 '13 at 3:23
• @KevinCarlson I would be grateful if you(or anyone else) post the answer to the title question with a detailed proof. – Makoto Kato Aug 9 '13 at 3:30
• I haven't studied the rng case before, but I'll be back before long if no one else gets here. – Kevin Carlson Aug 9 '13 at 3:36

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 $\Bbb F_2$.

• Why is $R$ finitely generated in $\hat{R}$? – Martin Brandenburg Aug 11 '13 at 15:55
• Any good definition of "Noetherian rng $R$" should at least imply that $R$ is finitely generated over itself, and hence over $\hat{R}$. @MartinBrandenburg – rschwieb Aug 11 '13 at 17:39

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

• Could you explain why $\tilde{A}$ is Noetherian? – Makoto Kato Aug 9 '13 at 20:57
• Could you explain why $\tilde{A}$ is Noetherian when $A$ is so? – Makoto Kato Aug 10 '13 at 9:44

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.

• Could you elaborate on why a Noetherian Boolean ring with unit is finite? – Makoto Kato Aug 9 '13 at 7:13
• It's Artinian, because it's Noetherian and primes are maximal; Artinian rings have finitely many maximals; their nilradical=Jacobson radical is nilpotent; thus the product=intersection of some powers of all the maximals is zero; apply CRT to get a decomposition of an Artin ring into finitely many local Artinian quotients; quotients of Boolean rings are Boolean and the only local Boolean ring is $\Bbb{Z}_2$ so every Noetherian Boolean ring is a finite product of $\Bbb{Z}_2$s. This is all done in detail in Pete Clark's notes, section 9. – Kevin Carlson Aug 9 '13 at 7:42
• Thanks. Let $A$ be a Noetherian Boolean ring which may not have a unity. Since $A$ is finitely generated as an ideal, there exists an element $a$ such that $A = aA$. Clearly $a$ is a unity. – Makoto Kato Aug 9 '13 at 9:30
• QUOTE The non-compact Boolean spaces (i.e. non-compact totally disconnected Hausdorff spaces) are exactly the non-closed open subsets of compact Boolean spaces. END QUOTE A compact Hausdorff space is normal. Hence an open subset of a compact Hausdorff space is locally compact. However, it seems to me that a totally disconnected Hausdorff space cannot be always locally compact. – Makoto Kato Aug 9 '13 at 19:00
• Yes, thanks, Boolean spaces are also required to be locally compact. – Kevin Carlson Aug 9 '13 at 19:33