Do “zero-divisor-closures” of a commutative ring always contain idempotents?

Question

Let $R$ be a commutative ring which is not necessarily a domain, and let $P(X)$ denote the power set of a set $X$.

Galois connections

Recall that for any $\rho \subseteq M\times N$, folloowing the notion of a galois connection, we can associate maps $$ \phi\colon P(M)\to P(N),\quad S\mapsto\{n\in N\mid s\rho n \forall s\in S\},\\ \psi\colon P(N)\to P(M),\quad T\mapsto\{m\in M\mid m\rho t \forall t\in T\}. $$ It is a classic exercise – and not too hard – to verify the following properties:

1. $\phi, \psi$ are antitone
2. $\phi\circ\psi$ and $\psi\circ\phi$ are monotone and extensive
3. $\phi\circ\psi\circ\phi=\phi$ and $\psi\circ\phi\circ\psi=\psi$
4. $\phi\circ\psi$ is a closure operator on $P(M)$ with closure system $\mathrm{Im(\phi)}$ and $\psi\circ\phi$ is a closure operator on $P(N)$ with closure system $\mathrm{Im(\psi)}$.

In particular, if $M=N$ and $\rho\subseteq M^2$ is symmetric, we get $\phi=\psi$ and $\phi$ monotonous, $\phi^3=\phi$.

Rings

Back to rings.

Let $α\colon P(R)\to P(R)$ denote the map associated to the “zero-divisor” relation $a\rho b \Leftrightarrow ab=0$.

LEM If $S$ is an arbitrary subset of $R$, $α(S)\unlhd R$.

Proof Indeed, let $a, b\in α(S)$. Then for all $s\in S$, we have $(a+b)s = as+bs=0$, hence $a+b\in α(S)$. Furthermore, if $r\in R$, then $(ra)s = r(as) = 0$ for all $s\in S$, hence $ra\in α(S)$. q.e.d.

Question

Do the sets $\mathrm{Im}(α)$ necessarily contain an idempotent element?

The motivation for this question comes from the prototypical example of $R=K^2$ with $K$ a field, where we can identify $\mathrm{Im}(α)$ to be $$ \{0, (K, 0), (0, K), K^2\}. $$ There, the nontrivial ideals $(K, 0)$ and $(0, K)$ contain idempotents $(0, 1)$ and $(1, 0)$, respectively – of course, these can be used to infer that the ring we started with indeed is a direct product of two smaller rings. I am wondering whether this can be recovered just by looking at this behavior of zero divisors.

Answer 1

Well, $\alpha(S)$ always contains $0$ which is an idempotent. But it might be nontrivial and contain no other idempotents, and in particular it may not be generated by idempotents as an ideal (which seems to be what you really want). For instance, if $R=\mathbb{Z}/4\mathbb{Z}$, then $\alpha(\{2\})=\{0,2\}$. Or, if $k$ is a field and $R=k[x,y]/(xy)$, then $\alpha(\{x\})=(y)$ which contains no idempotents other than $0$.

(By the way, the usual notation for your $\alpha$ is Ann (for annihilator) which may help you find more information about it.)