Simple example for a rng with an inverse semi-group as the multiplicative group

Question

I’m looking for a ring without an multiplicative identity, and in which every element $x$ has a weak inverse $y$ such that $xyx=x,yxy=y$ preferably simple to construct and or of finite size.
If it has an agreed upon name, please mention it.

Answer 1

A slight modification of paulinho's suggestion in the comments works: we can take the rng $M_{\infty}(\mathbb{R}) = \bigcup_n M_n(\mathbb{R})$ of $\mathbb{N} \times \mathbb{N}$ matrices such that all but finitely many entries are nonzero. Every element lies in some $M_n(\mathbb{R})$ and so has a Moore-Penrose pseudoinverse. The identity matrices $I_n \in M_n(\mathbb{R})$ provide a sequence of approximate identities but there is no global identity.

Edit: A simpler example in the same spirit: take an infinite direct sum $\bigoplus_i F_i$ of any sequence of fields. The weak inverse is given componentwise by returning either the inverse if it exists or zero otherwise. Again there is a sequence of approximate identities but no global identity. (This example was inspired by thinking about compactly supported functions on the real line and then simplified and generalized. Note also that $\bigoplus_i \mathbb{R}$ embeds into $M_{\infty}(\mathbb{R})$ as the subrng of diagonal matrices.)

Edit #2: Okay, this was fun. Let's call a rng satisfying the second condition (every element has a weak inverse) a weak division rng. Note that if such a rng has an identity then in particular it must be von Neumann regular and much is known about these. It's a nice exercise to show that every matrix ring over a field (probably over a division ring too but I haven't checked this) is a weak division rng.

Here are two claims:

Claim #1: Every finite weak division rng has an identity, and in fact is isomorphic to a finite product of matrix rings over finite fields (these are precisely the finite semisimple rings).

Claim #2: Every finitely generated commutative weak division rng has an identity.

You didn't ask about Claim #2 so I won't include the proof unless you want it; here's a proof of Claim 1. First, every finite rng $R$ has a canonical decomposition into a finite product

$$R = \prod_p R_p$$

where $R_p$ is the $p$-power part of $R$ ($p$ prime). Moreover, weak inverses exist in such a product iff they exist componentwise, so the problem reduces to the case that $R = R_p$ for some $p$; we assume this from now on, as well as that $R$ is a weak division rng.

Suppose $x \in R$ satisfies $p^2 x = 0$ but $px \neq 0$ (so $x$ has additive order $p^2$). By assumption $px$ has a weak inverse $y$, so that

$$(px)y(px) = px.$$

But $(px)y(px) = p^2 xyx = p^2 x = 0$, which is a contradiction; we conclude that $R$ has no such elements $x$. (This can also be phrased without contradiction but it's a bit more unintuitive.) So $R$ is an $\mathbb{F}_p$-vector space and hence an $\mathbb{F}_p$-algebra.

Next, consider the unitization $R \times \mathbb{F}_p$, where we freely adjoin a unit to $R$ as an $\mathbb{F}_p$-algebra.

Proposition: With the above hypotheses, $R \times \mathbb{F}_p$ has trivial Jacobson radical and hence is semiprimitive.

(Edit #4: This proof had a gap but I fixed it. Hopefully!)

Proof. We want to show that if $(x, c) \in J(R \times \mathbb{F}_p)$ is an element of the Jacobson radical then it must be equal to zero. The projection to $\mathbb{F}_p$ equips $\mathbb{F}_p$ with the structure of a simple $R \times \mathbb{F}_p$-module, so in order for $(x, c)$ to be in the Jacobson radical, which requires that it act trivially on every simple module, we must have $c = 0$.

If $c = 0$, recall that if $S$ is an artinian ring then the Jacobson radical $J(S)$ is a nilpotent ideal and that every element of $J(S)$ generates a nilpotent left ideal. But the left ideal generated by $(x, 0)$ contains $(yx, 0)$ where $y$ is the weak inverse of $x$, and since $yx$ is idempotent ($yxyx = yx$) it follows that the left ideal generated by $(x, 0)$ can't be nilpotent unless $yx = 0$, which gives $xyx = x = 0$. $\Box$

Now it follows as a corollary that the unitization $R \times \mathbb{F}_p$ is both semiprimitive and (finite, hence) artinian, and so must be semisimple. The semisimple $\mathbb{F}_p$-algebras are precisely the finite products of matrix rings over finite fields $\mathbb{F}_{p^i}$ (by the Artin-Wedderburn theorem plus Wedderburn's little theorem), so the unitization must be such a product. Furthermore it admits a quotient map to $\mathbb{F}_p$ (with kernel $R$), so one of the factors in this product must be $\mathbb{F}_p$, and the kernel must be another finite product of matrix rings over finite fields and hence in particular must have a unit.