A finite pseudo-ring such that $ab^2=b$ for some $(a,b)\in A$

Question

I have this exercise which I found very difficult:

$(A,+,.)$ is a finite pseudo-ring (a structure satisfying all the axioms of a ring except for the existence of a multiplicative identity), and $(a,b)\in A^2$ such that $ab^2=b$, prove that $bab=b$.

I used every possible combination of powers of $a$ and $b$ but didn't find anything, the idea I was using is the construction of a sequence $a^nb^m$ for example and there for there exist $n,m,n',m'$ such that $a^nb^m=a^{n'}b^{m'}$ and I use the given equality to reduce it, but it' s very difficult.

Can someone help me find the right sequence or the right method to solve this , thanks you

Answer 1

Here is the simplest form of the problem:

Let $M$ be a finite semigroup (written multiplicatively). Let $a$ and $b$ be elements of $M$ such that $ab^2 = b$. Then, prove that $bab = b$.

Solution. Let $G$ be the subset $\left\{ b \right\} \cup \left\{ bm \mid m \in M \right\}$ of $M$. Clearly, $G$ is a finite set. Moreover, a map $L_b : G \to G$ can be defined by $L_b \left(u\right) = bu$ for all $u \in G$.

Now, every $u \in G$ satisfies $a L_b\left(u\right) = u$. (Indeed, when $u = b$, this follows immediately from $a \underbrace{L_b\left(u\right)}_{=bu} = ab\underbrace{u}_{=b} = abb = ab^2 = b = u$. On the other hand, when $u = bm$ for some $m \in M$, this follows from $a \underbrace{L_b\left(u\right)}_{=bu} = ab\underbrace{u}_{=bm} = abbm = \underbrace{ab^2}_{=b}m = bm = u$.) This yields that every $u \in G$ can be uniquely reconstructed from $L_b\left(u\right)$. Hence, the map $L_b$ is injective. Thus, the map $L_b$ (being an injective map from the finite set $G$ to itself) must also be surjective. Hence, there exists some $u \in G$ such that $b = L_b\left(u\right)$ (since $b \in G$). Consider this $u$. We have $b = L_b\left(u\right) = bu$. But $u \in G$, so that either $u = b$, or $u = bm$ for some $m \in M$. Let us consider the second case. Thus, $u = bm$ for some $m \in M$. Then, $ab\underbrace{u}_{=bm} = abbm = \underbrace{ab^2}_{=b}m = bm = u$, so that $b\underbrace{abu}_{=u} = bu = b$, and therefore $b = ba\underbrace{bu}_{=b} = bab$. So we are done in the second case. In the first case, the same argument goes through, but with all the $m$'s removed. Thus, we are done in both cases.

Notice that we only used the finiteness of $G$, not the finiteness of $M$.

Answer 2

This is indeed a result of semigroup theory. Since darij-grinberg gave a nice detailed proof, I will just briefly indicate where this proof comes from, namely two results on finite semigroups:

In a finite semigroup, every element $s$ has an idempotent power (usually denoted by $s^\omega$).

In a finite semigroup, if $s$ and $s^2$ generate the same ideal, then $s^\omega s = s$.

Your question. If $abb = b$, then $b$ and $bb$ generate the same ideal and thus $b^\omega b = b$. Since $abb = b$ implies $abb^\omega = b^\omega$, we get $babb^\omega = bb^\omega$, that is $bab = b$.

Remark. One can prove that your result still holds in a compact semigroup.

All of this relies on the study of Green's relations (see any textbook on semigroup theory). The four Green's relations $\mathcal{R}$, $\mathcal{L}$, $\mathcal{J}$ and $\mathcal{H}$ are defined on $S$ as follows: \begin{align} s \mathrel{\mathcal{R}} t &\iff \text{$s$ and $t$ generate the same right ideal} \\ s \mathrel{\mathcal{L}} t &\iff \text{$s$ and $t$ generate the same left ideal} \\ s \mathrel{\mathcal{J}} t &\iff \text{$s$ and $t$ generate the same ideal} \\ s \mathrel{\mathcal{H}} t &\iff s \mathrel{\mathcal{R}} t \text{ and } s \mathrel{\mathcal{L}} t \\ \end{align}