TITLE: A finite pseudo-ring such that $ab^2=b$ for some $(a,b)\in A$
QUESTION [6 upvotes]: 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

REPLY [4 votes]: 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}