Let $S$ be a semigroup, a subset $I\subseteq S$ is called an ideal if $SI \subseteq I$ and $IS \subseteq S$. We denote by $S^1$ the semigroup $S$ adjoined with a identity if it does not contains one, and for $a \in S$ we set $J(a) = S^1aS^1$, the principal ideal generated by $a$. Also we set for $a,b \in S$
$$
a \mathcal J b :\Leftrightarrow S^1 a S^1 = S^1 b S^1.
$$
For an ideal we can form the Rees factor semigroup denotes by $S / I$, which essentially means collapsing everything in $I$ to a zero in the factor semigroup. Denote the $\mathcal J$-equivalence class of some $a \in S$ by $J_a$.
A principal factor is a factor semigroup of the form $J(a) / (J(a) \setminus J_a)$, see the Ecyclopdia of mathematics. The unique minimal ideal, called the kernel of $S$, is among the principal factors.
A null semigroup (or semigroup with zero multiplication) is a semigroup $S$ with a zero $0 \in S$ such that $ab = 0$ for all $a,b \in S$. A semigroup is called semisimple if none of the principal factors is a null semigroup.
I want to show that a semigroup $S$ is semisimple if and onyl if $A^2 = A$ for every ideal $A \subseteq S$.
This is an exercise from the book Fundamentals of Semigroup Theory by J. Howie, page 95.
What I do not understand is that if a semigroup contains a zero $0$, then the minimal ideal must be $\{0\}$, in particular this is a null semigroup. But I can very well find semigroups such that $A^2 = A$ for every ideal that have a zero. For example take any group $G$, adjoin a zero $0$ by setting $g0 = 0g =0$ and observe that the only ideals are $\{0\}$ and $G \cup\{0\}$ itself, and they both fulfill the condition on ideals…
So any hints on this exercise, or what I have understood wrong here?
EDIT: Maybe there is a typo in the exercise, and just the principal factors not equal to the kernel are meant. But the same exercise appears in the classic Algebraic Theory of Semigroups by Clifford/Preston, and the above link to the Encyclopedia does not excludes the kernel. But if we exclude the kernel, than $A^2 = A$ for every ideal $A$ iff no principal factor not equal the kernel is null. For $A^2 \subsetneq A$ choose $a \in A \setminus A^2$, then $J(a) / N(a)$ is null, for if $x,y \in J(a)$ with $xy \notin N(a)$ would imply $a = u(xy)v$ for some $u,v \in S^1$ and $(ux)(yv) \in A^2$. Conversely,
as $S^2 / I = (S/I)^2$ for every semigroup $S$ and ideal $I$ we have that $(J(a)/N(a))^2 = J(a)^2 / N(a) = J(a) / N(a)$ and as $|J(a)/N(a)| > 1$ if $N(a) \ne\emptyset$ those principal factors not equal the kernel are not null. But in case is is not a typo I ask for feedback/clarification…
Best Answer
Clifford and Preston give the following definition:
This is the right definition and it makes the exercise correct since the semigroup $0$ is simple.
Unfortunately, Howie modified the definition, leading to a wrong statement. The same wrong definition (and the same exercise) have been copied and pasted in Higgins' book Techniques of semigroup theory.