Is there a finite commutative semigroup $S$ with $S^2 = S$ which is not a monoid?

Question

Is there a finite commutative semigroup $S$ with $S^2 = S$ which is not a monoid? where $S^2=\{ab\mid a,b\in S\}$.

As is known, if such $S$ can be a ring with an addition then it is a monoid? So if there's any, its product is not a ring multiplication.

Answer 1

The two minimal examples are the semigroup $S = \{a, b\}$ defined by $aa = ba = a$, $ab = bb = b$ and its dual $\tilde S$, defined on the same support by $aa = ab = a$, $ba = bb = b$.

Answer 2

if such S can be a ring with an addition then it is a monoid?

In this case (a finite set $S$ such that $(S,\cdot,+)$ is a ring possibly without identity and $S^2=S$) then in can be proven that $S$ has a multiplicative identity. But without the requirement of an addition making it a rng, such a semigroup need not be a monoid, as the comments have shown.

The strategy appears in other solutions on this site.

1. Form the Dorroh extension $S_1=\Bbb Z\times S$. Thus $S_1$ is a ring with identity which has a copy of $S$ as an ideal such that $S^2=S$.
2. Notice that Nakayama's lemma applies to the ring $S_1$ and its finitely generated ideal $S$. Thus there exists $x\in S$ such that $(1-x)S=\{0\}$.
3. The last thing means that $(1-x)s=0$ for all $s\in S$, and hence $s=xs$,showing there is a left identity $x$ in $S$.
4. Applying the same reasoning on the other side of $S$, we get a left identity in $S$, which is necessarily the same as the right identity we found. That's it: $x$ is the identity.