Nonminimal idempotents

Question

Let $S$ be a semigroup with operation $+$. Note that we do not assume $S$ is abelian. We say that $I\subset S$ is a left ideal if and only if for all $s\in S$, $s+I\subseteq I$, and we say that an element $u\in S$ is an idempotent if and only if $u+u=u$. From now on we will omit the word 'left' from ideal, but implicitly assume it. We say that an ideal is minimal if there are no proper subsets of $I$ which are ideals, and we say that an idempotent is minimal if it is contained in some minimal ideal.

Do you know of a simple example of a nonminimal idempotent? I know that some such exist in the Stone-Čech compactification of a discrete semigroup under some additional assumptions, but I was wondering if there are any simple, concrete examples out there. I am not a specialist in Algebra, so please can you be specific?

Answer 1

If you take the set $\{0,1\}$ together with usual multiplication, you get a semigroup for which all elements are idempotents. $0$ is minimal, but $1$ is not.

In fact, in any semigroup admitting an absorbing element, this absorbing element will be the only minimal idempotent.

Answer 2

Consider the semigroup $\mathbb{N}^{\times}$ of non-negative integers under multiplication (it is important that $0 \in \mathbb{N}^{\times}$).

The subset $\{ 0 \} \subset \mathbb{N}^{\times}$ is an ideal. Furthermore, for any ideal $I \subseteq \mathbb{N}^{\times}$ we have $0 \in I$, as for any $n \in I$, we have $0 = 0.n$. Therefore $\{ 0 \} \subset I \subset \mathbb{N}^{\times}$, so $\{ 0 \}$ is the only minimal ideal of $\mathbb{N}^{\times}$.

But $\mathbb{N}^{\times}$ has another idempotent, namely $1$. Any ideal containing $1$ must also contain $\{ 0 \}$, so $1$ is a non-minimal idempotent.