## Exhibit the idempotents and principal ideals in a given semigroup

Denote the function $\mathsf{max}$ on $\mathbb{N}$ by $\wedge$. Let $S = \{0,1\} \times \mathbb{N}$, and define a binary operator $\star$ on $S$ by $(s,a) \star (t,b) = (0, a \wedge b)$ if $s = t$ and $(1, a \wedge b)$ if $s \neq t$. Show that $(S,\star)$ is a semigroup and exhibit all of its idempotents and principal ideals. Does $S$ have a kernel?

First, we argue that $\star$ is associative. To show this, we refer to the following tree diagram.

Associativity diagram

This diagram is to be read from left to right. Labels on an edge indicate an assumption that holds in all subsequent branches. Each path from the root to a leaf corresponds to a string of equalities, and together these imply that $\star$ is associative.

So $(S,\star)$ is a semigroup.

Suppose $(s,a)$ is idempotent. Then $(s,a) = (s,a) \star (s,a)$ $= (0, a)$, and we have $s = 0$. Conversely, $(0,a)$ is clearly idempotent for all $a \in \mathbb{N}$. So the idempotents in $S$ are precisely elements of the form $(0,a)$ with $a \in \mathbb{N}$.

Next we claim that $S$ is commutative. Indeed, if $s = t$, then $(s,a) \star (t,b) = (0,a \wedge b) = (0,b \wedge a)$ $= (t,b) \star (s,a)$ and if $s \neq t$ then $(s,a) \star (t,b) = (1, a \wedge b)$latex = (1, b \wedge a)\$ $= (t,b) \star (s,b)$.

We claim that the principal left ideal $L(s,a) = (s,a) \cup S(s,a)$ is $\{(t,b)\ |\ b \geq a\}$. Indeed, if $(t,b) \in L(s,a)$, then either $(t,b) = (s,a)$ or $(t,b) = (u,c)(s,a)$ for some $(u,a)$. But then $b = c \wedge a \geq a$. Conversely, consider $(t,b)$ with $b \geq a$, and let $u$ be 0 if $s =1$ and 1 otherwise. Now if $t = 0$, then $(t,b) = (s,b) \star (s,a)$ and if $t = 1$ then $(t,b) = (u,b) \star (s,a)$. So $L(s,a) = \{(t,b)\ |\ b \geq a\}$.

In particular, for every element $(s,a) \in S$, there is an ideal of $S$ not containing $(s,a)$ (for instance, $(s,a+1)$.) So $S$ has no kernel.