Let $S$ be a completely regular semigroup, expressed as a semilattice $Y$ of completely simple semigroups $S_{\alpha}$ ($\alpha \in{Y})$

Question

1. Show that, if $L$ is a left ideal of $S_{\alpha}$, then

$$L \cup [\cup \{S_{\beta} : \beta < \alpha\}]$$

is a left ideal of $S$.

2. Suppose now that $S= \mathcal{S}(Y;G_{\alpha};{\phi}_{\alpha,\beta})$ is a strong semi lattice of groups and that $L$ is a left ideal of $S$. Show that;

$$L \cap G_{\alpha} \neq \emptyset \Rightarrow \cup\{G_{\beta}: \beta \le \alpha\} \subseteq L$$

3. Show that $S$ is a Clifford semigroup if and only if it is regular and every one-sided ideal is a two sided ideal.

This is another question from Howies book. I struggle to understand what a semilattice is, and ideals are something I find hard to manipulate, so thought this would be a question for me to try and tackle (foolish I know!). Any help is obviously greatly appreciated, just trying to get a feel for how other people would approach this problem.

Cheers!

Answer 1

1) In a semilattice $\alpha\beta\leq \alpha$ and $\alpha\beta\leq \beta$, and so $\{S_{\beta}:\beta < \alpha\}$ is a (2-sided) ideal in $S$. A union of left ideals is a left ideal and so this part follows.

2) Suppose $L\cap G_{\alpha}\not=\varnothing$. Then there exists $g\in L\cap G_{\alpha}$. If $\beta \leq \alpha$, then $\beta\alpha = \beta$. Hence for every $h\in G_{\beta}$, $hg^{-1}\in G_{\beta}$ and so $h=hg^{-1}\cdot g\in L$. It follows that $G_{\beta}\subseteq L$ for all $\beta\leq \alpha$, as required.

3) ($\Leftarrow$) Since every left ideal is a right ideal and every right ideal is a left ideal, this is in particular true for principal left and right ideals. If $x\in S$ is arbitrary, then $S^1x$ is the least left ideal containing $x$ and $xS^1$ is a left ideal containing $x$ by our assumption. Hence $S^1x\subseteq xS^1$. Dually, $S^1x\supseteq xS^1$ and so $S^1x=xS^1$. In other words, $\mathscr{R}=\mathscr{L}$ in $S$. Since $\mathscr{H}=\mathscr{L}\cap\mathscr{R}$ and $\mathscr{D}=\mathscr{L}\circ \mathscr{R}$, it follows that $\mathscr{H}=\mathscr{D}=\mathscr{R}=\mathscr{L}$.

Since $S$ is regular, every $\mathscr{D}$-class of $S$ contains an idempotent. From $\mathscr{D}=\mathscr{H}$, every $\mathscr{H}$-class also contains an idempotent. It follows that every $\mathscr{R}$-class and every $\mathscr{L}$-class contain precisely one idempotent, and so $S$ is an inverse semigroup, and hence a Clifford semigroup.

($\Rightarrow$) Suppose $S$ is a Clifford semigroup. Then it is completely regular, and hence regular. It is also a strong semilattice of groups and so, without loss of generality we can assume that $S=\mathcal{S}(Y; G_{\alpha}:\phi_{\alpha, \beta})$ as in part 2.

We will show that every right ideal is a left ideal. A dual argument shows that every left ideal is a right ideal, and hence every one-sided ideal is a two-sided ideal.

Since every right ideal is a union of principal right ideals, it suffices to show that every principal left ideal is a right ideal. Let $x\in S$ and consider $L=S^1x$ (the principal left ideal generated by $x$). Since $S$ is a union of groups, it follows that $x\in G_{\alpha}$ for some $\alpha\in Y$. Hence by 2, $\bigcup\{G_{\beta}:\beta\leq \alpha\}$ is a subset of $L$. On the other hand, if $y\in L$, then $y=zx$ for some $z\in S^1$. But $z\in G_{\gamma}$ and so $y=zx\in G_{\gamma\alpha}\subseteq L$ (since $\gamma\alpha\leq \alpha$). Thus $L = \bigcup\{G_{\beta}:\beta\leq \alpha\}$.

So, if $y\in L$ and $s\in S$ are arbitrary, then $y\in G_{\beta}$, $\beta\leq \alpha$, and $s\in G_{\gamma}$, $\gamma\in Y$. This implies that $yz\in G_{\beta\gamma}\subseteq L$, since $\beta\gamma\leq \beta\leq \alpha$, and so $L$ is a right ideal as required.