Biordered sets of regular rings

By a semigroup, we mean a non-empty set $S$ with an associative binary operation, $(x,y)\mapsto xy$, from $S\times S\to S$. An element $x$ of a semigroup $S$ is said to be regular if there exists an element $x^{\prime}$ in $S$ with $xx^{\prime}x=x$. If $x^{\prime}$ satisfies the equation $x^{\prime}xx^{\prime}=x^{\prime}$ also, then $x^{\prime}$ is called a generalized inverse of $x$. It is not difficult to show that every regular element has a generalized inverse, for if $x^{\prime}$ satisfies the equation of regularity, then $x^{\prime\prime}=x^{\prime}xx^{\prime}$ satisfies both equations for generalized inverses. A semigroup in which all the elements are regular is called a regular semigroup. A ring $R$ is said to be a regular ring if its multiplicative semigroup is regular.

Any regular semigroup has a rich supply of idempotents, that is, elements $e$ for which $e^{2}=e$. In [2], the set of idempotents of a regular semigroups is given an abstract characterization as a partial algebra with two quasi-orders, which is termed a regular biordered set. We give here a few essential notions of biordered sets. Details can be found in [2]. Let $E$ be a non-empty set in which a partial binary operation is defined. (This means the product $ef$ is defined only for certain pairs $e,f$ of elements of $E$.) We define two relations $\omega^{r}$ and $\omega^{l}$ on $E$ by

$$\omega^{l}=\{(e,f)\in E\times E\colon ef=e\}\quad\text{and}\quad\omega^{r}=\{(e,f)\in E\times E\colon fe=e\}$$

One of the axioms of a biordered set is that these relations are quasi-orders, that is, they are reflexive and transitive. Note that this means the relation $\omega$ defined by

$$\omega=\omega^{l}\textstyle\bigcap\omega^{r}$$

is a partial order on $E$. For $e$ in $E$, we define

$$\omega^{l}(e)=\{f\in E\colon f\mathrel{\omega^{l}}e\}\quad\text{and}\quad\omega^{r}(e)=\{f\in E\colon f\mathrel{\omega^{r}}e\}$$

And for $e$, $f$ in $E$, we define

$$\mathsf{M}(e,f)=\omega^{l}(e)\textstyle\bigcap\omega^{r}(e)$$

Also, for $e$, $f$ in $E$, we define the sandwich set of $e$ and $f$ by

$$\mathsf{S}(e,f)=\{h\in\mathsf{M}(e,f)\colon g\preceq h\;\;\text{for all}\;\;g\in\mathsf{M}(e,f)\}$$

where $\preceq$ is the quasi-order defined by

$$g\preceq h\iff eg\mathrel{\omega^{r}}eh,gf\mathrel{\omega^{l}}hf$$

The regularity condition on a biordered set is that

$$\mathsf{S}(e,f)\neq\emptyset\;\;\text{for all $e$ and $f$ in $E$}$$

We first see how certain properties of the idempotents in a regular ring can be formulated in biorder terms and then later show that these properties actually characterize the biordered set of a ring of matrices.

Let $R$ be a ring with unity and let $E$ be the set of idempotents of $R$. It is easily seen that if $e$ is an idempotent in $R$, then $1-e$ is also an idempotent in $R$. Thus if we denote $1-e$ by ${e}^{\circ}$, then we have a map $e\mapsto{e}^{\circ}$ with ${e^{\circ\circ}}=e$. We first prove some elementary properties of this map in terms of the biorder relations of $E$.

The condition $fe=0$ can also be formulated in biorder terms in any regular semigroup with a zero element, using the idea of the $\mathsf{M}$-set defined earlier.

Using this, the second part of Proposition 2.1 can be reformulated as follows:

We next note that if $e$ and $f$ are idempotents in a ring with $ef=fe=0$, then $e+f$ is also an idempotent. This property can be characterized in biorder terms. We first note that the conditions $ef=fe=0$ are equivalent to the conditions $e\mathrel{\omega^{l}}{f}^{\circ}$ and $f\mathrel{\omega^{l}}{e}^{\circ}$, by Proposition 2.1(ii) and the relation $e\mathrel{\omega^{l}}{f}^{\circ}$ is equivalent to $f\mathrel{\omega^{r}}{e}^{\circ}$, by Proposition 2.1(i). It follows that $ef=fe=0$ iff $f\mathrel{\omega}{e}^{\circ}$. Also, if this condition holds, then the idempotent $e+f$ can be characterized in terms of sandwich sets. For this, we make use of the fact that if $E$ is the biordered set of idempotents of a regular semigroup $S$, then

$$\mathsf{S}(e,f)=\{h\in E\colon fhe=h\;\;\text{and}\;\;ehf=ef\;\;\text{in}\;\;S\}$$

(cf.Theorem 1.1 of [2]).

Another property of the biordered set of a regular ring is linked to the ideal theory of regular rings. It is well known that in a regular ring, every principal left or right ideal is generated by an idempotent, and that the set of principal left ideals and the set of principal right deals of a regular ring with unity form dually isomorphic complemented modular lattices (see [3], [1]). We next show that under certain conditions discussed above, a biordered set can be realized as the biordered set of a regular semigroup whose principal left ideals and principal right ideals form dually isomorphic complemented modular lattices.

We start by observing that in any semigroup $S$ with zero, we can define left annihilator of an element $s$ by

$$\mathsf{A}_{l}(s)=\{x\in S\colon xs=0\}$$

and the right annihilator of $s$ by

$$\mathsf{A}_{r}(s)=\{x\in S\colon sx=0\}$$

The semigroup $S$ is said to be a strongly regular Baer semigroup if the set of left annihilators of elements of $S$ is equal to the set of principal left ideals of $S$ and the set of right annihilators is equal to the set of principal right ideals of $S$. It can be shown that the multiplicative semigroup of a regular ring with unity is a strongly regular Baer semigroup ([1]). Also, a strongly regular Baer semigroup is a regular semigroup, in the sense defined earlier.

We now show that if a biordered set satisfies some of the properties discussed in the previous section, then it can be realized as the biordered set of a strongly regular Baer semigroup.

To prove this result, we make use of a couple of lemmas. First we show that in any biordered set satisfying the above conditions, the duals of these conditions also hold.

Now if $E$ is a regular biordered set, then there exists a regular semigroup $S$ with $E$ as its set of idempotents and which is idempotent generated, in the sense that every element of $S$ is a product of elements of $E$ (see Section 6 of [2]). It is easy to see that if $E$ is a regular biordered set satisfying (E1) and (E2), then 0 is the zero and 1 is the identity of every idempotent generated regular semigroup with $E$ as its set of idempotents. We next show that for each $e$ in $E$, the generator of the annihilators of $e$ in such a semigroup is ${e}^{\circ}$.

Now we can prove our theorem.

Now the set of principal left ideals and the set of principal right ideals of a strongly regular Baer semigroup can be shown to be complemented modular lattices which are dually isomorphic (see [1]). Also, the partially ordered set of principal left ideals and the partially ordered set of principal right ideals of any regular semigroup are isomorphic to the quotients of its biordered set by certain equivalence relations, as indicated below.

Let $S$ be a regular semigroup and let $E$ be the biordered set of its idempotents. We define the relations $\mathcal{L}$ and $\mathcal{R}$ on $E$ by

$$\mathcal{L}=\omega^{l}\textstyle\bigcap\,(\omega^{l})^{-1}\quad\text{and}\quad\mathcal{R}=\omega^{r}\textstyle\bigcap\,(\omega^{r})^{-1}$$

It is easily seen that the relations $\mathcal{L}$ and $\mathcal{R}$ are equivalences on $E$ and hence partition $E$. For each $e$ in $E$, we denote the $\mathcal{L}$-class containing $e$ by $\mathcal{L}(e)$ and the $\mathcal{R}$-class containing $e$ by $\mathcal{R}(e)$. The set of all $\mathcal{L}$-classes is denoted by $E/\mathcal{L}$ and the set of all $\mathcal{R}$-classes by $E/\mathcal{R}$. Now for $e$ and $f$ in $E$, if $e\mathrel{\omega^{l}}f$ and $e^{\prime}\in\mathcal{L}(e)$ and $f^{\prime}\in\mathcal{L}(f)$, then $e^{\prime}\mathrel{\omega^{l}}e\mathrel{\omega^{l}}f\mathrel{\omega^{l}}f^{\prime}$, so that $e^{\prime}\mathrel{\omega^{l}}f^{\prime}$, since $\omega^{l}$ is transitive. Hence we can unambiguously define a relation $\leq$ on $E/\mathcal{L}$ by

$$\mathcal{L}(e)\leq\mathcal{L}(f)\;\;\text{if and only if}\;\;e\mathrel{\omega^{l}}f$$

It is not difficult to see that this relation is a partial order on $E/\mathcal{L}$. Also, it can be easily seen that for $e$ and $f$ in $E$, we have $e\mathrel{\omega^{l}}f$ if and only if $Se\subseteq Sf$ and so $\mathcal{L}(e)\leq\mathcal{L}(f)$ if and only if $Se\subseteq Sf$. Thus the partially ordered set $E/\mathcal{L}$ is isomorphic with the partially ordered set of principal left ideals of $S$. Similarly, we can define a partial order $\leq$ on the set $E/\mathcal{R}$ of $\mathcal{R}$-classes in $E$ by

$$\mathcal{R}(e)\leq\mathcal{R}(f)\;\;\text{if and only if}\;\;e\mathrel{\omega^{r}}f$$

and this partially ordered set is isomorphic with the partially ordered set of principal right ideals of $S$.

Thus if $E$ is the biordered set of idempotents of a strongly regular Baer semigroup, then the quotients $E/\mathcal{L}$ and $E/\mathcal{R}$ are complemented modular lattices and they are dually isomorphic. In the following, we denote these lattices by $\mathbb{L}\,(E)$ and $\mathbb{R}\,(E)$ respectively. So, from Theorem 3.1, we get the following

We also note the following result on complements in $E/\mathcal{L}$

Since the multiplicative semigroup of a regular ring with unity is a strongly regular Baer semigroup, these results holds in particular for biordered sets of regular rings. Now in [3], it is shown that if $L$ is a complemented modular lattice satisfying certain conditions, then it can be realized as the lattice of principal left ideals of a matrix ring over a regular ring. To translate these conditions into biorder terms, we take a look at the idempotents in such a ring.

In [3], it is shown that if $R$ is a regular ring, then for every natural number $n$, the ring $R_{n}$ of $n\times n$ matrices over $R$ is also a regular ring. In this section, we look at some peculiarities of the biordered set of $R_{n}$. We first note that this ring contains a special class of idempotents. For each $i=1,2,\dotsc,n$ we define $\mathsfsl{E}_{i}$ to be the $n\times n$ matrix with a single 1 at the $i^{\text{th}}$ row and $i^{\text{th}}$ column and 0’s elsewhere. It easily follows from the usual rules of matrix multiplication that $\mathsfsl{E}_{i}$ is an idempotent for each $i$. Also, $\mathsfsl{E}_{i}\mathsfsl{E}_{j}=\mathsfsl{0}$ for $i\neq j$ and $\mathsfsl{E}_{1}+\mathsfsl{E}_{2}+\dotsb+\mathsfsl{E}_{n}=\mathsfsl{I}$, where $\mathsfsl{0}$ is the $n\times n$ zero matrix and $\mathsfsl{I}$ is the $n\times n$ identity matrix. We now see how the condition on the sum of these idempotents can be translated into biorder terms.

This discussion, together with Lemma 2.2, gives the following

Another property of these idempotents is that any pair of them generate principal left ideals which have a common complement (see the proof of Theorem 3.3, Part II,[3]). To describe this property in biorder terms, we introduce some terminology from [2] in a slightly modified form.

Let $e$ and $f$ be idempotents in a biordered set $E$. As in [2], by an $E$-sequence from $e$ to $f$, we mean a finite sequence $e_{0}=e,e_{1},e_{2},\dotsc,e_{n-1},e_{n}=f$ of elements of $E$ such that $e_{i-1}(\mathcal{L}\textstyle\bigcup\mathcal{R})e_{i}$ for $i=1,2,\dotsc,n$ and in this case, $n$ is called the length of the $E$-sequence. If there exists an $E$-sequence from $e$ to $f$, we define $d(e,f)$ to be the length of the shortest $E$-sequence from $e$ to $f$; also we define $d(e,e)=1$. If there is no $E$-sequence form $e$ to $f$,we define $d(e,f)=0$. For our purposes, we will have to distinguish between $E$-sequences starting with $\mathcal{L}$ and those starting with $\mathcal{R}$. For idempotents $e$ and $f$, we define $d_{l}(e,f)$ to be the length of the shortest $E$-sequence from $e$ to $f$, which start with the $\mathcal{L}$ relation and $d_{r}(e,f)$ to be the length of the shortest $E$-sequence from $e$ to $f$ which start with the $\mathcal{R}$ relation.

The condition for two principal left ideals of a ring to have a common complement can be described in terms of the $d_{l}$ function as follows. Following [3], two elements of a lattice which have a common complement are said to be in perspective.

We next show that any regular biordered set satisfying some of the conditions discussed so far can be realized as the set of idempotents of a ring of matrices over a regular ring.

In this section, we prove our main result:

To prove this result, we first look at some consequences of these conditions. We start by noting that in the case of a biordered set satisfying (E1), (E2) and (E3), there is exactly one element in $\mathsf{S}({e}^{\circ},{f}^{\circ})\textstyle\bigcap\mathsf{S}({f}^{\circ},{e}^{\circ})$.