In this section, \((S ;\cdot , \leqslant , 1)\) is a finite negative tomonoid. We are interested in the one-element (Rees) coextensions of S. To this end, we will define a pomonoid of which any of the desired coextensions is a quotient.
If \(S = \{1\}\), we call S the trivial tomonoid. In this case, S possesses exactly one one-element Rees-coextension. Indeed, the two element chain \(\{0, 1\}\) can be made into a tomonoid in only one way: by defining \(1 \cdot 1 = 1\) and \(0 \cdot 1 = 1 \cdot 0 = 0 \cdot 0 = 0\). We will assume from now on that S is non-trivial, that is, S possesses at least two elements.
Denoting by \(\dot{0}\) the smallest element of S, we define \(S^\star = S {\setminus } \{ \dot{0}\}\) and \(\bar{S}= S^\star \mathbin {\dot{\cup }}\{0,\alpha \}\), where 0 and \(\alpha \) are new elements. We extend the total order on \(S^\star \) to \(\bar{S}\), letting \(0 \leqslant \alpha \leqslant a\) for any \(a \in S^\star \).
We view \(S^\star \) as a partial monoid; cf., e.g., [11]. The multiplication on \(S^\star \) is defined partially, the product of two elements \(a, b \in S^\star \) being defined only if ab, calculated in S, is an element of \(S^\star \). Then \(1 \in S^\star \), the products \(1 \, a\) and \(a \, 1\) exist for all \(a \in S^\star \), and \(1 \, a = a \, 1 = a\). Furthermore, for \(a, b, c \in S^\star \), the products ab and (ab)c exist if and only if bc and a(bc) exist, in which case \((ab)c=a(bc)\).
Our aim is to determine all possible ways of endowing \(\bar{S}\) with a monoidal product such that \(\bar{S}\) becomes a one-element coextension of S. Let us clarify what this means. Let \((\bar{S}; \cdot , \leqslant , 1)\) be a finite, negative tomonoid and assume that S is the Rees quotient of \(\bar{S}\) by the ideal \(\{0,\alpha \}\). Note that S consists of the elements of \(S^\star \), which we identify with the singleton congruence classes, and of the element \(\dot{0}\), which corresponds to the congruence class \(\{0,\alpha \}\). The multiplication in S arises from the multiplication in \(\bar{S}\) as follows: for \(a, b \in S\), we have
$$\begin{aligned} a \cdot b \;=\; {\left\{ \begin{array}{ll} a \cdot b &{}\quad \text {if}\ a, b \ne \dot{0}\ \text {and},\ \text {in}\ \bar{S}, a \cdot b \notin \{0,\alpha \}, \\ \dot{0}&{}\quad \text {otherwise.} \end{array}\right. } \end{aligned}$$
(3)
In the present context, we consider \((S; \cdot , \leqslant , 1)\) as fixed and we intend to determine \((\bar{S}; \cdot , \leqslant , 1)\). From (3) we observe that the multiplication on S determines to a good extent the one on \(\bar{S}\): for any two elements \(a, b \in \bar{S}{\setminus } \{0,\alpha \}\) whose product is non-zero in S, this product is in \(\bar{S}\) the same as in S. We may say that the partial monoid \(S^\star \) is a substructure of \(\bar{S}\): whenever, for \(a, b \in S^\star \), ab is defined and equals c in \(S^\star \), we have that \(a b = c\) holds in \(\bar{S}\) as well. Consequently, our aim is the extension of the partial tomonoid \(S^\star \) to a (total) tomonoid based on the chain \(\bar{S}\).
For a pair \((a,b) \in \bar{S}^2\) such that \(a, b \in S^\star \) and ab is defined in \(S^\star \), the product of a and b is in \(\bar{S}\) thus determined from the outset. We denote the set of all remaining pairs of elements of \(\bar{S}\) as follows:
$$\begin{aligned} \mathcal {N}= & {} \{ (a,b) \in \bar{S}^2 :a, b > \alpha \ \text {and}\ a b = \dot{0}\ \text {in}\ S\} \\&\cup \;\; \{0,\alpha \} \times \bar{S}\;\;\cup \;\; \bar{S}\times \{0,\alpha \}. \end{aligned}$$
For a pair \((a,b) \in \mathcal {N}\), it is in general not clear what the product of a and b in \(\bar{S}\) is, but it is clear that ab equals either 0 or \(\alpha \). Indeed, if \(a, b > \alpha \), then \(a b = c > \alpha \) in \(\bar{S}\) would imply that we have \(a b = c\) in S as well and thus \(a b \ne \dot{0}\) in S. Moreover, if \(a \leqslant \alpha \) or \(b \leqslant \alpha \), it follows \(a b \leqslant \alpha \) by the negativity of \(\bar{S}\), that is, \(a b = 0\) or \(a b = \alpha \). Determining the coextension \(\bar{S}\) means accordingly that we have to suitably divide \(\mathcal {N}\) into a set of pairs mapped to 0 and a set of pairs mapped to \(\alpha \).
We add that even the products of certain pairs in \(\mathcal {N}\) are clear from the outset. Indeed, we have \(0 \, a = a \, 0 = 0\) for any \(a \in \bar{S}\) and \(1 \, \alpha = \alpha \, 1 = \alpha \). But the following examples show that in all remaining cases the product may really be either 0 or \(\alpha \).
Example 3.1
For \(a, b \in \bar{S}\), let us define
$$\begin{aligned} a \cdot b \;=\; {\left\{ \begin{array}{ll} a b &{}\quad \text {if}\ a, b \in S^\star \ \text {and}\ a b\ \text {exists in}\ S^\star , \\ \alpha &{}\quad \text {if}\ a = \alpha \ \text {and}\ b = 1, \ \text {or}\ a = 1\ \text {and}\ b = \alpha , \\ 0 &{}\quad \text {otherwise.} \end{array}\right. } \end{aligned}$$
(4)
We readily verify that \((\bar{S};\cdot , \leqslant , 1)\) is a one-element coextension of \((S ;\cdot , \leqslant , 1)\).
Example 3.2
A further one-element coextension is \((\bar{S};\cdot , \leqslant , 1)\), where, for \(a, b \in \bar{S}\),
$$\begin{aligned} a \cdot b \;=\; {\left\{ \begin{array}{ll} a b &{}\quad \text {if}\ a, b \in S^\star \ \text {and}\ a b\ \text {exists in}\ S^\star , \\ 0 &{}\quad \text {if}\ a = 0\ \text {or}\ b = 0, \\ \alpha &{}\quad \text {otherwise.} \end{array}\right. } \end{aligned}$$
(5)
These examples are, in a sense, the extreme cases. Indeed, in case of Example 3.1, all products that are not determined from the outset are defined to be 0, whereas in case of Example 3.2 all these are \(\alpha \).
We now provide the key definition on which the present paper is based.
Definition 3.3
Let \(\mathcal R(S)\) be the free pomonoid over the chain \(\bar{S}\), subject to the following conditions:
-
(a)
\(a \, b = c\) for any \(a, b, c \in S^\star \) fulfilling this equation in S,
-
(b)
\(\varepsilon = 1\).
-
(c)
\(a \, b \leqslant \alpha \) for any \((a,b) \in \mathcal {N}\),
-
(d)
\(0 \, a = 0\) for any \(a \in \bar{S}\),
We call \(\mathcal R(S)\) the free one-element Rees coextension, or simply the free one-element coextension of S.
Proposition 3.4
Let T be a one-element coextension of S. Then there is a congruence \(\mathbin {\theta }\) on \(\mathcal R(S)\) such that each \(\mathbin {\theta }\)-class contains exactly one \(a \in \bar{S}\) and \(\mathcal R(S)/\mathbin {\theta }\) is isomorphic to T.
Proof
Note that \(\bar{S}\) and T are chains of equal size; let \(f :\bar{S}\rightarrow T\) be the order isomorphism. By Proposition 2.4, f extends to a pomonoid homomorphism \(\bar{f} :\mathcal F(\bar{S}) \rightarrow T\).
Let us identify the equalities and inequalities (a)–(d) in Definition 3.3 with the binary relation \(\trianglelefteq \) on \(\mathcal F(\bar{S})\); for instance, by (c) we require \(a b \trianglelefteq \alpha \) for \((a,b) \in \mathcal {N}\). Then \(\mathcal R(S) = \mathcal F(\bar{S})/\mathbin {\Theta (\trianglelefteq )}\). It is obvious that, for any \(a, b \in \mathcal F(\bar{S})\), \(a \trianglelefteq b\) implies \(\bar{f}(a) \leqslant \bar{f}(b)\). Hence, by Proposition 2.3, there is a homomorphism \(\tilde{\bar{f}} :\mathcal R(S) \rightarrow T\) such that \(\tilde{\bar{f}}({a/\mathbin {\Theta (\trianglelefteq )}}) = \bar{f}(a)\) for any \(a \in \mathcal F(\bar{S})\). In particular, for \(a \in \bar{S}\), we have \(\tilde{\bar{f}}({a/\mathbin {\Theta (\trianglelefteq )}}) = f(a)\) and it follows that \(\tilde{\bar{f}}\) is order-determining.
Thus T is a homomorphic image of \(\mathcal R(S)\) and the assertions follow. \(\square \)
To shorten the subsequent statements, let us call a congruence on \(\mathcal R(S)\) that, in the way indicated in Proposition 3.4, leads to a one-element coextension of S, a coextension congruence. Our aim is therefore to characterise these particular congruences on \(\mathcal R(S)\).
We begin by describing \(\mathcal R(S)\) itself. From now on we will make use of the simplified notation announced earlier: an element of \(\bar{S}\) or a word in \(\mathcal F(\bar{S})\) will also denote its congruence class in \(\mathcal R(S)\).
Proposition 3.5
Let \(\beta \) be the bottom element of \(S^\star \), that is, the atom of S.
-
(i)
\(\mathcal R(S)\) is the union of the intervals \([0, \alpha ]\) and \([\beta , 1]\), and we have \(0< \alpha < \beta \). Moreover, \([0, \alpha ]\) is a poideal and \([\beta , 1]\) is a chain, consisting of the pairwise distinct elements \(a \in S^\star \).
-
(ii)
1 is the top element and the identity of \(\mathcal R(S)\); in particular, \(\mathcal R(S)\) is a negative pomonoid.
-
(iii)
Let \(a, b \in [\beta , 1]\). If \(a b = c\) holds in \(S^\star \), then we have \(a b = c\) in \(\mathcal R(S)\) as well. If ab is in \(S^\star \) undefined, then we have \(a b \leqslant \alpha \) in \(\mathcal R(S)\).
Proof
We first show that the elements of \(\bar{S}\) are in \(\mathcal R(S)\) pairwise distinct. Indeed, for any one-element coextension T there is by Proposition 3.4 a congruence \(\mathbin {\theta }\) on \(\mathcal R(S)\) such that \(a \in \bar{S}\) are all in distinct classes and T is isomorphic to the quotient. Moreover, by Examples 3.1 and 3.2, a one-element coextension always exists. We conclude that distinct elements of \(\bar{S}\) are indeed distinct in \(\mathcal R(S)\).
By the definition of \(\mathcal R(S)\), the elements of \(\bar{S}\) moreover form a chain in \(\mathcal R(S)\). In particular, we have \(0< \alpha < \beta \leqslant 1\), thus \([0,\alpha ]\) and \([\beta ,1]\) are disjoint intervals of \(\mathcal R(S)\).
We have \(\varepsilon = 1\) by the defining equality (b) of Definition 3.3, thus 1 is the identity of \(\mathcal R(S)\). We furthermore have \(a \leqslant 1\) for any \(a \in \bar{S}\), and \(1 \cdot 1 = 1\). For \(a_1, \ldots , a_k \in \bar{S}\), \(k \geqslant 1\), it follows \(a_1 \ldots a_k \leqslant 1 \cdot \cdots \cdot 1 = 1\). Hence 1 is the top element of \(\mathcal R(S)\).
Furthermore, equality (d) of Definition 3.3 and the fact that \(0 \leqslant a\) for any \(\bar{S}\) imply that 0 is the bottom element of \(\mathcal R(S)\). Hence \([0,\alpha ]\) is a downset and since \(\mathcal R(S)\) is negative, \([0,\alpha ]\) is actually a poideal.
Let now \(a, b \in S^\star \). Then \(a, b \in [\beta , 1]\). Assume first that in S we have \(a b = c \ne \dot{0}\). This means \(a b = c\) holds in \(S^\star \) and by equality (a) the same equality then holds in \(\mathcal R(S)\). Assume second that in S we have \(a b = \dot{0}\). This means that ab is in \(S^\star \) undefined and by inequality (c) we have \(a b \leqslant \alpha \).
As \(\mathcal R(S)\) is (as a monoid) generated by \(\bar{S}\) and \([0,\alpha ]\) is a poideal, we conclude that \([\beta ,1]\) consists of the elements of \(S^\star \) only and \(\mathcal R(S) = [0,\alpha ] \cup [\beta ,1]\). \(\square \)
We thus see that the free one-element coextension \(\mathcal R(S)\) consists of \(S^\star \) and, strictly below this set, the poideal \([0, \alpha ]\). We note that the latter set is infinite. Proposition 3.5 furthermore implies that \(\mathcal R(S)/[0,\alpha ]\) is (isomorphic to) S.
Assume that \(\mathbin {\theta }\) is a coextension congruence. Then each \(\mathbin {\theta }\)-class contains exactly one \(a \in \bar{S}\). Hence, by Proposition 3.5, the quotient consists of the singletons \({a/\mathbin {\theta }} = \{a\}\), \(a \in S^\star \), as well as \(0/\mathbin {\theta }\) and \(\alpha /\mathbin {\theta }\), which partition \([0,\alpha ]\). In particular, a coextension congruence is uniquely determined by the class \(0/\mathbin {\theta }\).
Proposition 3.6
Let \(Z\subseteq [0,\alpha ]\) be a downset of \(\mathcal R(S)\) that is non-empty but does not contain \(\alpha \). Assume that the following condition holds:
$$\begin{aligned} \begin{aligned} \text {For any}\ a, b \in [0,\alpha ] {\setminus } Z\ \text {and any}\ c \in \bar{S},&\quad a c \in Z\ \text {if and only if}\ b c \in Z, \\ \text {and}&\quad c a \in Z\ \text {if and only if}\ c b \in Z. \end{aligned} \end{aligned}$$
For \(a, b \in \mathcal R(S)\), let
$$\begin{aligned} a \theta _{Z}b \quad \text {if and only if}\quad a = b \text { or } a, b \in Z\text { or } a, b \in [0,\alpha ] {\setminus } Z. \end{aligned}$$
(6)
Then \(\theta _{Z}\) is a congruence on \(\mathcal R(S)\) such that \(Z= {0/\theta _{Z}}\), and \(\mathcal R(S)/\theta _{Z}\) is a one-element coextension of S.
Up to isomorphism, every one-element coextension of S arises in this way from a unique downset \(Z\) of \(\mathcal R(S)\).
Proof
Let \(Z\) be a downset of \([0,\alpha ]\) and let \(\theta _{Z}\) be given as indicated. The \(\theta _{Z}\)-classes are then the singletons \(\{a\}\) for each \(a \in S^\star \), as well as \(Z\) and \([0,\alpha ] {\setminus } Z\). To show that \(\theta _{Z}\) is a congruence, let \(a, b, c \in \mathcal R(S)\). Assume that \(a \theta _{Z}b\); we have to show that \(a c \theta _{Z}b c\). Obviously, we may restrict to the case \(c \in \bar{S}\). If \(a \in S^\star \), then \(a = b\) and the assertion is clear. If \(a \in Z\), then also \(b \in Z\), and because \(Z\) is an ideal, we have \(a c, b c \in Z\) and hence \(a c \theta _{Z}b c\). Let \(a \in {[}0,\alpha ] {\setminus } Z\). Then also \(b \in [0,\alpha ] {\setminus } Z\). Because \([0,\alpha ]\) is a poideal, we have that \(a c, b c \in {[}0,\alpha ]\). By assumption, ac and bc are either both in \(Z\) or both in \([0,\alpha ] {\setminus } Z\), thus \(a c \theta _{Z}b c\). A similar argument shows that also \(c a \theta _{Z}c b\) holds, and we conclude that \(\theta _{Z}\) is a monoid congruence.
To see that \(\theta _{Z}\) is in fact a pomonoid congruence, assume \(c_0 \leqslant c_1 \theta _{Z}c_2 \leqslant \ldots \theta _{Z}c_k = c_0\). As \(Z\) and \([0,\alpha ]\) are downsets, it follows that \(c_0, \ldots , c_k\) all lie in the same \(\theta _{Z}\)-class.
Consider now the Rees quotient of \(\mathcal R(S)/\theta _{Z}\) by the two-element poideal \([0/\theta _{Z}, \alpha /\theta _{Z}]\). The resulting tomonoid is obviously isomorphic to \(\mathcal R(S)/[0,\alpha ]\) and hence to S. This means that \(\mathcal R(S)/\theta _{Z}\) is a one-element coextension of S.
We now turn to the last assertion. Let T be a one-element coextension of S. By Proposition 3.4, there is a congruence \(\mathbin {\theta }\) on \(\mathcal R(S)\) such that \(\mathcal R(S)/\mathbin {\theta }\) is isomorphic with T. Identifying T with \(\bar{S}\), the elements of T are the classes \(a/\mathbin {\theta }\), \(a \in \bar{S}\), and we have \({a/\mathbin {\theta }} = \{a\}\) for \(a \in S^\star \) and \({0/\mathbin {\theta }} \cup {\alpha /\mathbin {\theta }} = [0,\alpha ]\).
Let \(Z= {0/\mathbin {\theta }}\). As 0 is the bottom element of \(\mathcal R(S)\) and the \(\mathbin {\theta }\)-classes are convex, \(Z\) is a downset of \(\mathcal R(S)\) such that \(\varnothing \subset Z\subset [0,\alpha ]\). Furthermore, we have \([0,\alpha ] {\setminus } Z= {\alpha /\mathbin {\theta }}\). It follows that \(\mathbin {\theta }= \theta _{Z}\) according to (6). Moreover, for any \(a, b \in [0,\alpha ] {\setminus } Z\), we have \(a \mathbin {\theta }b \mathbin {\theta }\alpha \) and hence \(ac \mathbin {\theta }bc \mathbin {\theta }\alpha c\). Since \(\alpha c \in [0,\alpha ]\) it follows that ac and bc are both in \(Z\) or both in \([0,\alpha ] {\setminus } Z\). This completes the proof that all one-element coextensions of S arise in the way asserted from a downset \(Z\) of \(\mathcal R(S)\).
It remains to show that the downset giving rise to a one-element coextension is unique. But this is clear from the fact that distinct downsets \(Z\subset [0,\alpha ]\) induce distinct congruences \(\mathbin {\theta }\), that is, distinct products on \(\mathcal R(S)/\mathbin {\theta }\). \(\square \)
Example 3.7
We consider an example illustrating Proposition 3.6. Here as well as in case of subsequent examples of tomonoids, we denote the elements of the base set by lower case Latin letters from the end of the alphabet, except for the bottom and top elements, which are denoted by \(\dot{0}\) and 1, respectively. Let S be the five-element chain, that is, let
$$\begin{aligned} S \;=\; \{ \dot{0}, x, y, z, 1 \}, \end{aligned}$$
understood to be totally ordered as indicated. We make S into a tomonoid by defining, for \(a, b \in S\),
$$\begin{aligned} a \mathbin {\cdot }b \;=\; {\left\{ \begin{array}{ll} b &{}\quad \text {if}\ a = 1, \\ a &{}\quad \text {if}\ b = 1, \\ x &{}\quad \text {if}\ a = b = z, \\ \dot{0}&{}\quad \text {otherwise.} \end{array}\right. } \end{aligned}$$
The free one-element coextension \(\mathcal R(S)\) is infinite; Fig. 1 shows a part of its poset reduct. Furthermore, an example of a coextension congruence is indicated; the congruence classes are encircled.
According to Proposition 3.6, it would now be natural to aim at a characterisation of the downset \(Z= 0/\mathbin {\theta }\), where \(\mathbin {\theta }\) is a coextension congruence. Although this is feasible, we proceed more conveniently as follows.
Lemma 3.8
Let \(\mathbin {\theta }\) be a coextension congruence on \(\mathcal R(S)\) and define
$$\begin{aligned} \mathcal {Z}_{\mathbin {\theta }} \;=\; \{ (a,b) \in \bar{S}^2 :a b \mathbin {\theta }0 \}. \end{aligned}$$
Then \(\mathcal {Z}_{\mathbin {\theta }} \subseteq \mathcal {N}\). Moreover, \(\mathbin {\theta }\) is the congruence generated by the equations
$$\begin{aligned} a \, b = 0 \;\text { for any}\ (a,b) \in \mathcal {Z}_{\mathbin {\theta }}, \quad a \, b = \alpha \;\text { for any}\ (a,b) \in \mathcal {N}{\setminus } \mathcal {Z}_{\mathbin {\theta }}. \end{aligned}$$
Proof
If \((a,b) \in \bar{S}^2 {\setminus } \mathcal {N}\), then \(ab \in S^\star \) and hence \((a,b) \notin \mathcal {Z}_{\mathbin {\theta }}\). That is, \(\mathcal {Z}_{\mathbin {\theta }} \subseteq \mathcal {N}\).
The only \(\mathbin {\theta }\)-classes that are not singletons are \(0/\mathbin {\theta }\) and \(\alpha /\mathbin {\theta }\). Let \(a = a_1 \ldots a_k \in \mathcal R(S)\), where \(a_1, \ldots , a_k \in \bar{S}\), \(k \geqslant 0\). If \(k = 0\), then \(a = 1\) and hence neither \(a \mathbin {\theta }0\) nor \(a \mathbin {\theta }\alpha \) holds. If \(k = 1\), then \(a \mathbin {\theta }0\) iff \(a = 0\), and \(a \mathbin {\theta }\alpha \) iff \(a = \alpha \). Let \(k = 2\). Then \(a \mathbin {\theta }0\) iff \((a_1,a_2) \in \mathcal {Z}_{\mathbin {\theta }}\) and \(a \mathbin {\theta }\alpha \) iff \((a_1,a_2) \in \mathcal {N}{\setminus } \mathcal {Z}_{\mathbin {\theta }}\). Finally, assume that \(k \geqslant 3\). Let \(a_1 \ldots a_{k-1} \mathbin {\theta }z\), where \(z \in \bar{S}\). Then again, \(a \mathbin {\theta }0\) iff \((z,a_k) \in \mathcal {Z}_{\mathbin {\theta }}\), and \(a \mathbin {\theta }\alpha \) iff \((z,a_k) \in \mathcal {N}{\setminus } \mathcal {Z}_{\mathbin {\theta }}\). Hence the assertion follows by an inductive argument. \(\square \)
We conclude that each coextension congruence \(\mathbin {\theta }\) on \(\mathcal R(S)\) is determined by the set \(\mathcal {Z}_{\mathbin {\theta }} \subseteq \mathcal {N}\). The following lemma makes explicit how the class \(0/\mathbin {\theta }\) is determined by \(\mathcal {Z}_{\mathbin {\theta }}\).
Lemma 3.9
Let \(\mathbin {\theta }\) be a coextension congruence on \(\mathcal R(S)\). Let \(a = a_1 \ldots a_k \in [0,\alpha ]\), where \(a_1, \ldots , a_k \in \bar{S}\), \(k \geqslant 1\). Let \(i \in \{1, \ldots , k \}\) be smallest such that the product \(a_1 \ldots a_i\) is not in \(S^\star \). We have \(a \mathbin {\theta }0\) if and only if either \(a_i = 0\), or \(i < k\) and \((\alpha , a_j) \in \mathcal {Z}_{\mathbin {\theta }}\) for some \(i < j \leqslant k\), or \(i \geqslant 2\) and \((a_1 \ldots a_{i-1}, a_i) \in \mathcal {Z}_{\mathbin {\theta }}\).
Proof
Let i be as indicated; then \(a_1 \ldots a_i \mathbin {\theta }0\) or \(a_1 \ldots a_i \mathbin {\theta }\alpha \). It is easily checked that \(a \mathbin {\theta }0\) holds under one of the indicated conditions.
Conversely, assume \(a \mathbin {\theta }0\). If \(a_1 \ldots a_i \mathbin {\theta }0\), then either \(i = 1\) and \(a_i = a_1 = 0\), or \(i \geqslant 2\) and \((a_1 \ldots a_{i-1}, a_i) \in \mathcal {Z}_{\mathbin {\theta }}\). If \(a_1 \ldots a_i \mathbin {\theta }\alpha \), then there is a smallest \(j \in \{ i+1, \ldots k \}\) such that \(a_1 \ldots a_{j-1} \mathbin {\theta }\alpha \), thus \(a_1 \ldots a_j \mathbin {\theta }0\) and consequently \((\alpha , a_j) \in \mathcal {Z}_{\mathbin {\theta }}\). \(\square \)
Our aim is the characterisation of the set \(\mathcal {Z}_{\mathbin {\theta }} \subseteq \mathcal {N}\). We need some preparations.
Definition 3.10
Let \(\mathbin {\theta }\) be a coextension congruence on \(\mathcal R(S)\). Then we call the smallest element \(\varepsilon _l\in S^\star \cup \{\alpha \}\) such that \(\varepsilon _l\alpha \mathbin {\theta }\alpha \) the left border for \(\mathbin {\theta }\). Similarly, we call the smallest element \(\varepsilon _r\in S^\star \cup \{\alpha \}\) such that \(\alpha \varepsilon _r\mathbin {\theta }\alpha \) the right border for \(\mathbin {\theta }\).
We will denote the left and right border for a coextension congruence \(\mathbin {\theta }\) in the sequel always by \(\varepsilon _l\) and \(\varepsilon _r\), respectively. The name is justified by the fact that \(\varepsilon _l\) and \(\varepsilon _r\) provide a limitation for the class \(0/\mathbin {\theta }\) and hence for \(\mathcal {Z}_{\mathbin {\theta }}\), as stated in the following remark.
Remark 3.11
Let \(\mathbin {\theta }\) be a coextension congruence on \(\mathcal R(S)\). For any \(a \in \bar{S}\), we have \(0 = 0 \, \alpha \leqslant a \, \alpha \leqslant 1 \, \alpha = \alpha \) and thus either \(a \, \alpha \mathbin {\theta }0\) or \(a \, \alpha \mathbin {\theta }\alpha \). Hence \(\varepsilon _l\) is the smallest element in \(\bar{S}\) such that \((\varepsilon _l, \alpha ) \notin \mathcal {Z}_{\mathbin {\theta }}\), that is, for \(a \in \bar{S}\) we have \((a, \alpha ) \in \mathcal {Z}_{\mathbin {\theta }}\) iff \(a < \varepsilon _l\). Similarly we may characterise \(\varepsilon _r\).
Note furthermore that, for any \(a, b \in \bar{S}\) such that \((a,b) \in \mathcal {Z}_{\mathbin {\theta }}\), we have \(a = 0\), or \(b = 0\), or \(a < \varepsilon _l\) and \(b < \varepsilon _r\).
Lemma 3.12
Let \(\mathbin {\theta }\) be a coextension congruence with the borders \(\varepsilon _l, \varepsilon _r\).
-
(i)
\(\varepsilon _l= \alpha \) if and only if \(\varepsilon _r= \alpha \). In this case, \({0/\mathbin {\theta }} = \{0\}\) and the quotient is isomorphic to \((\bar{S};\cdot , \leqslant , 1)\) from Example 3.2. In particular, \(\alpha /\mathbin {\theta }\) is idempotent.
-
(ii)
Let \(\varepsilon _l, \varepsilon _r\in S^\star \). Then both \(\varepsilon _l\) and \(\varepsilon _r\) are idempotent elements of S.
Proof
-
(i)
Assume that \(\varepsilon _l= \alpha \). This means \(\alpha ^2 \mathbin {\theta }\alpha \) and hence also \(\varepsilon _r= \alpha \). Similarly, we see that \(\varepsilon _r= \alpha \) implies \(\varepsilon _l= \alpha \).
Moreover, for any \(a, b \in \bar{S}{\setminus } \{ 0 \}\), we have in this case \({ab/\mathbin {\theta }} = {a/\mathbin {\theta }} \cdot {b/\mathbin {\theta }} \geqslant ({\alpha /\mathbin {\theta }})^2 = \alpha /\mathbin {\theta }\). It follows \({0/\mathbin {\theta }} = \{0\}\) and multiplication in \(\mathcal R(S)/\mathbin {\theta }\) is given according to (5).
-
(ii)
By definition, \(\varepsilon _l\alpha \mathbin {\theta }\alpha \) and hence \(\varepsilon _l^2 \alpha \mathbin {\theta }\varepsilon _l\alpha \mathbin {\theta }\alpha \). We claim that \(\varepsilon _l^2 \in S^\star \). Indeed, otherwise \(\varepsilon _l^2 \leqslant \alpha < \varepsilon _l\) would imply \(\varepsilon _l^2 \alpha \leqslant \alpha ^2 \mathbin {\theta }0\), a contradiction. From the minimality property of \(\varepsilon _l\) we conclude \(\varepsilon _l\leqslant \varepsilon _l^2\). As we have \(\varepsilon _l^2 \leqslant \varepsilon _l\) by the negativity of \(\mathcal R(S)\), we have that \(\varepsilon _l^2 = \varepsilon _l\) holds in \(\mathcal R(S)\) and hence also in S.
The assertion concerning \(\varepsilon _r\) is seen analogously.
\(\square \)
We conclude that the border elements \(\varepsilon _l, \varepsilon _r\) either both equal \(\alpha \) or are idempotent elements of the partial monoid \(S^\star \).
Note that since \(S^\star = S {\setminus } \{\dot{0}\}\) and \(\dot{0}\) is an idempotent element of S, we can identify the pair \(\varepsilon _l, \varepsilon _r\) with a pair of two idempotent elements of the original tomonoid S; in this case, \(\dot{0}\in S\) corresponds to \(\alpha \in \bar{S}\).
We are now ready to compile the characteristic properties of the set \(\mathcal {Z}_{\mathbin {\theta }}\).
Lemma 3.13
Let \(\mathbin {\theta }\) be a coextension congruence on \(\mathcal R(S)\) with the borders \(\varepsilon _l, \varepsilon _r\). Then \(\mathcal {Z}= \mathcal {Z}_{\mathbin {\theta }}\) is a subset of \(\mathcal {N}\) such that, for any \(a, b, c, d \in \bar{S}\), the following holds:
-
(Z1)
If \((b,d) \in \mathcal {Z}\) and \(a \leqslant b\), \(\; c \leqslant d\), then \((a,c) \in \mathcal {Z}\).
-
(Z2)
\((a,0), (0,b) \in \mathcal {Z}\) and \((\alpha ,1), (1,\alpha ) \notin \mathcal {Z}\).
-
(Z3)
Let \(ab, bc \in S^\star \). Then \((a,bc) \in \mathcal {Z}\) if and only if \((ab,c) \in \mathcal {Z}\).
-
(Z4)
If \((a,b), (b,c) \in \mathcal {N}\), \(a \geqslant \varepsilon _l\), and \(c < \varepsilon _r\), then \((b,c) \in \mathcal {Z}\).
If \((a,b), (b,c) \in \mathcal {N}\), \(a < \varepsilon _l\), and \(c \geqslant \varepsilon _r\), then \((a,b) \in \mathcal {Z}\).
-
(Z5)
Let \(bc \in S^\star \) or \(b=1\). If \((a,b) \in \mathcal {N}\), \(c < \varepsilon _r\), then \((a, bc) \in \mathcal {Z}\).
Let \(ab \in S^\star \) or \(b=1\). If \((b,c) \in \mathcal {N}\), \(a < \varepsilon _l\), then \((ab, c) \in \mathcal {Z}\).
-
(Z6)
Let \(bc \in S^\star \) or \(b=1\). If \((a,b) \in \mathcal {N}\), \(c \geqslant \varepsilon _r\), and \((a,bc) \in \mathcal {Z}\), then \((a,b) \in \mathcal {Z}\).
Let \(ab \in S^\star \) or \(b=1\). If \((b,c) \in \mathcal {N}\), \(a \geqslant \varepsilon _l\), and \((ab,c) \in \mathcal {Z}\), then \((b,c) \in \mathcal {Z}\).
Proof
(Z1), (Z2), and (Z3) are immediate from the definition of \(\mathcal {Z}_{\mathbin {\theta }}\).
In case of (Z4)–(Z6), we show the first halves only; the second ones are seen similarly.
Ad (Z4): Assume that \((a,b), (b,c) \in \mathcal {N}\), \(a \geqslant \varepsilon _l\), and \(c < \varepsilon _r\). Then \(a b \leqslant \alpha \) and hence \(a b c \mathbin {\theta }0\) by the definition of \(\varepsilon _r\). Moreover, \(b c \leqslant \alpha \) and hence \(b c \mathbin {\theta }0\) or \(b c \mathbin {\theta }\alpha \). In the latter case, it would follow \(a b c \mathbin {\theta }\alpha \) by the definition of \(\varepsilon _l\), hence \(b c \mathbin {\theta }0\).
Ad (Z5): Assume that \((a,b) \in \mathcal {N}\) and \(c < \varepsilon _r\). Then \(a b c \mathbin {\theta }0\) and hence \((a, bc) \in \mathcal {Z}\).
Ad (Z6): Assume that \((a,b) \in \mathcal {N}\) and \(c \geqslant \varepsilon _r\). Then \((a,b) \in \mathcal {N}{\setminus } \mathcal {Z}\) implies \(a b c \mathbin {\theta }\alpha \) and hence \((a,bc) \in \mathcal {N}{\setminus } \mathcal {Z}\). \(\square \)
We now turn to the converse direction: given a pair \(\varepsilon _l, \varepsilon _r\) of border elements, we describe the possible subsets \(\mathcal {Z}_{\mathbin {\theta }}\) of \(\mathcal {N}\) that determine a coextension congruence \(\mathbin {\theta }\).
Lemma 3.14
Let \(\varepsilon _l, \varepsilon _r\in S^\star \cup \{\alpha \}\) and let \(\mathcal {Z}\subseteq \mathcal {N}\) be such that, for any \(a, b, c, d \in \bar{S}\), the conditions (Z1)–(Z6) hold. Then there is a uniquely determined coextension congruence \(\mathbin {\theta }\) with the borders \(\varepsilon _l, \varepsilon _r\) such that \(\mathcal {Z}= \mathcal {Z}_{\mathbin {\theta }}\).
Proof
Before beginning the proof, let us show two additional properties of \(\mathcal {Z}\).
(Z7) For any \((a,b) \in \mathcal {Z}\) such that \(a, b \ne 0\), we have \(a < \varepsilon _l\) and \(b < \varepsilon _r\).
Indeed, we may apply the second half of (Z6) to \(b = 1\) and \(c=\alpha \), to conclude that \((a,\alpha ) \notin \mathcal {Z}\) if \(a \geqslant \varepsilon _l\). In view of (Z1), this shows one half of (Z7). The other one is seen similarly.
(Z8) \((a,\alpha ) \in \mathcal {Z}\) for any \(a < \varepsilon _l\), and \((a,\alpha ) \in \mathcal {N}{\setminus } \mathcal {Z}\) for any \(a \geqslant \varepsilon _l\). Similarly, \((\alpha ,b) \in \mathcal {Z}\) for any \(b < \varepsilon _r\), and \((\alpha ,b) \in \mathcal {N}{\setminus } \mathcal {Z}\) for any \(b \geqslant \varepsilon _r\).
Indeed, applying the second half of (Z5) to \(b=1\) and \(c=\alpha \), we see that \((a,\alpha ) \in \mathcal {Z}\) if \(a < \varepsilon _l\). Furthermore, if \(a \geqslant \varepsilon _l\), then \((a,\alpha ) \notin \mathcal {Z}\) by (Z7). This shows one half of (Z8); the other one is seen similarly.
For \(a, b \in \bar{S}\), let us define
$$\begin{aligned} a \cdot b \;=\; {\left\{ \begin{array}{ll} ab &{}\quad \text {if}\ a, b, ab \in S^\star , \\ 0 &{}\quad \text {if}\ (a,b) \in \mathcal {Z}, \\ \alpha &{}\quad \text {if}\ (a,b) \in \mathcal {N}{\setminus } \mathcal {Z}. \end{array}\right. } \end{aligned}$$
We shall show that the operation \(\cdot \) makes \(\bar{S}\) into a tomonoid. Indeed, 1 is an identity for \(\cdot \), and it is clear that \(\cdot \) is in both arguments order-preserving. It remains to prove the associativity.
Let \(a, b, c \in \bar{S}\); we have to show that \((a \cdot b) \cdot c = a \cdot (b \cdot c)\). We distinguish several cases.
Case 1. Assume that \(a b c \in S^\star \). Then \(a b, b c \in S^\star \) and hence \((a \cdot b) \cdot c = a b c = a \cdot (b \cdot c)\).
Case 2. Assume that \(a b, bc \in S^\star \) but \(a b c \notin S^\star \). Then \((ab, c), (a, bc) \in \mathcal {N}\) and, by (Z3), \((a \cdot b) \cdot c = ab \cdot c = a \cdot bc = a \cdot (b \cdot c)\).
Case 3. Assume that \((a,b) \in \mathcal {Z}\). Then \((a \cdot b) \cdot c = 0 \cdot c = 0\) by (Z2). Note, furthermore, that \(b \cdot c \leqslant b\). Hence \((a, b \cdot c) \in \mathcal {Z}\) by (Z1) and we conclude \(a \cdot (b \cdot c) = 0\).
Case 4. Assume that \((b,c) \in \mathcal {Z}\). We proceed analogously to Case 3.
Case 5. Assume that \((a,b), (b,c) \in \mathcal {N}{\setminus } \mathcal {Z}\). If \(a < \varepsilon _l\) and \(c \geqslant \varepsilon _r\), or \(a \geqslant \varepsilon _l\) and \(c < \varepsilon _r\), we get a contradiction by (Z4). Hence either \(a < \varepsilon _l\) and \(c < \varepsilon _r\), or \(a \geqslant \varepsilon _l\) and \(c \geqslant \varepsilon _r\). From (Z8), we derive \((a \cdot b) \cdot c = \alpha \cdot c = a \cdot \alpha = a \cdot (b \cdot c)\).
Case 6. Assume that \((a,b) \in \mathcal {N}{\setminus } \mathcal {Z}\) and \(b c \in S^\star \). If \(c < \varepsilon _r\), we have \((a \cdot b) \cdot c = \alpha \cdot c = 0\) and, by (Z5), \(a \cdot (b \cdot c) = a \cdot bc = 0\) as well. If \(c \geqslant \varepsilon _r\), we have \((a \cdot b) \cdot c = \alpha \cdot c = \alpha \) and, by (Z6), \(a \cdot (b \cdot c) = a \cdot bc = \alpha \) as well.
Case 7. Assume that \((b,c) \in \mathcal {N}{\setminus } \mathcal {Z}\) and \(a b \in S^\star \). We proceed analogously to Case 6.
We conclude that \((\bar{S}; \leqslant , \cdot , 1)\) is a tomonoid. Furthermore, \(\bar{S}/\alpha \) is isomorphic to S, that is, \(\bar{S}\) is a one-element coextension of S. Let \(\mathbin {\theta }\) be the coextension congruence on \(\mathcal R(S)\) such that \(\bar{S}\) is isomorphic to \(\mathcal R(S)/\mathbin {\theta }\). Then, for any \((a, b) \in \bar{S}^2\), we have
$$\begin{aligned} (a,b) \in \mathcal {Z}_{\mathbin {\theta }}\ \text {iff} \ a b \mathbin {\theta }0\ \text {iff} \ {a/\mathbin {\theta }} \cdot {b/\mathbin {\theta }} = {0/\mathbin {\theta }}\ \text {iff} \ a \cdot b = 0\ \text {in}\ \bar{S}\ \text {iff} \ (a,b) \in \mathcal {Z}. \end{aligned}$$
Hence \(\mathcal {Z}= \mathcal {Z}_{\mathbin {\theta }}\). By (Z8), the border elements of \(\mathbin {\theta }\) are \(\varepsilon _l,\varepsilon _r\).
Finally, let \(\mathbin {\theta }'\) be another coextension congruence on \(\mathcal R(S)\) such that \(\mathcal {Z}= \mathcal {Z}_{\mathbin {\theta }'}\). But then \(\mathcal {Z}_{\mathbin {\theta }} = \mathcal {Z}_{\mathbin {\theta }'}\) and hence, by Lemma 3.8, \(\mathbin {\theta }\) and \(\mathbin {\theta }'\) coincide. The uniqueness claim follows. \(\square \)
By a one-element\((\varepsilon _l, \varepsilon _r)\)-coextension of S, we mean the quotient of \(\mathcal R(S)\) induced by a coextension congruence with borders \(\varepsilon _l, \varepsilon _r\in S^\star \cup \{\alpha \}\). We may summarise our results as follows.
Theorem 3.15
Let \(\varepsilon _l, \varepsilon _r\in S^\star \cup \{\alpha \}\) and let \(\mathcal {Z}\subseteq \mathcal {N}\) be such that, for any \(a, b, c, d \in S^\star \), the conditions (Z1)–(Z6) hold. For \(a, b \in \bar{S}\), let
$$\begin{aligned} a \cdot b \;=\; {\left\{ \begin{array}{ll} ab &{}\quad \mathrm{if}\ a, b, ab \in S^\star , \\ 0 &{}\quad \mathrm{if}\ (a,b) \in \mathcal {Z}, \\ \alpha &{}\quad \mathrm{if}\ (a,b) \in \mathcal {N}{\setminus } \mathcal {Z}. \end{array}\right. } \end{aligned}$$
(7)
Then \((\bar{S}; \leqslant , \cdot , 1)\) is a one-element \((\varepsilon _l, \varepsilon _r)\)-coextension of S.
Up to isomorphism, any one-element \((\varepsilon _l, \varepsilon _r)\)-coextension arises in this way from a unique set \(\mathcal {Z}\subseteq \bar{S}^2\).
Proof
The first part holds by Lemma 3.14; the second part holds by Lemma 3.13. \(\square \)
In short, for a pair \(\varepsilon _l, \varepsilon _r\in S^\star \cup \{\alpha \}\), there is a one-to-one correspondence between the one-element \((\varepsilon _l, \varepsilon _r)\)-coextensions of S and the sets \(\mathcal {Z}\subseteq \mathcal {N}\) subject to the conditions (Z1)–(Z6).
We can formulate Theorem 3.15 as follows in an algorithmic fashion. At this point, the present work becomes comparable with the approach chosen in [10].
In what follows, we endow \(\mathcal {N}\) with the componentwise partial order \(\trianglelefteq \), that is, for \((a,b), (c,d) \in \mathcal {N}\), we write \((a,b) \trianglelefteq (c,d)\) if \(a \leqslant b\) and \(c \leqslant d\).
Theorem 3.16
Let \(\varepsilon _l, \varepsilon _r\in S^\star \cup \{\alpha \}\) such that that either \(\varepsilon _l= \varepsilon _r= \alpha \) or \(\varepsilon _l\), \(\varepsilon _r\) are idempotent elements of \(S^\star \). Let \(\mathbin {\sim }\) be the smallest equivalence relation on \(\mathcal {N}\) such that following holds:
-
(A1)
For any \(a,b \in \bar{S}\), we have \((a,0) \mathbin {\sim }(0,b) \mathbin {\sim }(1,0)\). Moreover, for any \(a \in \bar{S}\) such that \(a < \varepsilon _l\), we have \((a, \alpha ) \mathbin {\sim }(1,0)\). Similarly, for any \(a \in \bar{S}\) such that \(a < \varepsilon _r\), we have \((\alpha , a) \mathbin {\sim }(1,0)\).
-
(A2)
We have \((1,\alpha ) \mathbin {\sim }(\alpha ,1)\) and for any \((a,b) \in \mathcal {N}\) such that \(a, b \ne 0\) and \(a \geqslant \varepsilon _l\) or \(b \geqslant \varepsilon _r\), we have \((a,b) \mathbin {\sim }(1,\alpha )\).
-
(A3)
Let \(a, b, c, ab, bc \in S^\star \) and \((a,bc), (ab,c) \in \mathcal {N}\). Then \((a,bc) \mathbin {\sim }(ab,c)\).
-
(A4)
Let \((a,b), (b,c) \in \mathcal {N}\) such that \(a \geqslant \varepsilon _l\) and \(c < \varepsilon _r\). Then \((b,c) \mathbin {\sim }(1,0)\).
Similarly, let \((a,b), (b,c) \in \mathcal {N}\) such that \(a < \varepsilon _l\) and \(c \geqslant \varepsilon _r\). Then \((a,b) \mathbin {\sim }(1,0)\).
-
(A5)
Let \(a, b, c, bc \in S^\star \), \((a,b) \in \mathcal {N}\), and \(c < \varepsilon _r\). Then \((a, bc) \mathbin {\sim }(1,0)\).
Similarly, let \(a, b, c, ab \in S^\star \), \((b,c) \in \mathcal {N}\), and \(a < \varepsilon _l\). Then \((ab, c) \mathbin {\sim }(1,0)\).
-
(A6)
Let \(a, b, c, bc \in S^\star \), \((a,b) \in \mathcal {N}\), and \(c \geqslant \varepsilon _r\). Then \((a,b) \mathbin {\sim }(a,bc)\).
Similarly, let \(a, b, c, ab \in S^\star \), \((b,c) \in \mathcal {N}\), and \(a \geqslant \varepsilon _l\). Then \((b,c) \mathbin {\sim }(ab,c)\).
If \((1,0) \mathbin {\sim }(1,\alpha )\), a one-element \((\varepsilon _l, \varepsilon _r)\)-coextension of S does not exist. Otherwise, let \(\mathcal {Z}\subseteq \mathcal {N}\) be a union of \(\mathbin {\sim }\)-classes such that \((1,0) \in \mathcal {Z}\), \(\;(1,\alpha ) \notin \mathcal {Z}\), and \(\mathcal {Z}\) is a downset. Then \((\bar{S};\cdot , \leqslant , 1)\), where \(\cdot \) is defined by (7), is a one-element \((\varepsilon _l,\varepsilon _r)\)-coextension of S.
Every one-element coextension of S arises in this way.
Proof
Assume \((1,0) \mathbin {\not \sim }(1,\alpha )\) and let \(\mathcal {Z}\subseteq \mathcal {N}\) arise in the indicated fashion. Then \(\mathcal {Z}\) fulfils the properties (Z1)–(Z6) of Lemma 3.13; this is not difficult to check and we omit the details. Consequently, \(\mathcal {Z}\) gives rise to a one-element coextension as indicated in Theorem 3.15.
Conversely, let \(\mathcal {Z}= \mathcal {Z}_{\mathbin {\theta }}\), where \(\mathbin {\theta }\) is a coextension congruence. Then \(\mathcal {Z}\) is a subset of \(\mathcal {N}\) that fulfils, by Lemma 3.13 and the proof of Lemma 3.14, the properties (Z1)–(Z8). By (Z2), \((1,0) \in \mathcal {Z}\) but \((1,\alpha ) \notin \mathcal {Z}\). It is now easily checked that, for any \(a, b, c, d \in \bar{S}\) such that \((a,b) \mathbin {\sim }(c,d)\), either \((a,b), (c,d) \in \mathcal {Z}\) or \((a,b), (c,d) \notin \mathcal {Z}\). Hence \(\mathcal {Z}\) is a union of \(\mathbin {\sim }\)-classes. Since \(\mathcal {Z}\) contains (1, 0) but not \((1,\alpha )\), we have \((1,0) \mathbin {\not \sim }(1,\alpha )\). Finally, we have by (Z1) that \((c,d) \in \mathcal {Z}\), \(a \leqslant c\), \(b \leqslant d\) imply \((a,b) \in \mathcal {Z}\), that is, \(\mathcal {Z}\) is a downset. We conclude that any one-element coextension arises in the indicated way. \(\square \)
Example 3.17
Let us consider again the tomonoid S indicated in Example 3.7. Its multiplication table is given in Fig. 2(left). It is understood that the rows correspond to the first argument and the columns to the second one.
We determine the one-element coextensions of S for the case \(\varepsilon _l= \varepsilon _r= 1\). The extended base set is \(\bar{S}= \{ 0, \alpha , x, y, z, 1 \}\). In accordance with Theorem 3.16, we have to calculate the equivalence relation \(\mathbin {\sim }\) on \(\mathcal {N}\subseteq \bar{S}^2\).
The \(\mathbin {\sim }\)-classes are indicated in Fig. 2(right). Here as well as in the subsequent figures, \(\mathcal {N}\) is delimited against the remaining elements of \(\bar{S}^2\) by a bold line. Moreover, we denote the \(\mathbin {\sim }\)-class containing (1, 0) by 0, the \(\mathbin {\sim }\)-class containing \((1,\alpha )\) by \(\alpha \), and the remaining ones by Latin capital letters. It is indicated which element of \(\mathcal {N}\) belongs to which \(\mathbin {\sim }\)-class. We see that \(\mathcal {N}\) contains, apart from the classes 0 and \(\alpha \), the four further classes A, B, C, and D.
By Theorem 3.16, a one-element coextension of S corresponds to a union \(\mathcal {Z}\) of \(\mathbin {\sim }\)-classes, including the class 0 but excluding \(\alpha \) and such that \((c,d) \in \mathcal {Z}\), \(a \leqslant c\), and \(b \leqslant d\) imply \((a,b) \in \mathcal {Z}\). For instance, let \(\mathcal {Z}\) be the union of the classes 0, A, B, and D. Then we arrive at the coextension \((\bar{S}; \leqslant , \cdot , 1)\) that we have indicated in Fig. 1. The multiplication on \(\bar{S}\) is in this case defined as follows:
$$\begin{aligned} a \cdot b \;=\; {\left\{ \begin{array}{ll} b &{}\quad \text {if}\ a = 1, \\ a &{}\quad \text {if}\ b = 1, \\ x &{}\quad \text {if}\ a = b = z, \\ \alpha &{}\quad \text {if}\ a = z\ \text {and}\ b = y, \\ 0 &{}\quad \text {otherwise,} \end{array}\right. } \end{aligned}$$
where \(a, b \in \bar{S}\).