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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# Algebras of right ample semigroups

Junying Guo
/ Xiaojiang Guo
• Corresponding author
• College of Mathematics and Information Science, Jiangxi Normal University, Nanchang, Jiangxi 330022, China
• Email
• Other articles by this author:
• De Gruyter OnlineGoogle Scholar
Published Online: 2018-08-03 | DOI: https://doi.org/10.1515/math-2018-0075

## Abstract

Strict RA semigroups are common generalizations of ample semigroups and inverse semigroups. The aim of this paper is to study algebras of strict RA semigroups. It is proved that any algebra of strict RA semigroups with finite idempotents has a generalized matrix representation whose degree is equal to the number of non-zero regular 𝓓-classes. In particular, it is proved that any algebra of finite right ample semigroups has a generalized upper triangular matrix representation whose degree is equal to the number of non-zero regular 𝓓-classes. As its application, we determine when an algebra of strict RA semigroups (right ample monoids) is semiprimitive. Moreover, we prove that an algebra of strict RA semigroups (right ample monoids) is left self-injective iff it is right self-injective, iff it is Frobenius, and iff the semigroup is a finite inverse semigroup.

MSC 2010: 20M30; 16G60

## 1 Introduction

The mathematical structures which encode information about partial symmetries are certain generalizations of groups, called inverse semigroups. Abstractly, inverse semigroups are regular semigroups each of whose elements has exactly one inverse; equivalently, a regular semigroup is inverse if and only if its idempotents commute. Like groups, inverse semigroups first arose in questions concerned with the solutions of equations, but this time in Lie’s attempt to find the analogue of Galois theory for differential equations. The symmetries of such equations form what are now termed Lie pseudogroups, and inverse semigroups are, several times removed, the corresponding abstract structures. In addition to their early appearance in differential geometry, inverse semigroups have found a number of other applications in recent years including: C*-algebras; tilings, quasicrystals and solid-state physics; combinatorial group theory; model theory; and linear logic. Inverse semigroups have been widely investigated. For inverse semigroups, the readers are referred to the monographs of Petrich [1] and Lawson [2]. Because of the important role of inverse semigroups in the theory of semigroups, there are attempts to generalize inverse semigroups. (Left; Right) ample semigroups originally introduced by Fountain in [3] are generalizations of inverse semigroups in the range of (left pp semigroups; right pp semigroups) abundant semigroups.

Inverse semigroup algebras are a class of semigroup algebras which is widely investigated. For example, Crabb and Munn considered the semiprimitivity of combinatorial inverse semigroup algebras (see [4]); algebras of free inverse semigroups (see [5, 6]); nil-ideals of inverse semigroup algebras (see [7]). More results on inverse semigroup algebras are collected in a survey of Munn [8]. Recently, Steinberg [9, 10] investigated representations of finite inverse semigroups. Inverse semigroups are ample semigroups and any ample semigroup can be viewed as a subsemigroup of some inverse semigroup. Ample semigroups include cancellative monoids and path semigroups of quivers. In [11, 12], Okniński studied algebras of cancellative semigroups. Guo and Chen [13] proved that any algebra of finite ample semigroups has a generalized upper triangular matrix representation. Guo and Shum [14] established the construction of algebras of ample semigroups each of whose 𝓙*-classes contains a finite number of idempotents. For the related results on semigroup algebras, the reader can be referred to the books of Okniński [15], and Jesper and Okniński [16].

Frobenius algebras are algebras with non-degenerate bilinear mappings. They are closely related to the representation theory of groups which appear in many branches of algebras, algebraic geometry and combinatorics. Frobenius algebras and their generalizations, such as quasi-Frobenius algebras and right (left) self-injective algebras, play an important role and become a central topic in algebra. Wenger proved an important result (Wenger Theorem): an algebra of an inverse semigroup is left self-injective iff it is right self-injective; iff it is quasi-Frobenius; iff the inverse semigroup is finite (see [17]). In [18], Guo and Shum determined when an ample semigroup algebra is Frobenius and generalized Wenger Theorem to ample semigroup algebras.

Right (Left) ample semigroups are known as generalizations of ample semigroups and include left (right) cancellative monoids as proper subclass. So, it is natural to probe algebras of right ample semigroups. This is the aim of paper. Indeed, any ample semigroup is both a right ample semigroup and a left ample semigroup. The symmetry property is a “strict” one in the theory of semigroups. It is interesting whether the semigroup algebras have the “similar” properties if we destroy the “symmetry”. To be precise, for what “weak symmetry” assumption is the Wenger Theorem valid? By weakening the condition: S is a left ample semigroup, to the condition: Each 𝓛̂-class of S contains an idempotent, we introduce strict RA semigroups (dually, strict LA semigroups), which include ample semigroups and inverse semigroups as its proper subclasses. The following picture illustrates the relationship between strict RA (LA) semigroups and other classes of semigroups:

where 𝓔𝓒 is the class of EC-semigroups (that is, semigroups whose idempotents commute), for EC-semigroups, see [19, 20]; 𝓢𝓡𝓐 is the class of strict RA semigroups; 𝓢𝓛𝓐 is the class of strict LA semigroups; 𝓐 is the class of ample semigroups; 𝓘 is the class of inverse semigroups, and the symbolic “|” means that the upper class of semigroups includes properly the lower one.

Our ideas in this paper are somewhat similar as in [13, 18] and inspired by the references [21, 22]. In Section 2, we obtain some properties of strict RA semigroups. Section 3 is devoted to the representation theory of generalized matrices for algebras of strict RA semigroups. It is verified that any algebra of a strict RA semigroup with finite idempotents has a generalized matrix representation whose degree is equal to the number of non-zero regular 𝓓-classes (Theorem 3.2), extending the main result in [13]. In particular, we prove that any algebra of a finite right ample monoid has a generalized upper triangular matrix representation (Corollary 3.5). As applications of these representation theorems, we determine when an algebra of strict RA finite semigroups is semiprimitive (Theorem 3.10). This extends the related result of Guo and Chen in [13]. In particular, a sufficient and necessary condition for an algebra of right ample finite monoids to be semiprimitive is obtained (Corollary 3.11). In Section 4, we shall determine when an algebra of right ample semigroups is left self-injective. It is shown that an algebra of a strict RA semigroup is left self-injective, iff it is right self-injective, iff it is Frobenius and iff the Strict RA semigroup is a finite inverse semigroup (Theorem 4.11). This result extends the Wenger Theorem to the case for strict RA semigroups. Especially, it is verified that if an algebra of a right ample monoid is left (resp. right) self-injective, then the monoid is finite (Corollary 4.18). So, we give a positive answer to [15, Problem 6, p.328] for the case that S is a strict RA (LA) semigroup (especially, a right ample monoid). It is interesting to find that

• the “distance” between strict RA (LA) semigroups with finite inverse semigroups is the left (right) self-injectivity;

• for an algebra of strict RA (LA) semigroups, left (right) self-injectivity = quasi-Frobenoius = Frobenius.

In this paper we shall use the notions and notations of the monographs [23] and [15]. For right ample semigroups, the reader can be referred to [3].

## 2 Strict RA semigroups

Let S be a semigroup; we denote by E(S) the set of idempotents of S, by S1 the semigroup obtained from S by adjoining an identity if S does not have one.

To begin with, we recall some known results on Green’s relations. For any a, bS, define

$aLb⟺S1a=S1bi.e.a=xb,b=ya for some x,y∈S1;aRb⟺aS1=bS1i.e.a=bu,b=av for some u,v∈S1;H=L∩R;D=L∘R=R∘L.$

In general, 𝓛 is a right congruence and 𝓡 is a left congruence. a is regular in S if there exists xS such that axa = a. Equivalently, a is regular if and only if a𝓓e for some eE(S). And, S is called regular if each element of S is regular in S. We use Dx to denote the 𝓓-class of S containing x, and call a 𝓓-class containing a regular element a regular 𝓓-class. It is well known that any element of a regular 𝓓-class is regular and that the 𝓓-class containing the zero element 0 is just the set {0}.

As generalizations of Green’s 𝓛- and 𝓡-relations, we have 𝓛*- and 𝓡*-relations defined on S by

$aL∗bif(ax=ay⇔bx=byfor allx,y∈S1),aR∗bif(xa=ya⇔xb=ybfor allx,y∈S1).$

It is well known that 𝓛* is a right congruence and 𝓡* is a left congruence. In general, 𝓛 ⊆ 𝓛* and 𝓡 ⊆ 𝓡*. And, if a, b are regular, then a𝓛 (𝓡)b if and only if a𝓛* (𝓡*)b.

#### Definition 2.1

A semigroup S is right ample if

1. its idempotents commute;

2. every element a is 𝓛*-related a (unique) idempotent a*;

3. for any aS and eE(S), ea = a(ea)*.

Left ample semigroups are defined by duality. Moreover, a semigroup is ample if it is both left ample and right ample. Obviously, inverse semigroups are (left; right) ample semigroups.

As generalizations of 𝓛* and 𝓡*, we have 𝓛̂ and 𝓡̂ defined on S by

$aL^bif(a=ae⇔b=befor alle∈E(S1)),aR^bif(a=ea⇔b=ebfor alle∈E(S1)).$

In general, 𝓛̂ is not a right congruence and 𝓡̂ is not a left congruence. Clearly, 𝓛 ⊆ 𝓛* ⊆ 𝓛̂ and 𝓡 ⊆ 𝓡* ⊆ 𝓡̂.

#### Lemma 2.2

Let aS and eE(S). If a𝓡̂e, then ea = a. Moreover, if g, hE(S) and g 𝓡̂h, then g𝓡h.

#### Proof

Since e2 = e, the result follows from a𝓡̂e. The rest is trivial. □

By Lemma 2.2, it is evident that for regular elements a, b, a𝓛 (𝓡)b if and only if a𝓛̂ (𝓡̂)b. Especially, on a regular semigroup 𝓛 = 𝓛̂ and 𝓡 = 𝓡̂.

#### Definition 2.3

A semigroup S is strict RA if S is a right ample semigroup in which for any aS, there exists an idempotent e such that a𝓡̂e. Strict LA semigroups are defined by duality.

For a strict RA semigroup, each 𝓡̂-class contains exactly one idempotent; for, if e,f are idempotents and e𝓡̂f, then by Lemma 2.2, e𝓡f, so that e = f since idempotents of a strict RA semigroup commute. We shall denote by a# the unique idempotent in the 𝓡̂-class containing a. Obviously, ample semigroups are strict LA (strict RA) semigroups. So, strict RA semigroups (strict LA semigroups) are common generalizations of ample semigroups and inverse semigroups. Indeed, strict RA semigroups include ample semigroups as its proper subclass; for, it is easy to see that left cancellative monoids are strict RA semigroups but not all of left cancellative monoids are ample semigroups.

#### Example 2.4

Assume that Q = (V, E) is a quiver with vertices V = {1, 2,⋯,r}. Denote by (i1|i2|⋯|in) the path: $\begin{array}{}\stackrel{{i}_{1}}{\circ }\to \stackrel{{i}_{2}}{\circ }\to \cdots \to \stackrel{{i}_{n}}{\circ }.\end{array}$ In particular, we call the path of Q without vertices the empty path of Q, denoted by 0. For any iV, we appoint to have an empty path ei. Let P(Q) be the set of all paths of Q. On P(Q), define a multiplication by: for (i1|⋯|im),(j1|⋯|jn) ∈ P(Q),

$(i1|⋯|im)∘(j1|⋯|jn)=(i1|⋯|im|j2|⋯|jn) if im=j1;0 otherwise.$

Evidently, the path algebra RQ of the quiver Q over R is just the contracted semigroup R0[P(Q)]. By a routine computation, (P(Q),∘) is a semigroup in which

• 0 is the zero element of P(Q) and e1, e2, ⋯, er, 0 are all idempotents of P(Q);

• ei1 𝓡*(i1|⋯|im)𝓛*eim;

• eiej = 0 whenever ij.

It is easy to see that P(Q) is a strict RA semigroup having only finite idempotents.

#### Example 2.5

Let 𝓒 be a small category and 0 a symbol. On the set

$S(C):=⊔A,B∈Obj(C)Hom(A,B)⊔{0},$

define: for α, β ∈ ⊔A,BObj(𝓒)Hom(A, B),

$α∗β=α∘βifαandβcan be composed,0 otherwise$

and 0 * α = α * 0 = 0 * 0 = 0. It is a routine check that (S(𝓒),*) is a semigroup with zero 0, called the category semigroup of 𝓒. An arrow αHom(A, B) is an isomorphism if there exists βHom(B, A) such that αβ = 1A and βα = 1B. We call 𝓒 a groupoid if any arrow of 𝓒 is an isomorphism (for groupoids, see [22] and their references); and left (resp. right) cancellative if for any arrows α, β, γ of 𝓒, β = γ whenever αβ = αγ (resp. γα = γα). And, 𝓒 is cancellative if 𝓒 is both left cancellative and right cancellative. It is not difficult to see that any groupoid is a cancellative category. Leech categories and Clifford categories are left cancellative categories (for these kinds of categories, see [24]).

1. Assume that 𝓒 is a left cancellative category. By computation, in the semigroup S(𝓒), for any αHom(A, B),

• 1A𝓛*α𝓡̂1B;

• E(S(𝓒)) = {0} ⊔ {1C : CObj(𝓒)}.

It is not difficult to check that S(𝓒) is a strict RA semigroup. By dual arguments, if 𝓒 is cancellative, then S(𝓒) is an ample semigroup.

2. When 𝓒 is a groupoid. In this case, there exists βHom(B, A) such that αβ = 1A. It follows that αβα = α. This means that α is regular and hence S(𝓒) is a regular semigroup. Again by the foregoing arguments, S(𝓒) is indeed an inverse semigroup.

#### Lemma 2.6

Let S be a right ample semigroup and E(S) ⋅ S = S. If |E(S)| < ∞ then S is strict RA.

#### Proof

Let aS and denote by ea the minimal idempotent under the partial order ω on E(S) (that is, eωf if and only if e = ef = fe) such that eaa = a. Obviously, for any fE(S), fea = ea can imply that fa = a; conversely, if fa = a then feaa = a and fea = ea since feaωea. Thus a𝓡̂ea, so S is strict RA. □

#### Corollary 2.7

Let S be a right ample monoid. If S is finite, then S is strict RA.

For a right ample semigroup S, define: for a, bS,

$a≤rb if and only if a=be for some e∈E(S).$(1)

In (1), the idempotent e can be chosen as a*; for, by a = be, we have a = ae, so that a* = a*e, thereby a = aa* = bea* = ba*. So, arb if and only if a = ba*. It is not difficult to check that ≤r is a partial order on S. Moreover, we have

#### Lemma 2.8

If S is a right ample semigroup, then with respect tor, S is an ordered semigroup.

#### Proof

The proof is a routine check and we omit the detail. □

For an element a of a right ample semigroup S, we denote

$O(a)={x∈S:x≤ra}.$

If xr a then x = ax*, so that x = aa*x* = xa*, hence x* = x*a*, therefore x*ωa*, that is, x*ω(a*) where ω(a*) = {eE(S) : eωa*}. On the other hand, for f, hω(a*), if af = ah then f = a*f = a*h = h. Consequently, O(a) = {af : fω(a*)} and |O(a)| = |ω(a*)|.

#### Proposition 2.9

Let S be a strict RA semigroup and a, b, xS. If xr ab, then there exist uniquely u, vS such that

• (O1)

ur a, vr b;

• (O2)

u# = x#, u* = v# and x* = v*;

• (O3)

x = uv.

#### Proof

Denote v = (x#a)*bx* and u = x#av#. We have uv = x#av# ⋅ (x#a)*bx* = x#abx* = x. Because S is a right ample semigroup, we have x#a#r a# and a*v#r a*, so that u = x#a#aa*v#r a#aa* = a; and (x#a)*b#r b#, b*x*r b*, so that v = (x#a)*b#bb*x*r b#bb* = b.

Note that v = (x#a)*bx* = vx*, we observe v* = v*x*. On the other hand, since x = uv = uvv* = xv*, we get x* = x*v*. Thus x* = v* since E(S) commutes.

Indeed, by u = x#av#, we have u = uv#, so that u* = u*v#. It follows that u*r v#. Consider that 𝓛* is a right congruence, we have u*𝓛*u = x#av#𝓛*(x#a)*v# and u* = (x#a)*v# since each 𝓛*-class of S contains exactly one idempotent. So,

$v=v#v=v#(x#a)∗bx∗=v#(x#a)∗v=u∗v,$

further by definition, v# = u*v#. It follows that v#r u*. Therefore u* = v#.

By the definition of u, u = x#u and x#u# = u#. But u#u = u, so u#x = x. It follows that u#x# = x#. Therefore u# = x# since E(S) commutes.

Finally, we let m, n be elements of S satisfying the properties of u and those of v in (O1), (O2) and (O3), respectively. Then m* = n# and v* = x* = n*. By the arguments before Lemma 2.8, n = bn* = bv* = v and m = am* = an#. By the first equality, n# = v#. Thus u = x#av# = x#an# = x#m = m#m = m. This proves the uniqueness of u and v. □

Assume now that T is a strict RA semigroup with zero element 0. Let E be the set of nonzero idempotents of T. For any aT\{0}, (a)ij will denote the E × E matrix with entry a in the (i, j) position and zeros elsewhere. Also, we still use 0 to denote the E × E matrix each of whose entries is 0. Set

$M(T)={(a)ij:ia=a=aj}∪{0}$

and define an operation on M(T) by: A, BM(T),

$AB=0 ifA=0orB=0;0 ifA=(a)ij,B=(b)klandj≠k;(ab)il ifA=(a)ij,B=(b)klandj=k.$

It is easy to check that with respect to the above operation, M(T) is a semigroup with zero element 0. Moreover, we can observe

#### Lemma 2.10

The zero 0 and the elements (i)ii (iE) are all idempotents of M(T), and orthogonal each other.

Write X(T) = {0} ∪ {(a)ij : a# = i, a* = j}. We denote by PRA(T) the subsemigroup of M(T) generated by X(T). It is not difficult to see that

$RA(T)={(a)ij:there exist x1,x2,⋯,xn∈T such that a=x1x2⋯xn x1#=i,xn∗=j,xk∗=xk+1# for k=1,2,⋯,n−1}.$

In what follows, for any aT, we use a to denote (a)ij with a# = i, a* = j.

#### Proposition 2.11

In the semigroup PRA(T), for any (a)ijPRA(T), we have

1. a* = j.

2. (a)ij is regular if and only if a is regular in T. Moreover, if a is regular, then (a)ijX(T).

3. (a)ij𝓛*(j)jj.

#### Proof

1. By the definition of PRA(T), there exist x1, x2, ⋯, xnT such that a = x1x2xn, $\begin{array}{}{x}_{k}^{\ast }={x}_{k+1}^{\mathrm{#}}\end{array}$ where k = 1, 2, ⋯, n – 1, $\begin{array}{}{x}_{1}^{\mathrm{#}}=i\end{array}$ and $\begin{array}{}{x}_{n}^{\ast }=j.\end{array}$ Since 𝓛* is a left congruence, we have

$a=x1x2⋯xnL∗x1∗x2⋯xn=x2#x2⋯xnL∗⋯L∗xn−1∗xn=xnL∗xn∗=j$

and further since each 𝓛*-class of a left ample semigroup contains exactly one idempotent, we get a* = j.

2. Let (a)ij be regular. Then there exists (x)jiPRA(T) such that

$(a)ij(x)ji(a)ij=(a)ij.$

It follows that axa = a, whence a is regular in T.

Conversely, assume that a is regular in T and let y be an inverse of a in T. Since x1x2xnyx1x2xn = x1x2xn and x1𝓛*i1, we have x2xnyx1x2xn–1xn = x2xn–1xn, and hence

$x3⋯xn−1xn⋅yx1⋅x2⋯xn−1=x2∗x3⋯xn−1xn⋅yx1⋅x2⋯xn =x2∗x3⋯xn−1xn⋅yx1x2⋅x3⋯xn−1xn =x3#x3⋯xn−1xn =x3⋯xn.$

It follows that x3xn is regular. Continuing this process we have that xn is regular. By Lemma 2.2, $\begin{array}{}{x}_{n}^{\mathrm{#}}\mathcal{R}{x}_{n}.\end{array}$ Now, x1x2xnyx1x2xn = x1x2xn can imply that x1x2xnyx1x2xn–1 $\begin{array}{}{x}_{n}^{\mathrm{#}}\end{array}$ = x1x2xn–1 $\begin{array}{}{x}_{n}^{\mathrm{#}}\end{array}$, whence x1xn–1 = x1xn–1xnyx1x2xn–1 and further regular in T. By the foregoing proof, xn–1 is regular. Applying these arguments to x1xn–2, we know that xn–2 is regular, therefore xl is regular for l = 1, ⋯, n. But $\begin{array}{}{x}_{k}{\mathcal{L}}^{\ast }{x}_{k}^{\ast },\end{array}$ so $\begin{array}{}{x}_{k}\mathcal{L}{x}_{k}^{\ast }\end{array}$ and as 𝓛 is a right congruence,

$a=x1x2⋯xnLx1∗x2⋯xn=x2⋯xnL⋯Lxn−1∗xn=xnLxn∗=j$

and a* = j since each 𝓛*-class of a strict RA semigroup contains exactly one idempotent. On the other hand, since 𝓡 is a left congruence, we have

$a=x1x2⋯xnRx1x2⋯xn−1xn#=x1x2⋯xn−1R⋯Rx1x2#=x1Rx1#=i$

and a# = i since each 𝓡*-class of a left ample semigroup contains exactly one idempotent. We have now proved that (a)ijX(T). It follows that

$ay=a#=i and ya=a∗=j.$

Furthermore, (y)jiPRA(T). Consequently, (a)ij is regular since (a)ij(y)ji(a)ij = (aya)ij = (a)ij.

3. Now let (x)jk, (y)jlPRA(T). If (a)ij(x)jk = (a)ij(y)jl, then (ax)ik = (ay)il, hence k = l and ax = ay. By the second equality and applying (A), we have jx = jy, and further (j)jj(x)jk = (jx)ik = (jy)il = (j)jj(y)jl; if (a)ij(x)jk = (a)ij, then (a)ij(x)jk = (a)ij = (a)ij(j)jj, and by the foregoing proof, (j)jj(x)jk = (j)jj(j)jj. We have now proved that for all (x)jk, (y)jlPRA(T)1, we have (j)jj(x)jk = (j)jj(y)jl whenever (a)ij(x)jk = (a)jl. This and the equality (a)ij(j)jj = (a)ij derive that (a)ij𝓛*(j)jj. □

## 3 Generalized matrix representations

Let R1, R2, ⋯, Rn be associative rings (algebras) with identity and let Rij be a left Ri- right Rj-bimodule for i, j = 1,2, ⋯, n and i < j. We call the formal n × n matrix

$a1a12⋯a1na21a2⋯a2n⋮⋮⋱⋮an1an2⋯an$

with aiRi, aijRij for i, j = 1, 2, ⋯, n, a generalized n × n matrix. For 1 ≤ i, j, kn, there is a (Ri, Rk)-bimodule homomorphism ϕijk : RijRj RjkRik such that the square

is commutative. We denote (ab)ϕijk by ab. With respect to matrix addition and matrix multiplication, the set

$R1R12⋯R1nR21R2⋯R2n⋮⋮⋱⋮Rn1Rn2⋯Rn$

of all generalized matrices is an R-algebra, called a generalized matrix algebra of degree n. If Rij = 0 for any 1 ≤ jin, then we call the generalized matrix algebra a generalized upper matrix algebra of degree n.

#### Definition 3.1

A ring (An algebra) has a generalized matrix representation of degree n if there exists a ring (an algebra) isomorphism

$ϕ:A→R1R12⋯R1nR21R2⋯R2n⋮⋮⋱⋮Rn1Rn2⋯Rn.$

For a semigroup S with zero 0, we denote by Dreg(S) the number of nonzero regular 𝓓-classes of S. By [23, Proposition 3.2, p.45], Dreg(S) = |{De : eE(S)\{0}}| = |E(S)/𝓓S|– 1. We arrive at the main result of this section.

#### Theorem 3.2

(Generalized Matrix Representation Theorem). Let S be a strict RA semigroup and R a commutative ring with unity. If |E(S)| < ∞ then R0[S] has a generalized matrix representation of degree Dreg(S).

#### Proof

Define a map φ by

$φ:S→R0[PRA(S)];s↦∑u∈O(s)(u)u#,u∗$

and span this map linearly to R0[S]. By the arguments before Proposition 2.9, |O(s)| = |ω(s*)| ≤ |E(S)| < ∞ and hence φ is well defined.

For s, tS, by Proposition 2.9, there exist uniquely xO(s), yO(t) such that u = xy, x# = u#, x* = y# and y* = u*, for any uO(st). So,

$φ(st)=∑u∈O(st)(u)u#,u∗=∑u=xy,x#=u#,x∗=y#,y∗=u∗,x∈O(s),y∈O(t)(xy)x#,y∗=∑u=xy,x#=u#,x∗=y#,y∗=u∗,x∈O(s),y∈O(t)(x)x#,x∗(y)y#,y∗ by Proposition 2.9=∑x∈O(s)(x)x#,x∗∑y∈O(t)(y)y#,y∗=φ(s)φ(t)$

and φ is a homomorphism.

Let $\begin{array}{}x=\sum _{u\in supp\left(x\right)}{r}_{u}u,y=\sum _{v\in supp\left(y\right)}{r}_{v}v\in {R}_{0}\left[S\right]\mathrm{\setminus }\left\{0\right\}\end{array}$ and φ(x) = φ(y). Denote

Λ(x) = {a* : asupp(x)};

$\begin{array}{}m\left(x,i\right)=\sum _{u\in supp\left(x\right),{u}^{\ast }=i}{r}_{u}u,\end{array}$

and let Max(x) be the set of maximal elements of Λ(x) under ≤r. Let Max*(x) = {a* : aMax(x)} and M1(x) = ∑iMax*(x)m(x, i). Define recursively

$Mk(x)=∑l∈Max∗x−∑j=1k−1φ(Mj(x))mx−∑j=1k−1φ(Mj(x)),l,$

where M0(x) = 0 and Mi(0) = 0 for any positive integer i. By the definition of φ, there must be a positive integer n such that Mn(x) = 0. Let n(x) be the smallest integer such that Mn(x)(x) = 0. Now again by the definition of φ, it is not difficult to see that $\begin{array}{}x=\sum _{i=1}^{n\left(x\right)}{M}^{i}\left(x\right).\end{array}$ Because R0[S] is a free R-module with a basis S\{0}, φ(x) = φ(y) can imply that M1(x) = M1(y). It follows that φ(xM1(x)) = φ(yM1(y)). By the foregoing proof, M2(x) = M1(xM1(x)) = M1(yM1(y)) = M2(y). Continuing this process, we have n(x) = n(y) and Mk(x) = Mk(y) for 2 ≤ kn(x). Therefore

$x=∑i=1n(x)Mi(x)=∑i=1n(y)Mi(y)=y,$

and φ is injective.

Now let |E(S)| = n + 1. Then we may assume that E(S)\{0} = {e1, e2, ⋯, en}. For any (a)ijPRA(S), we have ia = a and a* = j. It follows that aj = a. Note that i = ek and j = el for some 1 ≤ k, ln. Thus (ek)ii(a)ij = (a)ij = (a)ij(el)jj and

$(∑p=1nep¯)⋅(a)ij=(ek)ii(a)ij=(a)ij=(a)ij(el)jj=(a)ij⋅(∑p=1nep¯).$

Therefore $\begin{array}{}1=\sum _{p=1}^{n}\overline{{e}_{p}}\end{array}$ is the identity of R0[PRA(S)]. So, R0[S] has an identity.

For convenience, we identify R0[S] with T := φ(R0[S]). If ekTelT, then there exist xek R0[T]el, yelR0[T]ek such that ek = xy, yx = el. Thus there are a ∈supp(x), b ∈supp(y) such that ek = ab, ba = el. As xekR0[T]el, we get eka = a. Now, ek𝓡a and a is a regular element of PRA(S). By Proposition 2.11, a = (u)pq for some a regular element u of S. But

$(ekuel)ek,el=(ek)ek,ek(u)pq(el)el,el=ek¯ael¯=a=(u)pq,$

so by the multiplication of PRA(S), ek = p, el = q and hence ek = u#, el = u*. Thus ek𝓡u𝓛el. In other words, ek𝓓el in the semigroup S. Now we prove that if ekTelT, then ek𝓓el in the semigroup S. Because the reverse is obvious, it is now verified that ekTelT if and only if ek𝓓el in the semigroup S.

Let $\begin{array}{}\pi ={\cup }_{i=1}^{r}{E}_{i}\end{array}$ be the partition of E(S)\{0} induced by 𝓓|E(S) and let {f1, f2, ⋯, fr} be representatives of this partition π. Moreover, we let nk = |E(Dfk)| where Dfk is the 𝓓-class of S containing fk. (Of course, r = Dreg(S).) Then by the foregoing proof,

$T=⊕i=1nei¯T≅⊕k=1rnkfk¯T$

and as fks are mutually orthogonal, T is isomorphic to the generalized matrix algebra

$Mn1(f1¯Tf1¯)Mn1,n2(f1¯Tf2¯)⋯Mn1,nr(f1¯Tfr¯)Mn2,n1(f2¯Tf1¯)Mn2(f2¯Tf2¯)⋯Mn2,nr(f2¯Tfr¯)...............................................Mnr,n1(fr¯Tf1¯)Mnr,n2(fr¯Tf2¯)⋯Mnr(fr¯Tfr¯).$(2)

By the construction of PRA(S), fkTfl = R0[Mkl]) where 𝓝 is the set of positive integers and

• $\begin{array}{}{M}_{ij}=\left\{\prod _{k=1}^{m}{x}_{i}:m\in \mathcal{N},{x}_{1},\cdots ,{x}_{k}\in S,{x}_{1}^{\mathrm{#}}={f}_{i},{x}_{m}^{\ast }={f}_{j},{x}_{k}^{\ast }={x}_{k+1}^{\ast }\text{\hspace{0.17em}for\hspace{0.17em}}1\le k\le m-1\right\};\end{array}$

• Mij = {(a)fi,fj : aMij}.

The proof is finished. □

#### Example 3.3

Let S be a semigroup each of whose 𝓛*-classes contains at least one idempotent, and assume that the idempotents of S are in the center of S. By [25], S is a strong semilattice Y of left cancellative monoids Mα with αY. A routine check can show that S is a strict RA semigroup in which for any aS, a# = a* = fα, where fα is the identity of Mα. It is not difficult to see that PRA(S) = ∑αY{(a)fα,fα : aMα}. Now let f1, f2, ⋯, fn be all nonzero idempotents of S and fi be the identity of the left cancellative monoid Mαi for i = 1, 2, ⋯, n. With notations in the proof of Theorem 3.2, Mi,j = ∅ when ij. So, by Theorem 3.2, R0[S] is isomorphic to the generalized matrix algebra diag(R[Mα1], R[Mα2], ⋯, R[Mαn]).

As in [26], a ring (an algebra) 𝕀 has a generalized upper triangular matrix representation of degree n if there exists a ring (algebra) isomorphism

$ϕ:A→R1R12⋯R1n0R2⋯R2n⋮⋮⋱⋮00⋯Rn.$

#### Theorem 3.4

(Generalized Triangular Matrix Representation Theorem). Let S be a strict RA semigroup and R a commutative ring. If

1. |E(S)| < ∞; and

2. for any eE(S), Me,e = {xS : ex = x, e = x*} is a subgroup of S,

then R0[S] has a generalized upper triangular matrix representation of degree Dreg(S).

#### Proof

Let us turn back to the proof of Theorem 3.2. By hypothesis, Mfi,fi is a subgroup of S, in other words, any element xMfi,fi is in a subgroup of S, and so x𝓗e for some eE(S). But x* = fi, now e𝓛fi and further e = fi. Thus x# = fi = x*. It follows that xMii and Mfi,fiMii. On the other hand, it is clear that MiiMfi,fi. Therefore Mii = Mfi,fi and is a subgroup of S.

We may claim:

• Claim A

For any 1 ≤ k, lr with kl, if Mkl ≠ ∅, then Mlk = ∅.

Indeed, if otherwise, we pick aMkl, bMlk, and (a)fk,fl,(b)fl,fkPRA(S). Also, a* = fl, fka = a, flb = b and b* = fk, hence ab𝓛*b, ba𝓛*a. It follows that (ab)fk,fk,(ba)fl,flPRA(S). Thus abMkk, baMll. But Mkk, Mll are both subgroups, so ab𝓗fk, ba𝓗fl. It follows that a and b are regular in S, and a𝓡fk𝓛b, a𝓛fk𝓡b. Thus fk𝓓fl. This is contrary to that {f1, f2, ⋯, fr} are representatives of the partition of E(S)\{0} induced by 𝓓|E(S). So, we prove Claim A.

We next verify:

• Claim B

For any 1 ≤ i, j, kr, if Mij ≠ ∅ and Mjk ≠ ∅, then Mik ≠ ∅.

Pick xMij, yMjk. So, x* = fj, fjy = y and xy𝓛*y𝓛*fk. It follows that (x)fi,fj(y)fj,fk = (xy)fi,fk. Hence xyMik and Mik ≠ ∅. This results in Claim B.

Consider the quiver Q whose vertex set is V = {1, 2, ⋯, r} and in which there is an edge from i to j if and only if Mij ≠ ∅. By Claim B, there is a path from i to l if and only if Mil ≠ ∅. Again by Claim A, the quiver Q has no cycles. Now by [41, Corollary, p.143], the vertices of Q can be labeled V = {1, 2, ⋯, r} in such a way that if there is an edge from i to j then i < j. This means that we can relabel V = {1, 2, ⋯, r} such that if Mij ≠ ∅ then i < j. In this case, the algebra (2) is a generalized upper triangular matrix algebra. The proof is completed. □

Let us turn back to the proof of Theorem 3.2 again. Assume now that S is finite. For x, yMii, we have x* = fi = y*, fix = x and fiy = y. So, xy𝓛*fiy = y𝓛*fi. Thus xyMii and Mii is a subsemigroup of S. For aMii, if ax = ay, then x = fix = fiy = y and Mii is left cancellative. But Mii is finite, so Mii is a subgroup. Now by Theorem 3.4, the following corollary is immediate.

#### Corollary 3.5

Let S be a strict RA semigroup and R a commutative ring. If S is finite, then R0[S] has a generalized upper triangular matrix representation of degree Dreg(S).

#### Remark 3.6

Note that ample semigroups are strict RA semigroups. By Corollary 3.5, any algebra of ample finite semigroups has a generalized upper matrix representation. So, Corollary 3.5 extends the triangular matrix representation theorem on algebras of ample finite semigroups [13, Theorem 4.5].

By Corollary 2.7, any right ample finite monoid is strict RA. So, we have the following corollary.

#### Corollary 3.7

Let S be a right ample monoid and R a commutative ring. If S is finite, then R0[S] has a generalized upper triangular matrix representation of degree Dreg(S).

#### Example 3.8

With notation in Example 2.4, by Theorem 3.2, R0[P(Q)] is isomorphic to the generalized matrix algebra

$R1R12⋯R1nR21R2⋯R2n⋮⋮⋱⋮Rn1Rn2⋯Rn,$

where Ri is the subalgebra of R0[P(Q)] generated by all paths from i to i, and Rij is the free module with the set of all paths from i to j as a base.

Now let Q have no loops. In this case, Mei,ei = {ei} is a subgroup of P(Q). By Theorem 3.4, R0[P(Q)] has a generalized upper triangular matrix representation.

#### Example 3.9

Let 𝓒 be a left cancellative category with |Obj(𝓒)| < ∞. By Example 2.5, S(𝓒) is a strict RA semigroup in which |E(S(𝓒))| < ∞. Obviously, the relation

$π={(1A,1B):A,B∈Obj(C) and there exist arrows α,β such that 1A=αβ,1B=βα}$

is an equivalence on the set Y := E(S(𝓒))\{0}. Let Y/π = {Y1, Y2, ⋯, Yr}. Moreover, we let Xi = {AObj(𝓒) : 1BYi} and ni = |Xi|. If pick AiXi, then by Theorem 3.2, R0[S(𝓒)] is isomorphic to the generalized matrix algebra

$Mn1(R[Hom(A1,A1)])Mn1,n2(R[Hom(A1,A2)])⋯Mn1,nr(R[Hom(A1,Ar)])Mn2,n1(R[Hom(A2,A1)])Mn2(R[Hom(A2,A2)])⋯Mn2,nr(R[Hom(A2,Ar)])⋮⋮⋱⋮Mnr,n1(R[Hom(Ar,A1)])Mnr,n2(R[Hom(Ar,A2)])⋯Mnr(R[Hom(Ar,Ar)]).$

Now let 𝓒 be a groupoid. In this case, Hom(Ai, Aj) = ∅ whenever ij, and further R0[S(𝓒)] is isomorphic to the generalized matrix algebra

$Mn1(R[Hom(A1,A1)])0⋯00Mn2(R[Hom(A2,A2)])⋯0⋮⋮⋱⋮00⋯Mnr(R[Hom(Ar,Ar)]).$

By definition, the category algebra R𝓒 of 𝓒 over R is just the contracted semigroup algebra R0[S(𝓒)]. For category algebras, see [33, 38, 39].

We conclude this section by giving a sufficient and necessary condition for the semigroup algebra of a strict RA finite semigroup to be semiprimitive, which is just [13, Theorem 5.5] when the semigroup is ample.

#### Theorem 3.10

Let S be a strict RA semigroup and R a commutative ring. If S is finite, then R0[S] is semiprimitive if and only if the following conditions are satisfied:

1. S is an inverse semigroup;

2. for every maximum subgroup G of S, R[G] is semiprimitive.

#### Proof

By [13, Theorem 5.4], it suffices to verify the sufficiency. To see this, we assume that R0[S] is semiprimitive. With the notations in the proof of Theorem 3.2, K0[S] is isomorphic to the generalized upper triangular matrix algebra

$Mn1(R0[M11¯])Mn1,n2(R0[M12¯])⋯Mn1,nr(R0[M1r¯])0Mn2(R0[M22¯])⋯Mn2,nr(R0[M2r¯])...............................................00⋯Mnr(R0[Mrr¯]).$

It follows that the Jacobson radical J(R0[S]) of R0[S] is equal to

$JMn1(R0[M11¯])R0[M12¯]⋯R0[M1r¯]0JMn2(R0[M22¯])⋯R0[M2r¯].........................................................00⋯JMnr(R0[Mrr¯]).$

So, J(Mni(R0[Mii])) = 0 and for i, j = 1, 2, ⋯, r, Mni,nj(R0[Mij]) = 0 if ij. Thus R0[Mii] is semiprimitive, so that R0[Mii] is semiprimitive; and Mij = ∅ if ij. Since S is finite, Mii is finite, so that Mii is a subgroup of S. But for all aMii, a𝓛fi giving a𝓛fi, so a𝓗fi since Mii is a subgroup of S, thus Mii is indeed a maximum subgroup of S. By the choice of fi’s, we know that any nonzero idempotent of S is 𝓓-related to some fi, thereby any nonzero maximum subgroup of S is isomorphic to some Mii. Therefore the condition (B) is satisfied.

For the condition (A), we prove only that any nonzero element of S is regular. Let aS\{0} and a# = e, a = f. Let e𝓓fi and f𝓓fj. Then there exist ueSfk, vfkSe, xfjSf, yfSfj such that e = uv, fi = vu, f = yx, fj = xy. It follows that fi𝓡v𝓛e and f𝓡y𝓛fj. Thus vayMij since (v)fi,e(a)e,f(y)f,fj = (vay)fi,fj. We consider the following two cases:

• If fi = fj, then vayMii and so as Mii is a subgroup of S, there exists bS such that vay = vaybvay, so that a = uvayx = uvaybvayx = uvaybvayx = aybva. It follows that a is regular.

• Assume fifj. By the foregoing proof, Mij = ∅, contrary to the fact: vayMij.

Consequently, a is regular, as required. □

Based on Corollary 2.7, any right ample finite monoid is strict RA. Again by Theorem 3.10, the following corollary is obvious.

#### Corollary 3.11

Let S be a right ample monoid. If S is finite, then K0[S] is semiprimitive if and only if the following conditions are satisfied:

1. S is an inverse semigroup;

2. for every maximum subgroup G of S, R[G] is semiprimitive.

Note that inverse semigroups are strict RA. By Theorem 3.10, we can re-obtain the well-known result on inverse semigroup algebras as follows:

#### Corollary 3.12

Let S be an inverse semigroup. If S is finite, then K0[S] is semiprimitive if and only if for every maximum subgroup G of S, R[G] is semiprimitive.

## 4 Self-injective algebras

Recall that an algebra 𝔄 (possibly without unity) is right (respectively, left) self-injective if 𝔄 is an injective right (respectively, left) 𝔄-module. Okiniński pointed out that for a semigroup S and a field K, the algebra K[S] is right (left) self-injective if and only if so is K0[S] (see [15, the arguments before Lemma 3, p.188]). So, in this section we always assume that the semigroup has zero element θ. The aim of this section is to answer when the semigroup algebra of a strict RA semigroup is left self-injective.

To begin with, we recall a known result on left (right) perfect rings, which follows from [27, Theorem (23.20), p.354 and Corollary (24.19), p.365].

#### Lemma 4.1

Let R be a ring with unity. If R is left (right) perfect, then

1. R does not contain an infinite orthogonal set of nonzero idempotents.

2. Any quotient of R is left (right) perfect.

We need some known facts on left self-injective semigroup algebras. By the dual of [28, Theorem 1], any left self-injective algebra is a left perfect algebra. Note that, by the argument in [29, Remark], whenever K0[S] is left perfect, there exist ideals Si, i = 0, 1, ⋯, n, such that

$θ=S0⊏S1⊏⋯⊏Sn=S$

and the Rees quotients Si/Si+1 are completely 0-simple or T-nilpotent. So, the following lemma is straight.

#### Lemma 4.2

Let S be an arbitrary semigroup and K a field. If K0[S] is a left self-injective K-algebra, then

1. There exist ideals Si, i = 0, 1, ⋯, n, such that θ = S0S1 ⊏ ⋯ ⊏ Sn = S and the Rees quotients Si/Si+1 are completely 0-simple or T-nilpotent.

2. ([15, Lemmas 9 and 10, p.192]) S satisfies the descending chain condition on principal right ideals and has no infinite subgroups.

3. [28, Theorem 1] K0[S] is left perfect.

4. [28, Lemma 1] $\begin{array}{}\frac{{K}_{0}\left[S\right]}{Rad\left({K}_{0}\left[S\right]\right)}\end{array}$ is regular, where Rad(K0[S]) is the Jacobson radical of K0[S].

Moreover, we have

#### Lemma 4.3

Let S be a right ample semigroup. If K0[S] is left self-injective, then

1. |E(S)| < ∞.

2. For any eE(S)\{0}, the subset Me,e = {xS: ex = x, x = e} is a finite subgroup of S.

#### Proof

1. By Lemma 4.2, we assume that Si, i = 0, 1, ⋯, n, are ideals of S satisfying the conditions:

• θ = S0S1 ⊏ ⋯ ⊏ Sn = S; and

• the Rees quotients Si/Si+1 are completely 0-simple or T-nilpotent.

So, S = {θ} ⊔ $\begin{array}{}\left({\bigsqcup }_{i=1}^{n}{S}_{i}\mathrm{\setminus }{S}_{i-1}\right).\end{array}$ This shows that for any eE(S)\{θ}, there exists i ≥ 1 such that Si/Si−1 is completely 0-simple and eSi/Si−1. Now, to verify that E(S) is finite, it suffices to prove that any completely 0-simple semigroup Si/Si−1 is finite.

Now let Si/Si−1 be a completely 0-simple semigroup. By the definition of Rees quotient, any nonzero idempotents of Si/Si−1 is an idempotent of S. But E(S) is a semilattice, so Si/Si−1 is an inverse semigroup. Thus Si/Si−1 is a Brandt semigroup. By the structure theorem of Brandt semigroups in [30], Si/Si−1 is isomorphic to the semigroup T = I × G × I ⊔ {θ} whose multiplication is defined by

$(a,g,x)(b,h,y)=(a,gh,y) if x=b;θ otherwise,$

where I is a nonempty set and G is a subgroup of Si/Si−1 Clearly, G is a subgroup of S. By Lemma 4.2(B), G is a finite group. By computation, any nonzero idempotent of T is of the form: (a, 1G, a) where 1G is the identity of G.

For convenience, we identify Si/Si−1 with T. Now, we need only to show that |I| < ∞. By Lemma 4.2, K0[S] is left perfect, and so $\begin{array}{}\frac{{K}_{0}\left[S\right]}{Rad\left({K}_{0}\left[S\right]\right)}\end{array}$ is semisimple. It follows that $\begin{array}{}\frac{{K}_{0}\left[S\right]}{Rad\left({K}_{0}\left[S\right]\right)}\end{array}$ has an identity. It is easy to see that K0[Si−1] is an ideal of K0[S]. Consider the algebra $\begin{array}{}W:=\frac{{K}_{0}\left[S\right]}{Rad\left({K}_{0}\left[S\right]\right)+{K}_{0}\left[{S}_{i-1}\right]}.\end{array}$ Note that

$K0[S]Rad(K0[S])Rad(K0[S])+K0[Si−1]Rad(K0[S])≅K0[S]Rad(K0[S])+K0[Si−1]≅K0[S]K0[Si−1]Rad(K0[S])+K0[Si−1]K0[Si−1],$

we can observe that

• (a, 1G, a)(b, 1G, b) = θ whenever ab;

• W has an unity;

• (a, 1G, a) + Rad(K0[S]) + K0[Si−1] ≢ (b, 1G, b) + Rad(K0[S]) + K0[Si−1] whenever ab.

Again by the property that (a, 1G, a)(b, 1G, b) = θ whenever ab, we get that the set

$X:={(a,1G,a)+Rad(K0[S])+K0[Si−1]:a∈I}$

is an orthogonal set of nonzero idempotents of W. On the other hand, since $\begin{array}{}\frac{{K}_{0}\left[S\right]}{Rad\left({K}_{0}\left[S\right]\right)}\end{array}$ is semisimple, we know that $\begin{array}{}\frac{{K}_{0}\left[S\right]}{Rad\left({K}_{0}\left[S\right]\right)}\end{array}$ is left perfect, and so by Lemma 4.1, W is left perfect. Again by Lemma 4.1, this shows that W does not contain an infinite orthogonal set of nonzero idempotents. It follows that |X| < ∞. Therefore |I| < ∞ since |X| = |I|. Consequently, Si/Si−1 is finite, and so |E(S)| < ∞.

2. Let aMe,e. Consider the chain of principal right ideals of S:

$⋯anS1⊆an−1S1⊆⋯⊆a2S1⊆aS1.$

By Lemma 4.2, there exists a positive integer n such that anS1 = an+1S1. That is, there is xS1 such that an = an+1x. Hence an−1 = aan−1 = aanx = anx. Continuing this process, we can obtain a = ax. But a = aa, now a𝓡a. Thus a is regular. Therefore a𝓗a since a𝓛a. So, Me,e is a subgroup of S and further by Lemma 4.2, Me,e is finite. □

For convenience, in the rest of this section, we always let S be a strict RA semigroup and K a field. Assume that K0[S] is a left self-injective K-algebra. Let us turn back to the proof of Theorem 3.2. We know that $\begin{array}{}\sum _{i=1}^{n}\overline{{e}_{i}}\end{array}$ is the identity of K0[PRA(S)]. By Theorem 3.4, we have that Mij = ∅ if j < i, and

$K0[S]:=B≅Mn1(K0[M11¯])Mn1,n2(K0[M12¯])⋯Mn1,nr(K0[M1r¯])0Mn2(K0[M22¯])⋯Mn2,nr(K0[M2r¯])..................................................00⋯Mnr(K0[Mrr¯]).$

Since our aim is to show that S is finite, for convenience, we may assume ni = 1 for i = 1, 2, ⋯, r. So, we let

$B=K0[M11¯]K0[M12¯]⋯K0[M1r¯]0K0[M22¯]⋯K0[M2r¯]..............................00⋯K0[Mrr¯].$

For 1 ≤ jr, we denote by mj the smallest positive integer in the set {i:Mij ≠ ∅}. By definition, mjj for any i.

#### Lemma 4.4

MklMmj,j = 0 whenever MklMmj,mj.

#### Proof

By definition, Mk,mjMmj,jMk,j and

$Mkl¯⋅Mmj,j¯=0 if l≠mi⊆Mk,j¯ if l=mj.$

In the second case, kmj. This shows that k = mj by the minimality of mj. □

|Mmj,j| < ∞.

#### Proof

Lemma 4.3 results the case for mj = j. Assume now that mj < j. By Lemma 4.4, the algebra

$C:=0⋯000⋯0⋮⋮⋮⋮⋮⋮⋮0⋯0K0[Mmj,j¯]0⋯0⋮⋮⋮⋮⋮⋮⋮0⋯000⋯0$

is a left ideal of 𝔅. Pick wMmj,j and define

$θ:Mmj,j→Mmj,j;x→x if x∈Mmj,mjw0 if otherwise,$

and span linearly to K0[Mi,mi]. Further define a map ζ of ℭ into 𝔅 by

$X=0⋯000⋯0⋮⋮⋮⋮⋮⋮⋮0⋯0(x)mj,j0⋯0⋮⋮⋮⋮⋮⋮⋮0⋯000⋯0→ζ(X)=0⋯000⋯0⋮⋮⋮⋮⋮⋮⋮0⋯0(θ(x))mj,j0⋯0⋮⋮⋮⋮⋮⋮⋮0⋯000⋯0.$

By Lemma 4.4, a routine computation shows that ζ is a 𝔅-module homomorphism. But 𝔅 is left self-injective, now by Baer condition, there exists U ∈ 𝔅 such that ζ(A) = AU for any A ∈ ℭ. Especially, ζ(X) = XU. Now let U = (ukl). Then ujj = $\begin{array}{}\sum _{k=1}^{n}\end{array}$ rk ak where rkK, akMjj. Since ζ(X) = XU, we have (θ(x))mj,j = (θ(x)mj,j ujj = ( $\begin{array}{}\sum _{k=1}^{n}\end{array}$ rkxak)mj,j and

$x=∑k=1nrkxak.$(3)

So, we may let n1, n2, ⋯, ns be positive integers such that

• n1 + n2 + ⋯ + ns = n;

• (∗) xal = x, for l = 1, 2, ⋯, n1;

• (∗∗) xanq+1 = xanq+2 = ⋯ = xanq+nq+1−1 = bq for q = 1,2, ⋯, s, where b1, b2, ⋯, bs are different elements in Mjj.

Now by Eq. (3), we get r1 + r2 + ⋯ rn1 = 1 and rnl+1+rnl+2 + ⋯ rnl+nl+1−1 = 0 for l = 1, 2, ⋯, s. Note that fj = x 𝓛x. Therefore by Eq. (∗), al = fjal = xal = x = fj, for l = 1, 2, ⋯, n1; and by Eq. (∗∗), xanq+1 = xanq+2 = ⋯ = xanq+nq+1−1 for q = 1, 2, ⋯, s, so that as x = fj, anq+1 = anq+2 = ⋯ = anq+nq+1−1 for q = 1, 2, ⋯, s. Consequently, ujj = fj.

Now, for any xMmj,j, (θ(x))mj,j = (x)mj,jujj = (xfj)mj,j = (x)mj,j. This shows that θ(x) = x and Mmj,j = Mmj,mjw since θ(x) = 0 for any xMmj,j\Mmj,mjw. It follows that |Mmj,j| < ∞ since Mmj,mj = MfjMfj,fj and by Lemma 4.3 (B), is a finite subgroup. □

|Mij| < ∞.

#### Proof

If i = mj, nothing is to prove.

If mji and Mij ≠ ∅, then mj < i. Let i0 be the smallest positive integer of the set

$Y={k:there existk=k0,k1,k2,⋯,km=isuch thatMKl−1,kl≠∅forl=1,2,⋯,m−1}.$

We shall prove i0 = mj. Assume on the contrary that i0mj. By definition, Mk0j ≠ ∅ and specially Mi0j ≠ ∅. Thus mj < i0. By the minimality of i0, Mki = ∅ for 1 ≤ k < i0, and further Mki0Mi0j = 0. This and Lemma 4.4 show that

$D=0⋯000⋯0⋮⋮⋮⋮⋮⋮⋮0⋯0K0[Mmj,j¯]0⋯0⋮⋮⋮⋮⋮⋮⋮0⋯0K0[Mi0,j¯]0⋯0⋮⋮⋮⋮⋮⋮⋮0⋯000⋯0$

is a left ideal of 𝔅 and that the map η defined by

$A=0⋯000⋯0⋮⋮⋮⋮⋮⋮⋮0⋯0(a)mj,j0⋯0⋮⋮⋮⋮⋮⋮⋮0⋯0(b)i0,j0⋯0⋮⋮⋮⋮⋮⋮⋮0⋯000⋯0→η(A)=0⋯000⋯0⋮⋮⋮⋮⋮⋮⋮0⋯0(a)mj,j0⋯0⋮⋮⋮⋮⋮⋮⋮0⋯000⋯0⋮⋮⋮⋮⋮⋮⋮0⋯000⋯0$

is a 𝔅-module homomorphism. Since 𝔅 is left self-injective, it follows from Baer condition that there exists V ∈ 𝔅 such that η(A) = AV for any A ∈ 𝔇. Hence (a)mj,j = (a)mj,jvjj and 0 = (b)mj,jvjj where V = (vij). By the first equality, avjj = a and further by a similar arguments as proving ujj = fj in the proof of Lemma 4.5, fj = vjj ≠ 0; by the second equality, 0 = (b)mj,jvjj = (bfj)mj,j = (b)mj,j and b = 0, so that as b = fj, we get 0 = fj, thus vjj = 0, contrary to the foregoing proof: vjj ≠ 0. Therefore mj = i0. We have now proved that Mi0iMijMmj,j. This shows that cMijMmj,j for some cMi0i. Note that for w, zMij, if cw = cz, then fiw = fiz, so that w = z. We can observe that |cMij| = |Mij|, and |Mij| ≤ |Mmj,j| < ∞. □

Mij = ∅ if ij.

#### Proof

By Lemmas 4.2 and 4.6, 𝔅 is a finite dimensional algebra and further is quasi-Frobenius. Assume on the contrary that there exists Mij ≠ ∅. Let n be the biggest number such that Min ≠ ∅ for some in, and further let m be the biggest number such that Mmn ≠ ∅. Obviously, m < n. By definition, Mkm = ∅ if m < k (if not, then as Mmn ≠ ∅ and by definition, Mkn ≠ ∅, contrary to the maximality of m); and Mnl = ∅ if n < l (if not, then as Min ≠ ∅ and by definition, Mil ≠ ∅, contrary to the maximality of n). By these, a routine computation shows that

$E=0⋯0K0[M1n¯]0⋯00⋯0K0[M2n¯]0⋯0⋮⋮⋮⋮⋮⋮⋮0⋯0K0[Mmn¯]0⋯00⋯000⋯0⋮⋮⋮⋮⋮⋮⋮0⋯000⋯0$

is a left ideal of 𝔅.

Let xK0[Mjl] and Mij ≠ ∅. If K0[Mij]x = 0, then ax = 0 for some aMij, so that by a similar argument as proving ujj = fj in the proof of Lemma 4.5, ax = 0. By a = fj, x = fjx = ax = 0. So, K0[Mij]x = 0 if and only if x = 0. Because Mmn ≠ ∅, this can show that the right annihilator annr(𝔈) of 𝔈 in 𝔅 is equal to

$K0[M11¯]⋯K0[M1,n−1¯]K0[M1n¯]K0[M1,n+1¯]⋯K0[M1r¯]⋮⋱⋮⋮⋮⋯⋮0⋯K0[Mn−1,n−1¯]K0[Mn−1,n¯]K0[Mn−1,n+1¯]⋯K0[Mn−1,r¯]0⋯000⋯00⋯00K0[Mn+1,n+1¯]⋯K0[Mn+1,r¯]⋮⋯⋮⋮⋮⋱⋮0⋯000⋯K0[Mrr¯].$

But the left annihilator ann(annr(𝔈)) of annr(𝔈) in 𝔅 includes

$0⋯0K0[M1n¯]0⋯00⋯0K0[M2n¯]0⋯0⋮⋮⋮⋮⋮⋮⋮0⋯0K0[Mnn¯]0⋯00⋯000⋯0⋮⋮⋮⋮⋮⋮⋮0⋯000⋯0.$

It is clear that ann(annr(𝔈)) ≠ 𝔈. But 𝔅 is a quasi-Frobenius algebra, so that by [31, Theorem 30.7, p.333] we should have ann(annr(𝔈)) = 𝔈, a contradiction. □

#### Theorem 4.8

Let S be a strict RA semigroup and K a field. Then K0[S] is left self-injective if and only if S is a finite inverse semigroup.

#### Proof

To verify Theorem 4.8, by [18, Theorem 4.1], it suffices to prove that if K0[S] is left self-injective, then S is regular. Assume now that K0[S] is left self-injective. By Theorem 3.4, K0[S] is isomorphic to

$Mn1(K0[M11¯])Mn1,n2(K0[M12¯])⋯Mn1,nr(K0[M1r¯])Mn2,n1(K0[M21¯])Mn2(K0[M22¯])⋯Mn2,nr(K0[M2r¯])......................................................Mnr,n1(K0[Mr1¯])Mnr,n2(K0[Mr2¯])⋯Mnr(K0[Mrr¯]).$

By Theorem 3.4 and Lemma 4.7, whenever ij, Mni,nj(K0[Mij]) = 0, so that Mij = ∅, thus {xS : x# = fi, x = fj} = ∅. Note that fi’s are representatives of the partition $\begin{array}{}\pi ={\cup }_{i=1}^{r}{E}_{r}\end{array}$ of E(S)\{0} induced by 𝓓. Therefore {xS : x# = e, x = h} = ∅ for all eEi, hEj with ij. This means that for all xS, x#𝓓x.

On the other hand, by Lemma 4.3(B), for any eE(S)\{0}, Me,e = {xS : x# = e, x = e} is a subgroup of S and so Mii is a subgroup of S. Let fE(S)\{0} and e 𝓓 f. Let x, yS with x# = e, y# = f, x = f and y = e, and of course x, yMe,f. Then as 𝓛 is a right congruence, we have xy𝓛fy = y, so that y ≠ 0. Similarly, yx ≠ 0. Note that (x)e,f, (y)f,ePRA(S). We observe that (xy)e,e = (x)e,f(y)f,ePRA(S)\{0}, and hence xyMe,e. Thus there is aMe,e such that e = axy since Me,e is a subgroup of S with identity e. This and ye = y imply that e𝓛y. Therefore y is regular. We have now proved that y is regular whenever y#𝓓y.

However, any element of S is regular and S is regular, as required. □

It is a natural problem whether Theorem 4.8 is valid for the case for right self-injectivity. We answer this question next. Firstly, we prove the following lemma.

#### Lemma 4.9

Let S be a right ample semigroup and K0[S] have an unity 1. If K0[S] is right self-injective, then S is a finite inverse semigroup.

#### Proof

For any aS, S1a is a left ideal of S, and so K0[S1a] is a left ideal of K0[S]. Since K0[S] is right self-injective, and by [31, Lemma 30.9, p.334], we have ann(annr(K0[S1a])) = K0[S1a]. By the definition of 𝓛, annr(K0[S1a]) = annr(a) = annr(a) = (1 − a)K0[S] and so

$K0[S1a]=annℓ(annr(K0[S1a])=annℓ((1−a∗)K0[S])=K0[S]a∗=K0[S1a∗].$

It follows that S1a = S1a. Thus a𝓛a so that a is a regular element of S. Therefore S is an inverse semigroup, and by the Wenger Theorem, S is a finite inverse semigroup. □

Based on Lemma 4.9, we can verify the following theorem, which illuminates that Theorem 4.8 is valid for right self-injectivity.

#### Theorem 4.10

Let S be a strict RA semigroup. Then K0[S] is right self-injective if and only if S is a finite inverse semigroup.

#### Proof

By Theorem 4.8, it suffices to verify the necessity. By Lemma 4.9, we need only to show that K0[S] has an identity. Now let us turn back to the proof of Lemma 4.3(A). We notice that the proof is valid for right self-injectivity and so |E(S)| < ∞ when K0[S] is right self-injective. By Theorem 3.4, we have that K0[S] is isomorphic to the generalized upper triangular matrix algebra

$Mn1(R0[M11¯])Mn1,n2(R0[M12¯])⋯Mn1,nr(R0[M1r¯])0Mn2(R0[M22¯])⋯Mn2,nr(R0[M2r¯]).................................................00⋯Mnr(R0[Mrr¯]).$

It is easy to see that K0[S] has an identity. We complete the proof. □

We now arrive at the main result of this section, which follows immediately from Theorems 4.8 and 4.10.

#### Theorem 4.11

Let S be a strict RA semigroup and K a field. Then the following statements are equivalent:

1. K0[S] is left self-injective;

2. K0[S] is right self-injective;

3. S is a finite inverse semigroup;

4. K0[S] is quasi-Frobenius;

5. K0[S] is Frobenius.

Note that ample semigroups are both strict RA and strict LA. By Theorem 4.11 and its dual, the following corollary is immediate, which is the main result of [18] (see, [18, Theorem 4.1]).

#### Corollary 4.12

Let S be an ample semigroup and K a field. Then the following statements are equivalent:

1. K0[S] is left self-injective;

2. S is a finite inverse semigroup;

3. K0[S] is quasi-Frobenius;

4. K0[S] is Frobenius;

5. K0[S] is right self-injective.

Let S be a right ample monoid. Note that E(S) ⋅ S = S. By Corollary 2.7, S is a strict RA semigroup. If K0[S] is left self-injective, then S has only finite idempotents. Now, by Theorem 4.11, the following corollary is immediate.

#### Corollary 4.13

Let S be a right ample monoid. Then the following statements are equivalent:

1. K0[S] is left self-injective;

2. S is a finite inverse semigroup;

3. K0[S] is quasi-Frobenius;

4. K0[S] is Frobenius;

5. K0[S] is right self-injective.

Recall that if 𝓐 is a ring and 𝓐1 is the standard extension of 𝓐 to a ring with unity, then 𝓐 is left self-injective if and only if the left 𝓐1-module 𝓐 satisfies the Baer condition (c.f. [32, Chapter 1]). The following example, due to Okniński [28], shows that in Theorem 4.8, the assumption that E(S) is a semilattice, i.e., all idempotents commute, is essential.

#### Example 4.14

Let S = {g, h} be the semigroup of left zeros, and ℚ the field of rational numbers. Consider the algebra ℚ[S] = ℚ0[S] and the standard extension ℚ[S]1 of ℚ[S] to a ℚ-algebra with unity. It may be shown that for any left ideal I of ℚ[S]1, any homomorphism of left ℚ[S]1-modules I → ℚ[S], extends to a homomorphism of ℚ[S]1-modules Iℚ[S]1 → ℚ[S]. Moreover, by computing the right ideals of ℚ[S]1, one can easily check that ℚ[S]1 satisfies Baer’s condition. Hence, ℚ[S] satisfies Baer’s condition as ℚ[S]1-module which means that ℚ[S] is left self-injective. On the other hand, since ℚ[S] has no right identities, it is obvious that ℚ[S] is not right self-injective.

Obviously, the polynomial algebra K[x] is indeed the semigroup algebra K[M], where M = {1, x, x2, ⋯}. It is easy to see that M is a cancellative monoid, and of course, a strict RA semigroup. By Theorem 4.11 and its dual, K[x] is neither left self-injective nor right self-injective. Recall that a category 𝓒 is said to be finite if |Obj(𝓒)| < ∞ and |Hom(A, B)| < ∞, for all A, BObj(𝓒). For the self-injectivity of path algebras and category algebras, we have the following proposition.

#### Proposition 4.15

1. Let Q be a quiver and K a field. Then the path algebra KQ is left self-injective if and only if Q has neither edges nor loops; if and only if KQ is right self-injective.

2. Let 𝓒 be a left cancellative category. Then the category algebra K𝓒 is left self-injective if and only if 𝓒 is a finite groupoid; if and only if K𝓒 is right self-injective.

#### Proof

1. We only need to prove the necessity. If KQ is left (resp. right) self-injective, then S(Q) is a finite inverse semigroup. It follows that Q has no circles, and of course, no loops. On the other hand, since the idempotents of S(Q) are empty paths, it is easy to show that any edge is not regular in the semigroup S(Q), thus Q has no edges.

2. By Example 2.5, S(𝓒) is an inverse semigroup whenever 𝓒 is a groupoid. So, it suffices to verify the necessity. Assume that K𝓒 is left self-injective, then by Theorem 4.8, S(𝓒) is a finite inverse semigroup. It follows that 𝓒 is a finite groupoid. □

Recall from [27] that an algebra 𝔄 with unity is semisimple if and only if every left 𝔄-module is injective. So, any semisimple algebra is left self-injective. By Theorem 4.11 and Corollary 3.12, we immediately have

#### Theorem 4.16

Let S be a strict RA semigroup and K be a field. If K0[S] has a unity, then K0[S] is semisimple if and only if S is a finite inverse semigroup and the order of any maximum subgroup of S is not divided by the characteristic of K.

By Theorem 4.11, the following proposition is immediate, which answers positively [15, Problem 6, p.328] for strict RA semigroups:

$DoesthefactthatK[S]isaright(respectively,left)self−injectiveimplythatSisfinite?$

#### Proposition 4.17

Let S be a strict RA semigroup. If K0[S] is left (respectively, right) self-injective, then S is finite.

Moreover, we have the following corollary.

#### Corollary 4.18

Let S be a right ample monoid and K a field. If K0[S] is left (respectively, right) self-injective, then S is finite.

## Acknowledgement

This research is jointly supported by the National Natural Science Foundation of China (grant: 11361027; 11761034; 11661042); the Natural Science Foundation of Jiangxi Province (grant: 20161BAB201018) and the Science Foundation of the Education Department of Jiangxi Province, China (grant: GJJ14251).

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## About the article

Received: 2017-11-01

Accepted: 2018-05-22

Published Online: 2018-08-03

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 842–861, ISSN (Online) 2391-5455,

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© 2018 Guo and Guo, published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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