Algebras of right ample semigroups

Definition 2.1

A semigroupSis right ample if

1. its idempotents commute;

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

3. for anya ∈ S ande ∈ E(S), ea = a(ea)^*.

Lemma 2.2

Leta ∈ Sande ∈ E(S). If a𝓡̂e, then ea = a. Moreover, if g, h ∈ E(S) and g 𝓡̂h, then g𝓡h.

Proof

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

Definition 2.3

A semigroupS is strict RA ifSis a right ample semigroup in which for anya ∈ S, there exists an idempotentesuch that a𝓡̂e. Strict LA semigroups are defined by duality.

Example 2.4

Assume that Q = (V, E) is a quiver with vertices V = {1, 2,⋯,r}. Denote by (i₁|i₂|⋯|i_n) the path: ∘i1→∘i2→⋯→∘in. In particular, we call the path of Q without vertices the empty path of Q, denoted by 0. For any i ∈ V, we appoint to have an empty path e_i. Let P(Q) be the set of all paths of Q. On P(Q), define a multiplication by: for (i₁|⋯|i_m),(j₁|⋯|j_n) ∈ P(Q),

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

1. 0 is the zero element of P(Q) and e₁, e₂, ⋯, e_r, 0 are all idempotents of P(Q);

2. e_i1 𝓡^*(i₁|⋯|i_m)𝓛^*e_im;

3. e_ie_j = 0 whenever i ≠ j.

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

define: for α, β ∈ ⊔_{A,B∈Obj(𝓒)}Hom(A, B),

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 α ∈ Hom(A, B) is an isomorphism if there exists β ∈ Hom(B, A) such that αβ = 1_A and βα = 1_B. 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 α ∈ Hom(A, B),

   1. 1_A𝓛^*α𝓡̂1_B;

   2. E(S(𝓒)) = {0} ⊔ {1_C : C ∈ Obj(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 β ∈ Hom(B, A) such that αβ = 1_A. 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

LetSbe a right ample semigroup andE(S) ⋅ S = S. If |E(S)| < ∞ thenSis strict RA.

Proof

Let a ∈ S and denote by e_a the minimal idempotent under the partial order ω on E(S) (that is, eωf if and only if e = ef = fe) such that e_aa = a. Obviously, for any f ∈ E(S), fe_a = e_a can imply that fa = a; conversely, if fa = a then fe_a ⋅ a = a and fe_a = e_a since fe_aωe_a. Thus a𝓡̂e_a, so S is strict RA. □

Corollary 2.7

LetSbe a right ample monoid. IfSis finite, thenSis strict RA.

Lemma 2.8

IfSis a right ample semigroup, then with respect to ≤_r, Sis an ordered semigroup.

Proof

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

Proposition 2.9

LetSbe a strict RA semigroup and a, b, x ∈ S. If x ≤_rab, then there exist uniquely u, v ∈ Ssuch that

1. u ≤_r a, v ≤_r b;

2. u^# = x^#, u^* = v^#andx^* = v^*;

3. 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^# ≤_ra^# and a^*v^# ≤_ra^*, so that u = x^#a^# ⋅ a ⋅ a^*v^# ≤_ra^# ⋅ a ⋅ a^* = a; and (x^#a)^*b^# ≤_rb^#, b^*x^* ≤_rb^*, so that v = (x^#a)^*b^# ⋅ b ⋅ b^*x^* ≤_rb^#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^* ≤_rv^#. 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,

further by definition, v^# = u^*v^#. It follows that v^# ≤_ru^*. 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. □

Lemma 2.10

The zero 0 and the elements (i)_ii(i ∈ E) are all idempotents ofM(T), and orthogonal each other.

Proposition 2.11

In the semigroupPRA(T), for any (a)_ij ∈ PRA(T), we have

1. a^* = j.

2. (a)_ijis regular if and only ifais regular inT. Moreover, ifais regular, then (a)_ij ∈ X(T).

3. (a)_ij𝓛^*(j)_jj.

Proof

1. By the definition of PRA(T), there exist x₁, x₂, ⋯, x_n ∈ T such that a = x₁x₂⋯x_n, xk∗=xk+1# where k = 1, 2, ⋯, n – 1, x1#=i and xn∗=j. 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)_ji ∈ PRA(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 x₁x₂⋯x_nyx₁x₂⋯x_n = x₁x₂⋯x_n and x₁𝓛^*i₁, we have x₂⋯x_nyx₁x₂⋯x_n–1x_n = x₂⋯x_n–1x_n, 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 x₃⋯x_n is regular. Continuing this process we have that x_n is regular. By Lemma 2.2, xn#Rxn. Now, x₁x₂⋯x_nyx₁x₂⋯x_n = x₁x₂⋯x_n can imply that x₁x₂⋯x_nyx₁x₂⋯x_n–1xn# = x₁x₂⋯x_n–1xn#, whence x₁⋯x_n–1 = x₁⋯x_n–1 ∘ x_ny ∘ x₁x₂⋯x_n–1 and further regular in T. By the foregoing proof, x_n–1 is regular. Applying these arguments to x₁⋯x_n–2, we know that x_n–2 is regular, therefore x_l is regular for l = 1, ⋯, n. But xkL∗xk∗, so xkLxk∗ 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)_ij ∈ X(T). It follows that

   ay=a#=i and ya=a∗=j.

   Furthermore, (y)_ji ∈ PRA(T). Consequently, (a)_ij is regular since (a)_ij(y)_ji(a)_ij = (aya)_ij = (a)_ij.

3. Now let (x)_jk, (y)_jl ∈ PRA(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)_jl ∈ PRA(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. □

Definition 3.1

A ring (An algebra) has ageneralized matrix representation of degreenif there exists a ring (an algebra) isomorphism

Theorem 3.2

(Generalized Matrix Representation Theorem). LetSbe a strict RA semigroup andRa commutative ring with unity. If |E(S)| < ∞ thenR₀[S] has a generalized matrix representation of degreeD_reg(S).

Proof

Define a map φ by

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, t ∈ S, by Proposition 2.9, there exist uniquely x ∈ O(s), y ∈ O(t) such that u = xy, x^# = u^#, x^* = y^# and y^* = u^*, for any u ∈ O(st). So,

and φ is a homomorphism.

Let x=∑u∈supp(x)ruu,y=∑v∈supp(y)rvv∈R0[S]∖{0} and φ(x) = φ(y). Denote

Λ(x) = {a^* : a ∈ supp(x)};

m(x,i)=∑u∈supp(x),u∗=iruu,

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

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 M^n(x)(x) = 0. Now again by the definition of φ, it is not difficult to see that x=∑i=1n(x)Mi(x). 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(x – M¹(x)) = φ(y – M¹(y)). By the foregoing proof, M²(x) = M¹(x – M¹(x)) = M¹(y – M¹(y)) = M²(y). Continuing this process, we have n(x) = n(y) and M^k(x) = M^k(y) for 2 ≤ k ≤ n(x). Therefore

and φ is injective.

Now let |E(S)| = n + 1. Then we may assume that E(S)\{0} = {e₁, e₂, ⋯, e_n}. For any (a)_ij ∈ PRA(S), we have ia = a and a^* = j. It follows that aj = a. Note that i = e_k and j = e_l for some 1 ≤ k, l ≤ n. Thus (e_k)_ii(a)_ij = (a)_ij = (a)_ij(e_l)_jj and

Therefore 1=∑p=1nep¯ 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 e_kT ≅ e_lT, then there exist x ∈ e_kR₀[T]e_l, y ∈ e_lR₀[T]e_k such that e_k = xy, yx = e_l. Thus there are a ∈supp(x), b ∈supp(y) such that e_k = ab, ba = e_l. As x ∈ e_kR₀[T]e_l, we get e_ka = a. Now, e_k𝓡a 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

so by the multiplication of PRA(S), e_k = p, e_l = q and hence e_k = u^#, e_l = u^*. Thus e_k𝓡u𝓛e_l. In other words, e_k𝓓e_l in the semigroup S. Now we prove that if e_kT ≅ e_lT, then e_k𝓓e_l in the semigroup S. Because the reverse is obvious, it is now verified that e_kT ≅ e_lT if and only if e_k𝓓e_l in the semigroup S.

Let π=∪i=1rEi be the partition of E(S)\{0} induced by 𝓓|_E(S) and let {f₁, f₂, ⋯, f_r} be representatives of this partition π. Moreover, we let n_k = |E(D_fk)| where D_fk is the 𝓓-class of S containing f_k. (Of course, r = D_reg(S).) Then by the foregoing proof,

and as f_k’s are mutually orthogonal, T is isomorphic to the generalized matrix algebra

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

1. Mij=∏k=1mxi:m∈N,x1,⋯,xk∈S,x1#=fi,xm∗=fj,xk∗=xk+1∗ for 1≤k≤m−1;

2. M_ij = {(a)_fi,fj : a ∈ M_ij}.

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 M_α with α ∈ Y. A routine check can show that S is a strict RA semigroup in which for any a ∈ S, a^# = a^* = f_α, where f_α is the identity of M_α. It is not difficult to see that PRA(S) = ∑_α∈Y{(a)_fα,fα : a ∈ M_α}. Now let f₁, f₂, ⋯, f_n be all nonzero idempotents of S and f_i be the identity of the left cancellative monoid M_αi for i = 1, 2, ⋯, n. With notations in the proof of Theorem 3.2, M_i,j = ∅ when i ≠ j. So, by Theorem 3.2, R₀[S] is isomorphic to the generalized matrix algebra diag(R[M_α1], R[M_{α2], ⋯, R[Mαn}]).

Theorem 3.4

(Generalized Triangular Matrix Representation Theorem). LetSbe a strict RA semigroup andRa commutative ring. If

1. |E(S)| < ∞; and

2. for anye ∈ E(S), M_e,e = {x ∈ S : ex = x, e = x^*} is a subgroup of S,

thenR₀[S] has a generalized upper triangular matrix representation of degreeD_reg(S).

Proof

Let us turn back to the proof of Theorem 3.2. By hypothesis, M_fi,fi is a subgroup of S, in other words, any element x ∈ M_fi,fi is in a subgroup of S, and so x𝓗e for some e ∈ E(S). But x^* = f_i, now e𝓛f_i and further e = f_i. Thus x^# = f_i = x^*. It follows that x ∈ M_ii and M_fi,fi ⊆ M_ii. On the other hand, it is clear that M_ii ⊆ M_fi,fi. Therefore M_ii = M_fi,fi and is a subgroup of S.

We may claim:

1. For any 1 ≤ k, l ≤ r with k ≠ l, ifM_kl ≠ ∅, thenM_lk = ∅.

   Indeed, if otherwise, we pick a ∈ M_kl, b ∈ M_lk, and (a)_fk,fl,(b)_fl,fk ∈ PRA(S). Also, a^* = f_l, f_ka = a, f_lb = b and b^* = f_k, hence ab𝓛^*b, ba𝓛^*a. It follows that (ab)_fk,fk,(ba)_fl,fl ∈ PRA(S). Thus ab ∈ M_kk, ba ∈ M_ll. But M_kk, M_ll are both subgroups, so ab𝓗f_k, ba𝓗f_l. It follows that a and b are regular in S, and a𝓡f_k𝓛b, a𝓛f_k𝓡b. Thus f_k𝓓f_l. This is contrary to that {f₁, f₂, ⋯, f_r} are representatives of the partition of E(S)\{0} induced by 𝓓|_E(S). So, we prove Claim A.

   We next verify:

2. For any 1 ≤ i, j, k ≤ r, ifM_ij ≠ ∅ andM_jk ≠ ∅, thenM_ik ≠ ∅.

   Pick x ∈ M_ij, y ∈ M_jk. So, x^* = f_j, f_jy = y and xy𝓛^*y𝓛^*f_k. It follows that (x)_fi,fj(y)_fj,fk = (xy)_fi,fk. Hence xy ∈ M_ik and M_ik ≠ ∅. 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 M_ij ≠ ∅. By Claim B, there is a path from i to l if and only if M_il ≠ ∅. 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 M_ij ≠ ∅ then i < j. In this case, the algebra (2) is a generalized upper triangular matrix algebra. The proof is completed. □

Corollary 3.5

LetSbe a strict RA semigroup andRa commutative ring. IfSis finite, thenR₀[S] has a generalized upper triangular matrix representation of degreeD_reg(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].

Corollary 3.7

LetSbe a right ample monoid andRa commutative ring. IfSis finite, thenR₀[S] has a generalized upper triangular matrix representation of degreeD_reg(S).

Example 3.8

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

where R_i is the subalgebra of R₀[P(Q)] generated by all paths from i to i, and R_ij 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, M_ei,ei = {e_i} 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

is an equivalence on the set Y := E(S(𝓒))\{0}. Let Y/π = {Y₁, Y₂, ⋯, Y_r}. Moreover, we let X_i = {A ∈ Obj(𝓒) : 1_B ∈ Y_i} and n_i = |X_i|. If pick A_i ∈ X_i, then by Theorem 3.2, R₀[S(𝓒)] is isomorphic to the generalized matrix algebra

Now let 𝓒 be a groupoid. In this case, Hom(A_i, A_j) = ∅ whenever i ≠ j, and further R₀[S(𝓒)] is isomorphic to the generalized matrix algebra

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].

Theorem 3.10

LetSbe a strict RA semigroup andRa commutative ring. IfSis finite, thenR₀[S] is semiprimitive if and only if the following conditions are satisfied:

1. Sis an inverse semigroup;

2. for every maximum subgroupGofS, 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

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

So, J(M_ni(R₀[M_ii])) = 0 and for i, j = 1, 2, ⋯, r, M_ni,nj(R₀[M_ij]) = 0 if i ≠ j. Thus R₀[M_ii] is semiprimitive, so that R₀[M_ii] is semiprimitive; and M_ij = ∅ if i ≠ j. Since S is finite, M_ii is finite, so that M_ii is a subgroup of S. But for all a ∈ M_ii, a𝓛^∗f_i giving a𝓛f_i, so a𝓗f_i since M_ii is a subgroup of S, thus M_ii is indeed a maximum subgroup of S. By the choice of f_i’s, we know that any nonzero idempotent of S is 𝓓-related to some f_i, thereby any nonzero maximum subgroup of S is isomorphic to some M_ii. Therefore the condition (B) is satisfied.

For the condition (A), we prove only that any nonzero element of S is regular. Let a ∈ S\{0} and a^# = e, a^∗ = f. Let e𝓓f_i and f𝓓f_j. Then there exist u ∈ eSf_k, v ∈ f_kSe, x ∈ f_jSf, y ∈ fSf_j such that e = uv, f_i = vu, f = yx, f_j = xy. It follows that f_i𝓡v𝓛e and f𝓡y𝓛f_j. Thus vay ∈ M_ij since (v)_fi,e(a)_e,f(y)_f,fj = (vay)_fi,fj. We consider the following two cases:

1. If f_i = f_j, then vay ∈ M_ii and so as M_ii is a subgroup of S, there exists b ∈ S such that vay = vaybvay, so that a = u ⋅ vay ⋅ x = u ⋅ vaybvay ⋅ x = uv ⋅ aybva ⋅ yx = a ⋅ ybv ⋅ a. It follows that a is regular.

2. Assume f_i ≠ f_j. By the foregoing proof, M_ij = ∅, contrary to the fact: vay ∈ M_ij.

Consequently, a is regular, as required. □

Corollary 3.11

LetSbe a right ample monoid. IfSis finite, thenK₀[S] is semiprimitive if and only if the following conditions are satisfied:

1. Sis an inverse semigroup;

2. for every maximum subgroupGofS, R[G] is semiprimitive.

Corollary 3.12

LetSbe an inverse semigroup. IfSis finite, thenK₀[S] is semiprimitive if and only if for every maximum subgroupG of S, R[G] is semiprimitive.

Lemma 4.1

LetRbe a ring with unity. IfRis left (right) perfect, then

1. Rdoes not contain an infinite orthogonal set of nonzero idempotents.

2. Any quotient ofRis left (right) perfect.

Lemma 4.2

LetSbe an arbitrary semigroup andKa field. IfK₀[S] is a left self-injectiveK-algebra, then

1. There exist idealsS_i, i = 0, 1, ⋯, n, such thatθ = S₀ ⊏ S₁ ⊏ ⋯ ⊏ S_n = Sand the Rees quotientsS_i/S_i+1are completely 0-simple orT-nilpotent.

2. ([15, Lemmas 9 and 10, p.192]) Ssatisfies 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] K0[S]Rad(K0[S])is regular, where Rad(K₀[S]) is the Jacobson radical ofK₀[S].

Lemma 4.3

LetSbe a right ample semigroup. IfK₀[S] is left self-injective, then

1. |E(S)| < ∞.

2. For anye ∈ E(S)\{0}, the subsetM_e,e = {x ∈ S: ex = x, x^∗ = e} is a finite subgroup ofS.

Proof

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

   1. θ = S₀ ⊏ S₁ ⊏ ⋯ ⊏ S_n = S; and

   2. the Rees quotients S_i/S_i+1 are completely 0-simple or T-nilpotent.

   So, S = {θ} ⊔ ⊔i=1nSi∖Si−1. This shows that for any e ∈ E(S)\{θ}, there exists i ≥ 1 such that S_i/S_i−1 is completely 0-simple and e ∈ S_i/S_i−1. Now, to verify that E(S) is finite, it suffices to prove that any completely 0-simple semigroup S_i/S_i−1 is finite.

   Now let S_i/S_i−1 be a completely 0-simple semigroup. By the definition of Rees quotient, any nonzero idempotents of S_i/S_i−1 is an idempotent of S. But E(S) is a semilattice, so S_i/S_i−1 is an inverse semigroup. Thus S_i/S_i−1 is a Brandt semigroup. By the structure theorem of Brandt semigroups in [30], S_i/S_i−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 S_i/S_i−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, 1_G, a) where 1_G is the identity of G.

   For convenience, we identify S_i/S_i−1 with T. Now, we need only to show that |I| < ∞. By Lemma 4.2, K₀[S] is left perfect, and so K0[S]Rad(K0[S]) is semisimple. It follows that K0[S]Rad(K0[S]) has an identity. It is easy to see that K₀[S_i−1] is an ideal of K₀[S]. Consider the algebra W:=K0[S]Rad(K0[S])+K0[Si−1]. 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

   1. (a, 1_G, a)(b, 1_G, b) = θ whenever a ≠ b;

   2. W has an unity;

   3. (a, 1_G, a) + Rad(K₀[S]) + K₀[S_i−1] ≢ (b, 1_G, b) + Rad(K₀[S]) + K₀[S_i−1] whenever a ≠ b.

   Again by the property that (a, 1_G, a)(b, 1_G, b) = θ whenever a ≠ b, 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 K0[S]Rad(K0[S]) is semisimple, we know that K0[S]Rad(K0[S]) 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, S_i/S_i−1 is finite, and so |E(S)| < ∞.

2. Let a ∈ M_e,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 x ∈ S¹ such that aⁿ = aⁿ⁺¹x. Hence aⁿ⁻¹ = a^∗aⁿ⁻¹ = a^∗aⁿx = aⁿx. Continuing this process, we can obtain a^∗ = ax. But a = aa^∗, now a𝓡a^∗. Thus a is regular. Therefore a𝓗a^∗ since a𝓛^∗a^∗. So, M_e,e is a subgroup of S and further by Lemma 4.2, M_e,e is finite. □