Self-injectivity of semigroup algebras

1 Introduction

Recall that an algebra (possibly without unity) R is right self-injective if R is an injective right R-module. Dually, left self-injective algebra is defined. Equivalently, then R is right self-injective if and only if the right R ¹-module R satisfies the Baer condition, where R ¹ is the standard extension of R to an algebra with unity (see [1, Chap. 1]). In this case, R has a left unity. (Left; right) Self-injective algebras are known as the generalizations of Frobenius algebras. These classes of algebras play an important role and have become a central topic in algebras.

For group algebras, it is well known that the group algebra K[G] of a group G over the field K is right self-injective if and only if the group G is finite; if and only if K[G] is Frobenius (for detail, see [2, Theorem 3.2.8]). Along this direction, self-injective and Frobenius semigroup algebras of finite semigroups have been investigated by many authors (for references, see [3,4]). In particular, it was proved that in some cases, the finiteness of the semigroup is a necessary condition for the semigroup algebra to be right (respectively, left) self-injective; for example, the semigroup is an inverse semigroup, a countable semigroup, a regular semigroup, etc. (cf. [3,4,5,6,7,8,9,10]). So, Okniński raised a problem: does that K[S] is a right (respectively, left) self-injective imply that S is finite? (see [11, Problem 6, p. 328]).

A semigroup S is called right principal projective (rpp), if for any a ∈ S, the right principal ideal aS ¹, regarded as a right S ¹-system, is projective. We can dually define left principal projective semigroup (lpp semigroup). As in [12], an abundant semigroup is defined as a semigroup being both rpp and lpp. Moreover, El Qallali and Fountain [13] defined idempotent-connected (IC) abundant semigroups. Indeed, IC abundant semigroups become a large class of semigroups including the class of regular semigroups and that of cancellative monoids as its proper subclasses. These three classes of semigroups have relationships as follows:

As known, rpp semigroups come from rpp ring. In precise, a ring R is (lpp; rpp) pp if and only if the multiplicative semigroup of R is (lpp; rpp) pp.

Recently, Guo and Shum [14] proved that the semigroup K[S] of an ample semigroup S is right self-injective; if and only if K[S] is left self-injective; if and only if K[S] is quasi-Frobenius; if and only if K[S] is Frobenius; if and only if S is a finite inverse semigroup. These results show that the “distance” between the class of finite inverse semigroups and that of ample semigroups is the right (left) self-injectivity of semigroup algebras. So-called an ample semigroup is an IC abundant semigroup whose set of regular elements forms an inverse subsemigroup. The class of ample semigroups contains properly the class of inverse semigroups. Indeed, for the self-injectivity, the symmetry of ℒ ⁎ − ℛ ⁎ in semigroups need not be so important. For semigroup algebras of finite ample semigroups, see ref. [15]. Guo and Guo [16] pointed out that the mentioned results as above in [14] is valid when the ample semigroup is weakened into a strict RA semigroup or a strict LA semigroup, especially, a right (left) ample monoid.

By inspiring the result of Okniński in [7]: for a regular semigroup S, if K[S] is right (left) self-injective then S is finite, we have a natural problem: whether the Okniński problem is valid for IC abundant semigroups? This is the main aim of this paper. It is worthy to record here that K[S] is right (respectively, left) self-injective if and only if so is K ₀[S] (see [11, p. 188]). We shall prove the following result:

As its applications, we determine when the algebra of IC abundant semigroups (respectively, primitively semisimple semigroups; primitive abundant semigroups) is semismiple (Theorem 6.2).

2 Preliminaries

Throughout this paper, we shall use the notions and notations of the monographs of Okniński [11] and Kelarev [17]. For semigroups, the readers can be referred to the textbooks of Clifford [18] and Howie [19]. Let S be a semigroup; we denote the set of idempotents of S as E(S), and the semigroup obtained from S by adjoining an identity if S does not have one by S ¹.

3 Primitive abundant semigroup algebras

In this section, we determine when a primitive abundant semigroup algebra is right (left) self-injective. We first recall some known results.

(i) [7,8] There exist ideals S _i , i = 0, 1,…, n, such that θ = S 0 ⊏ S 1 ⊏ ⋯ ⊏ S n = S and the Rees quotients S _i /S _i+1 are completely 0-simple or T-nilpotent.

(ii) [11, Lemmas 9 and 10, p. 192] S satisfies the descending chain condition on principal left ideals and has no infinite subgroups.

(iii) [11, Lemma 1, p. 187] K ₀[S] has a left identity.

(i) [11, Lemma 3, p. 188] Let J be an ideal of A . If J ² = 0 and A / J is finite dimensional, then A is finite dimensional.

(ii) [11, Lemma 1, pp. 187–188] ann ℓ ( ann r ( W ) ) ⊆ W + ann ℓ ( A ) for any finitely generated left ideal W of A , where ann r ( X ) ( ann ℓ ( X ) ) stands for the right (left) annihilators of a subset X of A in A .

(i) | I | < ∞ and | Γ | < ∞ .

(ii) T _α is a finite group for any α.

For any ( u ) i λ ∈ ℳ , since ε(u) _iλ = (u) _iλ , there is (a) _kμ ∈ supp(ε) such that (a) _kμ (u) _iλ = (u) _iλ . It follows that k = i. So, i ∈ I _ε and i ⊆ I _ε . Thus, |I| < ∞ since |I _ε | < ∞. Note that, by definition, ⊔ α ∈ Γ I α is a partition of I. We can observe that |Γ| ≤ |I| < ∞.

(ii) Let x ∈ T _α and j ∈ I _α . By Condition (U), there exists ξ ∈ Λ _α such that the entry p _ξj of the sandwich matrix P is a unit of T _α . Because

there exists n such that

by Lemma 3.1. Thus, there exists ( v ) k γ ∈ ℳ such that ((xp _ξj ) ⁿ⁻¹ x) _jξ = (v) _kγ ((xp _ξj ) ⁿ x) _jξ . It follows that

and ( x p ξ j ) n − 1 x = p ξ j − 1 p ξ k v p γ j ( x p ξ j ) n x . Therefore, by Condition (C), 1 α = p ξ j − 1 p ξ k v p γ j ⋅ x p ξ j since T _α is a cancellative monoid and

So, xp _ξj is a unit in T _α , whereby x is invertible in T _α since p _ξj is a unit in T _α . Consequently, T _α is a group and by Lemma 3.1, T _α is finite.□

Fact (†). If β is a sink of Q ( ℳ ) and M _αβ ≠ ∅, then |M _αβ | = |T _β | < ∞.

Pick a ∈ M _αβ . We shall prove that M _αβ = aT _β . If not, there exists b ∈ M _αβ /aT _β . Obviously, bT _β ⊆ M _αβ and a T β ⊓ b T β = ∅ since T _β is a group. Take some i ₀ ∈ I _α and set

Again by β is a sink of Q ( ℳ ) , we know that M _βγ ≠ ∅ for any γ ≠ β. By definition, we have that p λ j ∈ M β γ ⊔ { 0 } for all j ∈ I _γ , so that

Fact (‡). The entry p _λj of P is equal to 0 whenever λ ∈ Λ _β , j ∈ I _γ .

By Fact (‡), a routine check shows that C := A ⊔ ℬ ⊔ { 0 } is a right ideal of ℳ . Moreover, K 0 [ C ] is a right ideal of K 0 [ ℳ ] . By Fact (‡), a routine computation shows that the mapping φ is defined by the linear span of the mapping:

is a homomorphism of K 0 [ C ] into K 0 [ ℳ ] . But K 0 [ ℳ ] is injective, so by Baer condition, there exists z ∈ K 0 [ ℳ ] such that

By Condition (U), there exists η ∈ Λ _α such that p η i 0 (the entry of the sandwich matrix P) is a unit of T _α . It is easy to see that

(a) ( p η i 0 − 1 ) i 0 η x = x for any x ∈ A ⊔ ℬ .

(b) ( p η i 0 − 1 ) i 0 η z ( p η i 0 − 1 ) i 0 η = ∑ w ∈ J k w ( w ) i 0 η k where J ⊆ T α .

For any ( a ) i 0 λ ∈ A , since, by Eq. (1),

and, by Condition (C), w 1 p η i 0 a ≠ w 2 p η i 0 a for any w ₁, w ₂ ∈ T _α and w ₁ ≠ w ₂, we get that w p η i 0 a = a for any w ∈ J. So, w = p η i 0 − 1 for any w ∈ J. Again by Eq. (2), ( a ) i 0 λ = ( ∑ w ∈ J k w ) ⋅ ( a ) i 0 λ and ∑ w ∈ J k w = 1 (the unity of K). Therefore, ( p η i 0 − 1 ) i 0 η z ( p η i 0 − 1 ) i 0 η = ( p η i 0 − 1 ) i 0 η . Let y ∈ ℬ . Then, by Eq. (1),

and ℬ = 0 , contrary to the definition of ℬ . Thus, M _αβ = aT _β and we prove Fact (†).

Now, let M _πζ ≠ ∅. If ζ is not a sink of Q ( ℳ ) , then as Q ( ℳ ) is acyclic and has only finite vertices, there exists a path ζ = α ₀ → α ₂ → ⋯ → α _n such that α _n is a sink of Q ( ℳ ) . So, by Condition (M),

For d ∈ M α 0 α n , since M π ζ d ⊆ M π α n , we have | M π ζ d | ≤ | M π α n | . By Fact (†), | M π α n | < ∞ and |M _πζ d| < ∞. But, by Condition (C), |M _πζ | = |M _πζ d|, now |M _πζ | < ∞. This proves the lemma.□

(a) K 0 [ ℳ ] = U ⊕ V (regarded as K-vector spaces);

(b) V 2 = 0 ;

(c) V is an ideal of K 0 [ ℳ ] .

So, dim K ( K 0 [ ℳ ] / V ) = dim K ( U ) . Take X = { ( a ) i λ ∈ ℳ : i ∈ I ε , λ ∈ Λ ε } . Notice that |I _ε | < ∞ and |Λ _ε | < ∞. By Lemma 3.4, we get |X| < ∞, it follows that dim K ( X ) < ∞ where X is a K-space with a basis X. On the other hand, for any u ∈ U , we know that εuε = u, and further for any (v) _kl ∈ supp(u), there exist (x) _iλ , (y) _jμ ∈ supp(ε), (z) _mn ∈ supp(u) such that (v) _kl = (x) _iλ (z) _mn (y) _jμ = (xp _λm zp _nj y) _iμ . Hence, k = i ∈ I _ε and l = μ ∈ Λ _ε . Thus, (v) _kl ∈ X and so u ∈ X . It follows that U ⊆ X . This means that U is a subspace of X . Therefore, dim K ( U ) ≤ dim K ( X ) and U is finite dimensional. Now by Lemma 3.2 (i), K 0 [ ℳ ] is finite dimensional. Consequently, ℳ is a finite semigroup.□

(a) Γ = {1, 2,…,n};

(b) For any 1 ≤ i, j ≤ n and i ≠ j, whenever M _ij ≠ ∅, we have i < j;

(c) T _i is a finite group, for any 1 ≤ i ≤ n. Moreover, we let |I _i | = p _i and |Λ _i | = q _i for any i. By computation,

where M p i × q j ( K [ M i j ] ) is the set consisting of all p _i × q _j matrices over K[M _ij ], and

where P i j ∈ M q i × p j ( M i j ⊔ { 0 } ) . By multiplying with ℳ , we easily know that the contracted semigroup algebra K 0 [ ℳ ] is an algebra with the usual matrix addition and the multiplication is defined by: for A , B ∈ K 0 [ ℳ ]

where the right side is the usual matrix multiplication of A, P and B.

Let

be a left identity of K 0 [ ℳ ] . Since,

we have

especially, ε _n P _nn ε _n = ε _n ≠ 0. So,

and

We next verify that

(*) All of M _in with 1 ≤ i ≤ n − 1 are empty sets.

We assume, on the contrary, that not all M _in with 1 ≤ i ≤ n − 1 are empty sets. Without loss of generality, assume that M _1n ≠ ∅ and u ∈ M _1n . By a routine check,

is a left ideal of K 0 [ ℳ ] generated by the finite set

(since all M _ij are finite). If

i.e.,

then UP _nn X _n = 0. It is not difficult to know that P n n X n ∈ M q n × q n ( K [ T n ] ) . Let U = (u)₁₁ and u ∈ M ₁₂. Let

We have

and ua _1j = 0 for any 1 ≤ j ≤ q _n . Now let J = supp(a _1j ) and a 1 j = ∑ v l ∈ J r l v l ( r l ∈ K ) . Note that by Condition (C) ux ≠ uy for any x ≠ y ∈ supp(a _1j ). So, ∑ l ∈ J r l ( u v l ) = u a 1 j = 0 implies that r _l = 0. Thus, a _1j = 0. If U = (u) _i1, then by applying a similar argument as above, we may obtain that a _ij = 0 for any 1 ≤ i, j ≤ q _n . So, P _nn X _n = 0. Now, by Eq. (4), X _n = ε _n P _nn X _n = 0. It follows that

and a routine check shows the reverse inclusion. Thus,

On the other hand, by computation and Lemma 3.2, we have