Self-injectivity of semigroup algebras

Abstract It is proved that for an IC abundant semigroup (a primitive abundant semigroup; a primitively semisimple semigroup) S and a field K, if K 0[S] is right (left) self-injective, then S is a finite regular semigroup. This extends and enriches the related results of Okniński on self-injective algebras of regular semigroups, and affirmatively answers Okniński’s problem: does that a semigroup algebra K[S] is a right (respectively, left) self-injective imply that S is finite? (Semigroup Algebras, Marcel Dekker, 1990), for IC abundant semigroups (primitively semisimple semigroups; primitive abundant semigroups). Moreover, we determine the structure of K 0[S] being right (left) self-injective when K 0[S] has a unity. As their applications, we determine some sufficient and necessary conditions for the algebra of an IC abundant semigroup (a primitively semisimple semigroup; a primitive abundant semigroup) over a field to be semisimple.


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: IC abundant semigroups Abundant semigroups .
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 selfinjective; 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 − ⁎ ⁎ 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: Theorem. Let K be a field and S be in one of the following cases: (a) primitive abundant semigroups; (b) IC abundant semigroups; and (c) primitively semisimple semigroups. If K 0 [S] is right (left) self-injective, then S is a finite regular semigroup.
As its applications, we determine when the algebra of IC abundant semigroups (respectively, primitively semisimple semigroups; primitive abundant semigroups) is semismiple (Theorem 6.2).

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 .

IC abundant semigroups
The Green's relations on a semigroup are well known; see for example [19,Chapter II]. As generalizations of Green'sand -relations, we have ⁎and ⁎ -relations defined by  [12].
right *-ideal is defined. Moreover, an ideal of S is a *-ideal of S if it is both a left *-ideal and a right *-ideal. For a ∈ S, we denote by J*(a) the smallest *-ideal of S containing a. Following Fountain [12], we define = ⊓ Let I, Λ be nonempty sets and let Γ be a nonempty set indexing partitions P(I) = {I α :α ∈ Γ}, P(Λ) = {Λ α :α ∈ Γ} of I and Λ, respectively. For each pair (α,β) ∈ Γ × Γ, let M αβ be a set such that for each α ∈ Γ, T α := M αα is a monoid and for α ≠ β, either M αβ = ∅ or M αβ is a (T α ,T β )-bisystem. Let 0 be a symbol not in any M αβ . By the (α,β)-block of an I × Λ matrix we mean those (i,λ) positions with i ∈ I α , λ ∈ Λ β . The (α,α)-blocks are called the diagonal blocks of the matrix. Following the usual convention, we use (a) iλ to denote the I × Λ matrix with entry a in the (i,λ) position and zeros elsewhere, and denote by 0 the I × Λ matrix all of whose entries are 0. Let P = (p λi ) be an Λ × I sandwich matrix where a non-zero entry in the (α,β)-block is a member of M αβ . Suppose that the following conditions are satisfied: is commutative, where id αβ is the identity mapping on M αβ . (C) (In what follows, we simply denote (a ⊗ b)ϕ αβγ by ab, for a ∈ M αβ , b ∈ M αγ ). If a, a 1 , a 2 ∈ M αβ , b, b 1 , b 2 ∈ M αγ , then ab 1 = ab 2 implies b 1 = b 2 ; a 1 b = a 2 b implies a 1 = a 2 . Clearly, each T α is a cancellative monoid.
(U) For each a ∈ Γ and each λ ∈ Λ α (i ∈ I α ), there is a member i of I α (λ of Λ α ) such that p λi is a unit of T α . and A, B ∈ S. If A = 0 or B = 0, then APB = 0. Assume that A = (a) iλ and B = (b) jμ are non-zero. If p λj = 0, then (ap λj )b = 0 = a(p λj b) so that APB = 0. Assume that p λj ≠ 0 and let (i, λ) ∈ I α × Λ β , (j, μ) ∈ I γ × Λ δ . Then, a ∈ M αβ , b ∈ M γδ and p λj ∈ M βγ so that by Condition (M),we see that (ap λj )b = a(p λj b) is a well-defined member of M αδ and (ap λj b) iμ ∈ S. Thus, we have a product ∘ defined on S by A ∘ B = APB. We can easily check that (S,∘) is an abundant semigroup which is primitive, and called the PA blocked Rees matrix semigroup with the sandwich matrix P. For the sake of convenience, we denote this semigroup by  αβ satisfy the following conditions: Q has sources and sinks; (iii) The vertex set Γ of ( ) Q can be labeled Γ = {1, 2,…, n} in such a way that (i,j) ∈ E implies i < j.
Proof. We first prove Claim (*): For any α, β ∈ Γ such that α ≠ β, at most one of M αβ and M βα is not empty. If not, take some u ∈ M αβ , some v ∈ M βα , some k ∈ I α , and some λ ∈ Λ β . By the definition of blocked Rees matrix semigroup and Condition (U), there exist μ ∈ Λ β , l ∈ I β such that the entry p μl of the sandwich matrix P is in But T α is a group, so there exists a ∈ T α such that up μl va = 1 α . It follows that u·p μl va·u = 1 α u = u, contrary to Condition (R). We prove Claim (*). Let us consider the quiver ( ) Q . By Claim (*), we know that ( ) Q is acyclic and by [

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.  αβ be a primitive abundant semigroup and K be a field.
s u p p a n d : (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 since T α is a cancellative monoid and αβ be a primitive abundant semigroup and K be a field.
Proof. We first verify 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 ⊓ = ∅ aT bT β β since T β is a group. Take some i 0 ∈ I α and set , , a n d : Again by β is a sink of ( ) Q , we know that M βγ ≠ ∅ for any γ ≠ β. By definition, we have that Fact ( ‡). The entry p λj of P is equal to 0 whenever λ ∈ Λ β , j ∈ I γ .
, a routine computation shows that the mapping φ is defined by the linear span of the mapping: and, by Condition (C), ≠ w p a w p a ηi ηi 1 2 0 0 for any w 1 , w 2 ∈ T α and w 1 ≠ w 2 , we get that = wp a a ηi 0 for any contrary to the definition of . Thus, M αβ = aT β and we prove Fact ( †). Now, let M πζ ≠ ∅. If ζ is not a sink of ( ) Q , then as ( ) Q 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 . So, by Condition (M), and |M πζ d| < ∞. But, by Condition (C), |M πζ | = |M πζ d|, now |M πζ | < ∞. This proves the lemma. □ Proof. Let ε be a left identity, and I ε and Λ ε have the same meaning as in Lemma 3.3. Set It is easy to know that where is a K-space with a basis X. On the other hand, for any ∈ 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 . It follows that ⊆ . This means that is a subspace of . Therefore, (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, . By multiplying with , we easily know that the contracted semigroup algebra is an algebra with the usual matrix addition and the multiplication is defined by: for where the right side is the usual matrix multiplication of A, P and B. Let especially, ε n P nn ε n = ε n ≠ 0. So,  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 Obviously, , contrary to Eq. (7). Thus, M in = ∅ for any i < n. Moreover, P in = 0 in the sandwich matrix P, for any i < n.

Algebras of IC abundant semigroups
In this section, we shall research the self-injectivity of algebras of IC abundant semigroups. (ii) We claim: under ≤, S has a minimal nonzero idempotent e 0 ; for, if no, S has a chain of nonzero idempotents: e 1 > e 2 > ⋯ > e n > ⋯, so ⋯Se n ⊂ ⋯ ⊂ Se 2 ⊂ Se 1 , contrary to Lemma 3.1 (ii). It is not difficult to know that e 0 is primitive. □  Proof. We first prove that for any α ∈ Γ, |I α | < ∞. Indeed, if I α is infinite and choose elements i 1 , i 2 ,… from I α , then by Condition (U), there exist elements λ 1 , λ 2 ,… of Λ α such that p λ i j j (the entry of the sandwich matrix P) is a unity of T α , for j = 1, 2,…. By a routine check, every ( ) (c) By computation, and so is infinite. It follows that This is contrary to Lemma 4.2. Thus, |I α | < ∞ for any α. On the other hand, by Lemma 4.1, S has finite regular -classes. So, J has finite regular -classes. But, by [[12], Proposition 2.6 (6) and Proposition 4.1], the number of regular -classes of J is equal to |Γ| + 1, now |Γ| < ∞. Thus, |I| < ∞ since = ⊔ ∈ I I α Γ α . For any i ∈ I, by Condition (U), there exists λ i ∈ Λ such that p λ i i (the entry of the sandwich matrix P) is a unit of some T α . So, ( ) − p λ i iλ has a left identity. In addition, by the same reason as Lemma 3.3 (ii), we can prove (iii). We omit the detail. □ The following lemma is a key result to research the self-injective algebras of IC abundant semigroups, which may be proved by revising the proof of Theorem 3.6. For the completeness, we give the proof. (ii) For any 1 ≤ i, j ≤ n and i ≠ j, whenever M ij ≠ ∅, we have i < j; (iii) 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. We shall use the notations in the proof of Theorem 3.6. By (5), and of course, not in . Notice that by (4)    in contradiction to (5). We have now proved that M in = ∅ for i = 1, 2,…,n − 1.
By applying the similar arguments as above to Proof. Assume that = . As pointed out in the Introduction, by hypothesis that is right self-injective, has a left identity and let ε be a left identity of . Then, = ⊕ ( − ) e ε e as right 1 -modules. Thus, ( − ) ε e is an injective right 1 -module. Hence, / ≅ ( − ) ε e and is an injective right 1 -module. Now, let J be a right ideal of ( / ) 1 and ϕ a ( / ) 1 -module homomorphism of J into / . Observe that the 1 -module and 1 is an injective 1 -module homomorphism and ϕ also an 1 -module homomorphism of J into / . By / is an injective right 1 -module, there exists an 1 -module homomorphism φ of ( / ) 1 into / such that ϕ = φι. On the other hand, for any + ∈ / x , it follows that there is + ∈ / u such that , it follows that φ is indeed an ( / ) 1 -module homomorphism. Therefore, / is a right injective ( / ) 1 -module, when / is right self-injective. □ Lemma 4.6. Let S be a semigroup and U an ideal. Then S is a regular semigroup if and only if U and the Rees quotient S/U are both regular.
Proof. We only verify the sufficiency. To the end, we assume that U and S/U are both regular. For any a ∈ S, if a ∈ U, then a is regular in S; if a ∈ S/U, then as S/U is regular, there is b ∈ S/U such that a ∘ b ∘ a = a in S/U, in this case, by the definition of Rees quotient, a ∘ b ∘ a = aba and so aba = a in S, it follows that a is regular in S. However, a is regular in S. Thus, S is a regular semigroup. □ We arrive now at the main result of this section, which generalizes the main result of Okniński on right (respectively, left) self-injective algebras of a regular semigroup (see [7,Theorem 2], which answers affirmatively the Okniński's problem mentioned in the Introduction for the IC abundant semigroup case. Theorem 4.7. Let S be an IC abundant semigroup and K a field. If K 0 [S] is right (respectively, left) selfinjective, then S is a finite regular semigroup. In this case, K 0 [S] is Artinian.
Proof. By Lemma 4.1, we pick a primitive idempotent f 1 of S. By Lemma 2.2, S 1 = J*(f 1 ) is a primitive abundant ideal of S.
If S = S 1 , then by Theorem 3.6, S is a finite regular semigroup.
Suppose that S 1 is a proper ideal of S. Then by Lemma 4.4, S 1 is a regular subsemigroup of S; and by Lemmas 4.3 and 4.5, K 0 [S/S 1 ] is right self-injective.
Case (i). If S/S 1 is primitive, then by Theorem 3.6, S/S 1 is regular and by Lemma 4.6, S is regular.
Case (ii). Assume that T 1 := S/S 1 is not primitive. By Lemma 4.6, S is regular if and only if T 1 is regular. On the other hand, by the definition of Green's -relation, it is not difficult to see that for any ideal I of S, J a ⊆ I for all a ∈ I. This shows that D r (T 1 ) < D r (S) where D r (T) stands for the number of nonzero regular -classes of T. By applying the similar argument to T 1 , there exits a primitive abundant ideal S 2 of T 1 such that (1) S 2 is a regular semigroup; (2) T 2 = T 1 /S 2 is an IC abundant semigroup (by Lemma 2.1); (3) S is regular if and only if T 2 is regular (by Lemma 4.6); (4) K 0 [T 2 ] is right self-injective; and (5) D r (T 2 ) < D r (T 1 ).
This proceedings can continue only finite times since |D r (S)| < ∞ (by Lemma 4.1). So, there exists a positive integer r such that (a) T r is a primitive abundant semigroup; (b) K 0 [(T r )] is right self-injective; (c) S is regular if and only if so is T r . Again by Theorem 3.6, T r is a finite regular semigroup. Therefore, S is a regular semigroup. By the result of [7], for a regular semigroup S, if K 0 [S] is right self-injective, then S is finite, we get that S is a finite regular semigroup. We have finished the proof. Proof. (i). Suppose that ( ) ∈ a b , S ⁎ . Because ⁎ is the smallest equivalence containing ⁎ and ⁎ , it follows from [19, Proposition 5.14, p. 28] that there exist x 1 , x 2 ,…,x 2n−1 ∈ S such that