# A Note on Locally Inverse Semigroup Algebras

Authors: Xiaojiang Guo
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008-02-05)

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## Abstract

LetR be a commutative ring and S a finite locally inverse semigroup.
It is proved that the semigroup algebra R[S] is isomorphic to the direct
product of Munn algebras ℳ(R[GJ],mJ,nJ;PJ) with
J∈S/𝒥, where
mJ is the number of ℛ-classes in J, nJ the number of ℒ-classes in J, and
GJ a maximum subgroup of J. As applications, we obtain the sufficient
and necessary conditions for the semigroup algebra of a finite locally
inverse semigroup to be semisimple.


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## Body

## 1. Main Results

A regular
semigroupS is called a locally
inverse semigroup if for all idempotent e∈S, the local submonoid eSe is an inverse
semigroup under the multiplication of S. Inverse semigroups are locally inverse semigroups.
Inverse semigroup algebras are a class of semigroup algebras which is widely
investigated. One of fundamentally important results is that a finite inverse
semigroup algebra is the direct product of full matrix algebras over group
algebras of the maximum subgroups of this finite inverse semigroup. Consider
that all local submonoids of a locally inverse semigroup are inverse
semigroups, it is a very natural problem whether a finite locally inverse
semigroup algebra has a similar representation to inverse semigroup algebras.
This is the main topic of this note.Let𝒜 be an R-algebra. Let m and n be positive
integers, and let P be a fixed n×m matrix over 𝒜. Let ℳ:=ℳ(𝒜;m,n;P) be the vector
space of all m×n matrices over 𝒜. Define a product ∘ in ℳ by
(1.1)A∘B=APB(A,B∈ℳ),
where APB is the usual
matrix product of A, P, and B. Then ℳ is an algebra
over R. Following [1], we call ℳ the Munn m×n  matrix algebra
over  𝒜  with sandwich
matrix  P.By asemisimple semigroup, we mean a semigroup
each of whose principal factor is either a completely 0-simple semigroup or a
completely simple semigroup. It is well known that a finite regular semigroup
is semisimple. The Rees theorem tells us that any completely 0-simple
semigroup (completely simple semigroup) is isomorphic to some Rees matrix
semigroup ℳ0(G,I,Λ;P) (ℳ(G,I,Λ;P)), and vice
versa (for Rees matrix semigroups, refer to [1]). In what follows, by the
phrase “Let S=⋃J∈S/𝒥ℳ0(GJ;IJ,ΛJ;PJ) be a finite
regular semigroup,” we mean that S is a finite
regular semigroup in which the principal factor of S determined by
the 𝒥-class J is isomorphic
to the Rees matrix semigroup ℳ0(GJ;IJ,ΛJ;PJ) or ℳ(GJ;IJ,ΛJ;PJ) for any J∈S/𝒥.The following is the main result of this paper.Theorem 1.1.
LetS=⋃J∈S/𝒥ℳ0(GJ,IJ,ΛJ;PJ) be a finite
locally inverse semigroup. Then the semigroup algebra R[S]
is isomorphic
to the direct product of ℳ(R[GJ];|IJ|,|ΛJ|;PJ) with J∈S/𝒥.Based on Theorem1.1 and [1, Lemma 5.17, page 162, and Lemma 5.18, page 163], the following corollary is straightforward.Corollary 1.2.
LetS=⋃J∈S/𝒥ℳ0(GJ,IJ,ΛJ;PJ) be a finite
locally inverse semigroup. Then the semigroup algebra R[S]
has an identity
if and only if |IJ|=|ΛJ| and PJ is invertible
in the full matrix algebra M|IJ|(R[GJ]) for all J∈S/𝒥.Reference [1, Lemma 5.18, page 163] told us that ℳ(R[GJ],mJ,nJ;PJ) is isomorphic
to the full matrix algebra MnJ(R[GJ]) if ℳ(R[GJ],mJ,nJ;PJ) has an
identity. Now, we have the following.Corollary 1.3.
LetS=⋃J∈S/𝒥ℳ0(GJ,IJ,ΛJ;PJ) be a finite
locally inverse semigroup. If R[S]
has an
identity, then R[S]
is isomorphic
to the direct product of the full matrix algebras M|IJ|(R[GJ]) with J∈S/𝒥.The following corollary is a consequence of Corollary1.3.Corollary 1.4.
LetS=⋃J∈S/𝒥ℳ0(GJ,IJ,ΛJ;PJ) be a finite
locally inverse semigroup. Then the semigroup algebra R[S]
is semisimple
if and only if for all J∈S/𝒥,
(1)
|IJ|=|ΛJ|;(2)
PJ is invertible
in the full matrix algebra M|IJ|(R[GJ]);(3)
R[GJ] is
semisimple.


## 2. Proof of Theorem1.1

For our
purpose, we have the Möbius inversion theorem [2].Lemma 2.1.
Let (P,≤)
be a locally
finite partially ordered set (i.e., intervals are finite) in which each
principal ideal has a maximum and G
be an Abelian
group. Suppose that f:P→G
is a function
and define g:P→G by g(x)=∑y≤xf(y). Then f(x)=∑y≤xg(y)μ(x,y), where μ
is a Möbius
function.Now assume thatS is a regular
semigroup and a,b∈S. Define
(2.1)a≤b⟺thereexiste,f∈E(S)suchthata=eb=bf.
Then ≤ is a partial
order on S. Following [3], we call ≤ the natural
partial order on S. Equivalently, a≤b if and only if
for every (for some) f∈E(Rb)(f∈E(Lb)), there exists e∈E(Ra)(e∈E(La)) such that e≤f and a=eb(a=be). Moreover, Nambooripad [3, 4] proved that S is a locally
inverse semigroup if and only if the natural partial order ≤ is compatible
with respect to the multiplication of S.Lemma 2.2.
LetS
be a locally
inverse semigroup and a,b∈S. Then for any u≤ab, there exist x≤a and y≤b such that u=xy, x∈Ru and y∈Lu.Proof.
For anye∈E(Ra), we have ea=a and eab=ab. Let z be an inverse
of ab. Clearly, abz∈E(Rab). Note that eabz=abz. It is easy to check that abze∈E(S),abze≤e, and abzℛabze. Hence abzeℛab and there exists g∈E(S) such that u=gab and g≤abze(≤e). Thus ga≤a. On the other hand, since ℛ is a left
congruence and since abzeℛab, we have u=gabℛgabze=g; while since aℛe, we have gaℛge=g. These imply that uℛga. Dually, we have h∈E(S) such that u=abh, bh≤b and uℒbh. Since u=gab=abh=uh=(ga)(bh), we know that ga and bh are the
required elements x and y.Define a multiplication⊗ on S0=S∪{0} by
(2.2)x⊗y={xyifx≠0,y≠0,andy,xy∈Jx;0otherwise,
where xy is the product
of x and y in S. By the arguments of [4, page 9], (S0,⊗) is a semigroup.
We denote by S⊗ the semigroup (S0,⊗). For any J∈S/𝒥, we denote J0=J∪{0}. It is easy to check that (J0,⊗) is a
subsemigroup of S⊗, which is isomorphic to the principal factor of S determined by J. We will denote the semigroup (J0,⊗) by J⊗. By the definition of ⊗, it is easy to see that in the semigroup S⊗,(i)
Jx⊗⊗Jx⊗⊆Jx⊗ for all x∈S;(ii)
Jx⊗⊗Jy⊗=0 for all x,y∈S such that x∉Jy.
Thus R0[S⊗] is the direct
sum of the contracted semigroup algebras R0[J⊗] with J∈S/𝒥. Note that J⊗ is isomorphic
to some principal factor of S. We observe that J⊗ is a completely 0-simple
semigroup since S is a semisimple
semigroup, and thus J⊗ is isomorphic
to some Rees matrix semigroup ℳ0(GJ,IJ,ΛJ;PJ). By a result of [1], R0[ℳ0(GJ,IJ,ΛJ;PJ)] is isomorphic
to ℳ(R[GJ],|IJ|,|ΛJ|;PJ). Consequently, to verify Theorem 1.1, we need only to
prove that R[S] is isomorphic
to R0[S⊗].For the convenience of description, we introduce the
semigroupS¯. Put S¯={x¯∣x∈S}∪{0}. Define a multiplication on S¯ as follows:
(2.3)x¯∗y¯=x⊗y¯,
where we will identify 0¯ with 0. It is easy to see that S¯ is isomorphic
to S⊗. Hence the contracted semigroup algebra R0[S¯] is isomorphic
to the contracted semigroup algebra R0[S⊗]. For J∈S/𝒥, we denote J¯={x¯∣x∈J}∪{0}. It is easy to check that (J¯,∗) is a subsemigroup
of S¯ isomorphic to
the semigroup J⊗. So, for any J,K∈S/𝒥, we have
(2.4)J¯∗K¯{⊆J¯ifK=J,=0otherwise.For Theorem1.1, it remains to prove the following
lemma.Lemma 2.3.
R[S]≅R0[S¯].Proof.
We consider the mappingφ:R[S]→R0[S¯] given on the
basis by φ(s)=∑t≤st¯(s∈S). Clearly, φ is well
defined. Of course, φ and •¯ may be regarded
as the mappings of the ordered set (S,≤) into the
additive group of R0[S¯]. Now, by applying the Möbius inversion theorem to the
mappings φ and •¯, we have
(2.5)s¯=∑t≤sφ(t)μ(t,s)=φ(∑t≤stμ(t,s)),
where μ is the Möbius
function for (S,≤). Hence φ is surjective.
We will prove thatφ is injective.
For α0=∑x∈Spx0x∈R[S], we denote by supp(α0) the set {x∈S∣px0≠0} and by M(α0) the set of
maximal elements in the set supp(α0) with respect to
the partial order ≤. In recurrence, we define αn=αn−1−∑x∈M(αn−1)pxn−1x, where αn=∑x∈supp(αn)pxnx. Let βn=∑x∈supp(βn)qxnx with n=0,1,2,…. If φ(αn)=φ(βn), then by the definition of φ, ∑x∈M(αn)pxx¯+Γαn=φ(αn)=φ(βn)=∑y∈M(βn)qyny¯+Γβn, where Γαn=∑x∈M(αn)∑y∈S,y<xpyny¯ and Γβn=∑x∈M(βn)∑y∈S,y<xqyny¯, and hence ∑x∈M(αn)pxnx¯=∑x∈M(βn)qxnx¯, thus M(αn)=M(βn) and pxn=qxn for any x∈M(αn). This can imply the following.
Fact 2.4.
Ifφ(αn)=φ(βn), then M(αn)=M(βn) and by the
definition of φ, φ(αn+1)=φ(βn+1).
By the
definition ofφ, the following facts are immediate.
Fact 2.5.
αn=βn if and only if M(αn)=M(βn) and αn+1=βn+1.
Fact 2.6.
Ifφ(αn)=φ(βn) and M(αn)=supp(αn),M(βn)=supp(βn), then αn=βn.
Note that|supp(α0)|<∞ and supp(αn+1)⊆supp(αn). We thus have a smallest integer k such that M(αk)=supp(αk). Clearly, αk+1=0. This means that k is the smallest
integer t such that αt+1=0. Similarly, there exists the smallest integer l such that βl+1=0 and M(βl)=supp(βl). Now, assume φ(α0)=φ(β0). By using Fact 2.4 repeatedly,
(2.6)φ(α1)=φ(β1),φ(α2)=φ(β2),…,φ(αk+1)=φ(βk+1).
But φ(αk+1)=0, we have φ(βk+1)=0 and by the
definition of φ, βk+1=0. Thus k+1≥l+1 by the
minimality of l, and k≥l. Similarly, l≥k. Therefore k=l. Since φ(αk)=φ(βk), by Fact 2.6, we have αk=βk since M(αk)=supp(αk) and M(βl)=supp(βl). Again by the hypothesis φ(α0)=φ(β0), and by Fact 2.4, M(α0)=M(β0); and by (2.6), M(α1)=M(β1),M(α2)=M(β2),…,M(αk)=M(βk). By Fact 2.5, M(αk−1)=M(βk−1); and αk=βk imply αk−1=βk−1; moreover, by using Fact 2.5 repeatedly, αk−2=βk−2,…,α1=β1 and α0=β0. We have now proved that φ is injective.
Finally, for anys,t∈S, by (2.4), we have
(2.7)s¯*t¯={st¯ifs,t∈Jst,0otherwise,
and by Lemma 2.2,
(2.8)φ(s)*φ(t)=(∑x≤sx¯)*(∑y≤ty¯)=∑x∈Jst,x≤s∑y∈Jst,y≤tx¯*y¯=∑x∈Jst,x≤s∑y∈Jst,y≤txy¯.
Moreover, by Lemma 2.2, we have
(2.9)φ(st)=∑u≤stu¯=∑x∈Jst,x≤s∑y∈Jst,y≤txy¯=∑x≤s,x∈Jst∑y≤t,y∈Jstx¯*y¯=φ(s)*φ(t).
Thus φ is a
homomorphism of R[S] into R0[S¯]. Consequently, φ is an
isomorphism of R[S] onto R0[S¯].



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