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 *K*_{0}[*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*

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

*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 *K*_{0}[*S*] is left perfect, there exist ideals *S*_{i}, *i* = 0, 1, ⋯, *n*, such that

$$\begin{array}{}{\displaystyle \theta ={S}_{0}\u228f{S}_{1}\u228f\cdots \u228f{S}_{n}=S}\end{array}$$

and the Rees quotients *S*_{i}/*S*_{i+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* *K*_{0}[*S*] *is a left self*-*injective* *K*-*algebra*, *then*

*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*.

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

[28, Theorem 1] *K*_{0}[*S*] *is left perfect*.

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

Moreover, we have

#### Lemma 4.3

*Let* *S* *be a right ample semigroup. If* *K*_{0}[*S*] *is left self*-*injective, then*

|*E*(*S*)| < ∞.

*For any* *e* ∈ *E*(*S*)\{0}, *the subset* *M*_{e,e} = {*x* ∈ *S*: *ex* = *x*, *x*^{∗} = *e*} *is a finite subgroup of* *S*.

#### Proof

By Lemma 4.2, we assume that *S*_{i}, *i* = 0, 1, ⋯, *n*, are ideals of *S* satisfying the conditions:

So, *S* = {*θ*} ⊔
$\begin{array}{}{\displaystyle \left({\bigsqcup}_{i=1}^{n}{S}_{i}\mathrm{\setminus}{S}_{i-1}\right).}\end{array}$
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

$$\begin{array}{}{\displaystyle (a,g,x)(b,h,y)=\left\{\begin{array}{cl}(a,gh,y)& \text{\hspace{0.17em}if\hspace{0.17em}}x=b;\\ & \\ \theta & \text{\hspace{0.17em}otherwise,}\end{array}\right.}\end{array}$$

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, *K*_{0}[*S*] is left perfect, and so
$\begin{array}{}{\displaystyle \frac{{K}_{0}[S]}{Rad({K}_{0}[S])}}\end{array}$
is semisimple. It follows that
$\begin{array}{}{\displaystyle \frac{{K}_{0}[S]}{Rad({K}_{0}[S])}}\end{array}$
has an identity. It is easy to see that *K*_{0}[*S*_{i−1}] is an ideal of *K*_{0}[*S*]. Consider the algebra
$\begin{array}{}{\displaystyle W:=\frac{{K}_{0}[S]}{Rad({K}_{0}[S])+{K}_{0}[{S}_{i-1}]}.}\end{array}$
Note that

$$\begin{array}{}{\displaystyle \frac{\frac{{K}_{0}[S]}{Rad({K}_{0}[S])}}{\frac{Rad({K}_{0}[S])+{K}_{0}[{S}_{i-1}]}{Rad({K}_{0}[S])}}\cong \frac{{K}_{0}[S]}{Rad({K}_{0}[S])+{K}_{0}[{S}_{i-1}]}\cong \frac{\frac{{K}_{0}[S]}{{K}_{0}[{S}_{i-1}]}}{\frac{Rad({K}_{0}[S])+{K}_{0}[{S}_{i-1}]}{{K}_{0}[{S}_{i-1}]}},}\end{array}$$

we can observe that

–

(*a*, 1_{G}, *a*)(*b*, 1_{G}, *b*) = *θ* whenever *a* ≠ *b*;

–

*W* has an unity;

–

(*a*, 1_{G}, *a*) + *Rad*(*K*_{0}[*S*]) + *K*_{0}[*S*_{i−1}] ≢ (*b*, 1_{G}, *b*) + *Rad*(*K*_{0}[*S*]) + *K*_{0}[*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

$$\begin{array}{}{\displaystyle X:=\{(a,{1}_{G},a)+Rad({K}_{0}[S])+{K}_{0}[{S}_{i-1}]:a\in I\}}\end{array}$$

is an orthogonal set of nonzero idempotents of *W*. On the other hand, since
$\begin{array}{}{\displaystyle \frac{{K}_{0}[S]}{Rad({K}_{0}[S])}}\end{array}$
is semisimple, we know that
$\begin{array}{}{\displaystyle \frac{{K}_{0}[S]}{Rad({K}_{0}[S])}}\end{array}$
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, *S*_{i}/*S*_{i−1} is finite, and so |*E*(*S*)| < ∞.

Let *a* ∈ *M*_{e,e}. Consider the chain of principal right ideals of *S*:

$$\begin{array}{}{\displaystyle \cdots {a}^{n}{S}^{1}\subseteq {a}^{n-1}{S}^{1}\subseteq \cdots \subseteq {a}^{2}{S}^{1}\subseteq a{S}^{1}.}\end{array}$$

By Lemma 4.2, there exists a positive integer *n* such that *a*^{n}S^{1} = *a*^{n+1}*S*^{1}. That is, there is *x* ∈ *S*^{1} such that *a*^{n} = *a*^{n+1}*x*. Hence *a*^{n−1} = *a*^{∗}*a*^{n−1} = *a*^{∗}*a*^{n}x = *a*^{n}x. Continuing this process, we can obtain *a*^{∗} = *ax*. But *a* = *aa*^{∗}, now *a*𝓡*a*^{∗}. Thus *a* is regular. Therefore *a*𝓗*a*^{∗} since *a*𝓛^{∗}*a*^{∗}. So, *M*_{e,e} is a subgroup of *S* and further by Lemma 4.2, *M*_{e,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 *K*_{0}[*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 *K*_{0}[*PRA*(*S*)]. By Theorem 3.4, we have that *M*_{ij} = ∅ if *j* < *i*, and

$$\begin{array}{}{\displaystyle {K}_{0}[S]:=\mathfrak{B}\cong \left(\begin{array}{cccc}{M}_{{n}_{1}}({K}_{0}[\overline{{M}_{11}}])& {M}_{{n}_{1},{n}_{2}}({K}_{0}[\overline{{M}_{12}}])& \cdots & {M}_{{n}_{1},{n}_{r}}({K}_{0}[\overline{{M}_{1r}}])\\ 0& {M}_{{n}_{2}}({K}_{0}[\overline{{M}_{22}}])& \cdots & {M}_{{n}_{2},{n}_{r}}({K}_{0}[\overline{{M}_{2r}}])\\ ..............& ................& ...& .................\\ 0& 0& \cdots & {M}_{{n}_{r}}({K}_{0}[\overline{{M}_{rr}}])\end{array}\right).}\end{array}$$

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

$$\begin{array}{}{\displaystyle \mathfrak{B}=\left(\begin{array}{cccc}{K}_{0}[\overline{{M}_{11}}]& {K}_{0}[\overline{{M}_{12}}]& \cdots & {K}_{0}[\overline{{M}_{1r}}]\\ 0& {K}_{0}[\overline{{M}_{22}}]& \cdots & {K}_{0}[\overline{{M}_{2r}}]\\ .........& .........& ...& .........\\ 0& 0& \cdots & {K}_{0}[\overline{{M}_{rr}}]\end{array}\right).}\end{array}$$

For 1 ≤ *j* ≤ *r*, we denote by *m*_{j} the smallest positive integer in the set {*i*:*M*_{ij} ≠ ∅}. By definition, *m*_{j} ≤ *j* for any *i*.

#### Lemma 4.4

*M*_{kl} ⋅ *M*_{mj,j} = 0 *whenever* *M*_{kl} ≠ *M*_{mj,mj}.

#### Proof

By definition, *M*_{k,mj} ⋅ *M*_{mj,j} ⊆ *M*_{k,j} and

$$\begin{array}{}{\displaystyle \overline{{M}_{kl}}\cdot \overline{{M}_{{m}_{j},j}}\left\{\begin{array}{rll}=& 0& \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}if\hspace{0.17em}}l\ne {m}_{i}\\ \subseteq & \overline{{M}_{k,j}}& \phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}if\hspace{0.17em}}l={m}_{j}.\end{array}\right.}\end{array}$$

In the second case, *k* ≤ *m*_{j}. This shows that *k* = *m*_{j} by the minimality of *m*_{j}. □

#### Proof

Lemma 4.3 results the case for *m*_{j} = *j*. Assume now that *m*_{j} < *j*. By Lemma 4.4, the algebra

$$\begin{array}{}{\displaystyle \mathfrak{C}:=\left(\begin{array}{ccccccc}0& \cdots & 0& 0& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& {K}_{0}[\overline{{M}_{{m}_{j},j}}]& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& 0& 0& \cdots & 0\end{array}\right)}\end{array}$$

is a left ideal of 𝔅. Pick *w* ∈ *M*_{mj,j} and define

$$\begin{array}{}{\displaystyle \theta :{M}_{{m}_{j},j}\to {M}_{{m}_{j},j};\phantom{\rule{thickmathspace}{0ex}}x\to \left\{\begin{array}{rl}x& \text{\hspace{0.17em}if\hspace{0.17em}}x\in {M}_{{m}_{j},{m}_{j}}w\\ & \\ 0& \text{\hspace{0.17em}if otherwise,}\end{array}\right.}\end{array}$$

and span linearly to *K*_{0}[*M*_{i,mi}]. Further define a map *ζ* of ℭ into 𝔅 by

$$\begin{array}{}{\displaystyle X=\left(\begin{array}{ccccccc}0& \cdots & 0& 0& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& (x{)}_{{m}_{j},j}& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& 0& 0& \cdots & 0\end{array}\right)\to \zeta (X)=\left(\begin{array}{ccccccc}0& \cdots & 0& 0& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& (\theta (x){)}_{{m}_{j},j}& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& 0& 0& \cdots & 0\end{array}\right).}\end{array}$$

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* = (*u*_{kl}). Then *u*_{jj} =
$\begin{array}{}\sum _{k=1}^{n}\end{array}$
*r*_{k} *a*_{k} where *r*_{k} ∈ *K*, *a*_{k} ∈ *M*_{jj}. Since *ζ*(*X*) = *XU*, we have (*θ*(*x*))_{mj,j} = (*θ*(*x*)_{mj,j} *u*_{jj} = (
$\begin{array}{}\sum _{k=1}^{n}\end{array}$
*r*_{k}xa_{k})_{mj,j} and

$$\begin{array}{}{\displaystyle x=\sum _{k=1}^{n}{r}_{k}x{a}_{k}.}\end{array}$$(3)

So, we may let *n*_{1}, *n*_{2}, ⋯, *n*_{s} be positive integers such that

–

*n*_{1} + *n*_{2} + ⋯ + *n*_{s} = *n*;

–

(∗) *xa*_{l} = *x*, for *l* = 1, 2, ⋯, *n*_{1};

–

(∗∗) *xa*_{nq+1} = *xa*_{nq+2} = ⋯ = *xa*_{nq+nq+1−1} = *b*_{q} for *q* = 1,2, ⋯, *s*, where *b*_{1}, *b*_{2}, ⋯, *b*_{s} are different elements in *M*_{jj}.

Now by Eq. (3), we get *r*_{1} + *r*_{2} + ⋯ *r*_{n1} = 1 and *r*_{nl+1}+*r*_{nl+2} + ⋯ *r*_{nl+nl+1−1} = 0 for *l* = 1, 2, ⋯, *s*. Note that *f*_{j} = *x*^{∗} 𝓛^{∗}*x*. Therefore by Eq. (∗), *a*_{l} = *f*_{j}a_{l} = *x*^{∗}*a*_{l} = *x*^{∗} = *f*_{j}, for *l* = 1, 2, ⋯, *n*_{1}; and by Eq. (∗∗), *x*^{∗}*a*_{nq+1} = *x*^{∗}*a*_{nq+2} = ⋯ = *x*^{∗}*a*_{nq+nq+1−1} for *q* = 1, 2, ⋯, *s*, so that as *x*^{∗} = *f*_{j}, *a*_{nq+1} = *a*_{nq+2} = ⋯ = *a*_{nq+nq+1−1} for *q* = 1, 2, ⋯, *s*. Consequently, *u*_{jj} = *f*_{j}.

Now, for any *x* ∈ *M*_{mj,j}, (*θ*(*x*))_{mj,j} = (*x*)_{mj,j}*u*_{jj} = (*xf*_{j})_{mj,j} = (*x*)_{mj,j}. This shows that *θ*(*x*) = *x* and *M*_{mj,j} = *M*_{mj,mj}*w* since *θ*(*x*) = 0 for any *x* ∈ *M*_{mj,j}\*M*_{mj,mj}*w*. It follows that |*M*_{mj,j}| < ∞ since *M*_{mj,mj} = *M*_{fj} ⊆ *M*_{fj,fj} and by Lemma 4.3 (*B*), is a finite subgroup. □

#### Proof

If *i* = *m*_{j}, nothing is to prove.

If *m*_{j} ≠ *i* and *M*_{ij} ≠ ∅, then *m*_{j} < *i*. Let *i*_{0} be the smallest positive integer of the set

$$\begin{array}{}{\displaystyle Y=\{k:\text{there exist}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}k={k}_{0},{k}_{1},{k}_{2},\cdots ,{k}_{m}=i\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{such that}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{M}_{{K}_{l-1},{k}_{l}}\ne \mathrm{\varnothing}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}l=1,2,\cdots ,m-1\}.}\end{array}$$

We shall prove *i*_{0} = *m*_{j}. Assume on the contrary that *i*_{0} ≠ *m*_{j}. By definition, *M*_{k0j} ≠ ∅ and specially *M*_{i0j} ≠ ∅. Thus *m*_{j} < *i*_{0}. By the minimality of *i*_{0}, *M*_{ki} = ∅ for 1 ≤ *k* < *i*_{0}, and further *M*_{ki0} ⋅*M*_{i0j} = 0. This and Lemma 4.4 show that

$$\begin{array}{}{\displaystyle \mathfrak{D}=\left(\begin{array}{ccccccc}0& \cdots & 0& 0& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& {K}_{0}[\overline{{M}_{{m}_{j},j}}]& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& {K}_{0}[\overline{{M}_{{i}_{0},j}}]& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& 0& 0& \cdots & 0\end{array}\right)}\end{array}$$

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

$$\begin{array}{}{\displaystyle A=\left(\begin{array}{ccccccc}0& \cdots & 0& 0& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& (a{)}_{{m}_{j},j}& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& (b{)}_{{i}_{0},j}& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& 0& 0& \cdots & 0\end{array}\right)\to \eta (A)=\left(\begin{array}{ccccccc}0& \cdots & 0& 0& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& (a{)}_{{m}_{j},j}& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& 0& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& 0& 0& \cdots & 0\end{array}\right)}\end{array}$$

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,j}*v*_{jj} and 0 = (*b*)_{mj,j}*v*_{jj} where *V* = (*v*_{ij}). By the first equality, *av*_{jj} = *a* and further by a similar arguments as proving *u*_{jj} = *f*_{j} in the proof of Lemma 4.5, *f*_{j} = *v*_{jj} ≠ 0; by the second equality, 0 = (*b*)_{mj,j}*v*_{jj} = (*b**f*_{j})_{mj,j} = (*b*)_{mj,j} and *b* = 0, so that as *b*^{∗} = *f*_{j}, we get 0 = *f*_{j}, thus *v*_{jj} = 0, contrary to the foregoing proof: *v*_{jj} ≠ 0. Therefore *m*_{j} = *i*_{0}. We have now proved that *M*_{i0i}*M*_{ij} ⊆ *M*_{mj,j}. This shows that *cM*_{ij} ⊆ *M*_{mj,j} for some *c* ∈ *M*_{i0i}. Note that for *w*, *z* ∈ *M*_{ij}, if *cw* = *cz*, then *f*_{i}w = *f*_{i}z, so that *w* = *z*. We can observe that |*cM*_{ij}| = |*M*_{ij}|, and |*M*_{ij}| ≤ |*M*_{mj,j}| < ∞. □

#### Lemma 4.7

*M*_{ij} = ∅ *if* *i* ≠ *j*.

#### 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 *M*_{ij} ≠ ∅. Let *n* be the biggest number such that *M*_{in} ≠ ∅ for some *i* ≠ *n*, and further let *m* be the biggest number such that *M*_{mn} ≠ ∅. Obviously, *m* < *n*. By definition, *M*_{km} = ∅ if *m* < *k* (if not, then as *M*_{mn} ≠ ∅ and by definition, *M*_{kn} ≠ ∅, contrary to the maximality of *m*); and *M*_{nl} = ∅ if *n* < *l* (if not, then as *M*_{in} ≠ ∅ and by definition, *M*_{il} ≠ ∅, contrary to the maximality of *n*). By these, a routine computation shows that

$$\begin{array}{}{\displaystyle \mathfrak{E}=\left(\begin{array}{ccccccc}0& \cdots & 0& {K}_{0}[\overline{{M}_{1n}}]& 0& \cdots & 0\\ 0& \cdots & 0& {K}_{0}[\overline{{M}_{2n}}]& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& {K}_{0}[\overline{{M}_{mn}}]& 0& \cdots & 0\\ 0& \cdots & 0& 0& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& 0& 0& \cdots & 0\end{array}\right)}\end{array}$$

is a left ideal of 𝔅.

Let *x* ∈ *K*_{0}[*M*_{jl}] and *M*_{ij} ≠ ∅. If *K*_{0}[*M*_{ij}]*x* = 0, then *a**x* = 0 for some *a* ∈ *M*_{ij}, so that by a similar argument as proving *u*_{jj} = *f*_{j} in the proof of Lemma 4.5, *a*^{∗}*x* = 0. By *a*^{∗} = *f*_{j}, *x* = *f*_{j}*x* = *a*^{∗}*x* = 0. So, *K*_{0}[*M*_{ij}]*x* = 0 if and only if *x* = 0. Because *M*_{mn} ≠ ∅, this can show that the right annihilator *ann*_{r}(𝔈) of 𝔈 in 𝔅 is equal to

$$\begin{array}{}{\displaystyle \left(\begin{array}{ccccccc}{K}_{0}[\overline{{M}_{11}}]& \cdots & {K}_{0}[\overline{{M}_{1,n-1}}]& {K}_{0}[\overline{{M}_{1n}}]& {K}_{0}[\overline{{M}_{1,n+1}}]& \cdots & {K}_{0}[\overline{{M}_{1r}}]\\ \vdots & \ddots & \vdots & \vdots & \vdots & \cdots & \vdots \\ 0& \cdots & {K}_{0}[\overline{{M}_{n-1,n-1}}]& {K}_{0}[\overline{{M}_{n-1,n}}]& {K}_{0}[\overline{{M}_{n-1,n+1}}]& \cdots & {K}_{0}[\overline{{M}_{n-1,r}}]\\ 0& \cdots & 0& 0& 0& \cdots & 0\\ 0& \cdots & 0& 0& {K}_{0}[\overline{{M}_{n+1,n+1}}]& \cdots & {K}_{0}[\overline{{M}_{n+1,r}}]\\ \vdots & \cdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0& \cdots & 0& 0& 0& \cdots & {K}_{0}[\overline{{M}_{rr}}]\end{array}\right).}\end{array}$$

But the left annihilator *ann*_{ℓ}(*ann*_{r}(𝔈)) of *ann*_{r}(𝔈) in 𝔅 includes

$$\begin{array}{}{\displaystyle \left(\begin{array}{ccccccc}0& \cdots & 0& {K}_{0}[\overline{{M}_{1n}}]& 0& \cdots & 0\\ 0& \cdots & 0& {K}_{0}[\overline{{M}_{2n}}]& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& {K}_{0}[\overline{{M}_{nn}}]& 0& \cdots & 0\\ 0& \cdots & 0& 0& 0& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0& \cdots & 0& 0& 0& \cdots & 0\end{array}\right).}\end{array}$$

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

#### Theorem 4.8

*Let* *S* *be a strict RA semigroup and* *K* *a field. Then* *K*_{0}[*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 *K*_{0}[*S*] is left self-injective, then *S* is regular. Assume now that *K*_{0}[*S*] is left self-injective. By Theorem 3.4, *K*_{0}[*S*] is isomorphic to

$$\begin{array}{}{\displaystyle \left(\begin{array}{cccc}{M}_{{n}_{1}}({K}_{0}[\overline{{M}_{11}}])& {M}_{{n}_{1},{n}_{2}}({K}_{0}[\overline{{M}_{12}}])& \cdots & {M}_{{n}_{1},{n}_{r}}({K}_{0}[\overline{{M}_{1r}}])\\ {M}_{{n}_{2},{n}_{1}}({K}_{0}[\overline{{M}_{21}}])& {M}_{{n}_{2}}({K}_{0}[\overline{{M}_{22}}])& \cdots & {M}_{{n}_{2},{n}_{r}}({K}_{0}[\overline{{M}_{2r}}])\\ .................& .................& ...& .................\\ {M}_{{n}_{r},{n}_{1}}({K}_{0}[\overline{{M}_{r1}}])& {M}_{{n}_{r},{n}_{2}}({K}_{0}[\overline{{M}_{r2}}])& \cdots & {M}_{{n}_{r}}({K}_{0}[\overline{{M}_{rr}}])\end{array}\right).}\end{array}$$

By Theorem 3.4 and Lemma 4.7, whenever *i* ≠ *j*, *M*_{ni,nj}(*K*_{0}[*M*_{ij}]) = 0, so that *M*_{ij} = ∅, thus {*x* ∈ *S* : *x*^{#} = *f*_{i}, *x*^{∗} = *f*_{j}} = ∅. Note that *f*_{i}’s are representatives of the partition
$\begin{array}{}{\displaystyle \pi ={\cup}_{i=1}^{r}{E}_{r}}\end{array}$
of *E*(*S*)\{0} induced by 𝓓. Therefore {*x* ∈ *S* : *x*^{#} = *e*, *x*^{∗} = *h*} = ∅ for all *e* ∈ *E*_{i}, *h* ∈ *E*_{j} with *i* ≠ *j*. This means that for all *x* ∈ *S*, *x*^{#}𝓓*x*^{∗}.

On the other hand, by Lemma 4.3(*B*), for any *e* ∈ *E*(*S*)\{0}, *M*_{e,e} = {*x* ∈ *S* : *x*^{#} = *e*, *x*^{∗} = *e*} is a subgroup of *S* and so *M*_{ii} is a subgroup of *S*. Let *f* ∈ *E*(*S*)\{0} and *e* 𝓓 *f*. Let *x*, *y* ∈ *S* with *x*^{#} = *e*, *y*^{#} = *f*, *x*^{∗} = *f* and *y*^{∗} = *e*, and of course *x*, *y* ∈ *M*_{e,f}. Then as 𝓛^{∗} is a right congruence, we have *xy*𝓛^{∗}*fy* = *y*, so that *y* ≠ 0. Similarly, *yx* ≠ 0. Note that (*x*)_{e,f}, (*y*)_{f,e} ∈ *PRA*(*S*). We observe that (*xy*)_{e,e} = (*x*)_{e,f}(*y*)_{f,e} ∈ *PRA*(*S*)\{0}, and hence *xy* ∈ *M*_{e,e}. Thus there is *a* ∈ *M*_{e,e} such that *e* = *axy* since *M*_{e,e} is a subgroup of *S* with identity *e*. This and *ye* = *y* imply that *e*𝓛*y*. Therefore *y* is regular. We have now proved that *y* is regular whenever *y*^{#}𝓓*y*^{∗}.

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* *K*_{0}[*S*] *have an unity* 1. *If* *K*_{0}[*S*] *is right self*-*injective*, *then* *S* *is a finite inverse semigroup*.

#### Proof

For any *a* ∈ *S*, *S*^{1}*a* is a left ideal of *S*, and so *K*_{0}[*S*^{1}*a*] is a left ideal of *K*_{0}[*S*]. Since *K*_{0}[*S*] is right self-injective, and by [31, Lemma 30.9, p.334], we have *ann*_{ℓ}(*ann*_{r}(*K*_{0}[*S*^{1}*a*])) = *K*_{0}[*S*^{1}*a*]. By the definition of 𝓛^{∗}, *ann*_{r}(*K*_{0}[*S*^{1}*a*]) = *ann*_{r}(*a*) = *ann*_{r}(*a*^{∗}) = (1 − *a*^{∗})*K*_{0}[*S*] and so

$$\begin{array}{}{\displaystyle {K}_{0}[{S}^{1}a]=an{n}_{\ell}(an{n}_{r}({K}_{0}[{S}^{1}a])=an{n}_{\ell}((1-{a}^{\ast}){K}_{0}[S])={K}_{0}[S]{a}^{\ast}={K}_{0}[{S}^{1}{a}^{\ast}].}\end{array}$$

It follows that *S*^{1}*a* = *S*^{1}*a*^{∗}. Thus *a*𝓛*a*^{∗} 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* *K*_{0}[*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 *K*_{0}[*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 *K*_{0}[*S*] is right self-injective. By Theorem 3.4, we have that *K*_{0}[*S*] is isomorphic to the generalized upper triangular matrix algebra

$$\begin{array}{}{\displaystyle \left(\begin{array}{cccc}{M}_{{n}_{1}}({R}_{0}[\overline{{M}_{11}}])& {M}_{{n}_{1},{n}_{2}}({R}_{0}[\overline{{M}_{12}}])& \cdots & {M}_{{n}_{1},{n}_{r}}({R}_{0}[\overline{{M}_{1r}}])\\ 0& {M}_{{n}_{2}}({R}_{0}[\overline{{M}_{22}}])& \cdots & {M}_{{n}_{2},{n}_{r}}({R}_{0}[\overline{{M}_{2r}}])\\ ..............& ................& ...& ................\\ 0& 0& \cdots & {M}_{{n}_{r}}({R}_{0}[\overline{{M}_{rr}}])\end{array}\right).}\end{array}$$

It is easy to see that *K*_{0}[*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*:

*K*_{0}[*S*] *is left self*-*injective*;

*K*_{0}[*S*] *is right self*-*injective*;

*S* *is a finite inverse semigroup*;

*K*_{0}[*S*] *is quasi*-*Frobenius*;

*K*_{0}[*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*:

*K*_{0}[*S*] *is left self*-*injective*;

*S* *is a finite inverse semigroup*;

*K*_{0}[*S*] *is quasi*-*Frobenius*;

*K*_{0}[*S*] *is Frobenius*;

*K*_{0}[*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 *K*_{0}[*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*:

*K*_{0}[*S*] *is left self*-*injective*;

*S* *is a finite inverse semigroup*;

*K*_{0}[*S*] *is quasi*-*Frobenius*;

*K*_{0}[*S*] *is Frobenius*;

*K*_{0}[*S*] *is right self*-*injective*.

Recall that if 𝓐 is a ring and 𝓐^{1} is the standard extension of 𝓐 to a ring with unity, then 𝓐 is left self-injective if and only if the left 𝓐^{1}-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*] = ℚ_{0}[*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 *I*ℚ[*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*, *x*^{2}, ⋯}. 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*, *B* ∈ *Obj*(𝓒). For the self-injectivity of path algebras and category algebras, we have the following proposition.

#### Proposition 4.15

*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*.

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

#### Proof

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.

By Example 2.5, *S*(𝓒) is an inverse semigroup whenever 𝓒 is a groupoid. So, it suffices to verify the necessity. Assume that *K*𝓒 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* *K*_{0}[*S*] *has a unity*, *then* *K*_{0}[*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:

$$\begin{array}{}{\displaystyle Does\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}the\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}fact\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}that\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}K[S]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}is\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}a\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}right\mathit{(}respectively,left\mathit{)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}self-injective\phantom{\rule{thinmathspace}{0ex}}imply\phantom{\rule{thinmathspace}{0ex}}that\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}S\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}is\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}finite?}\end{array}$$

#### Proposition 4.17

*Let* *S* *be a strict RA semigroup. If* *K*_{0}[*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* *K*_{0}[*S*] *is left (respectively*, *right) self*-*injective*, *then* *S* *is finite*.

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