The aim of this paper is to characterize the representation ring of a non-pointed and noncocommutative bialgebra. First, the isomorphism classes of its indecomposable modules are classified. Then the tensor product of modules is established. Finally, its representation ring is described.
Citation: Huaqing Gong, Shilin Yang. The representation ring of a non-pointed bialgebra[J]. AIMS Mathematics, 2025, 10(3): 5110-5123. doi: 10.3934/math.2025234
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The aim of this paper is to characterize the representation ring of a non-pointed and noncocommutative bialgebra. First, the isomorphism classes of its indecomposable modules are classified. Then the tensor product of modules is established. Finally, its representation ring is described.
The concept of representation rings, also known as the Green ring, was first introduced by Green [1] in the 1960's while studying the representations of finite groups. Benson and Carlson further developed this work, and applied it to the representation theory of algebras (see [2,3]). Recently, the study of representations and representation rings of Hopf algebras and, more generally, bialgebras has received considerable attention. For example, Chen, Van Oystaeyen, and Zhang in [4] studied the representation rings of Taft algebras. Wang, Li, and Zhang in [5] determined the structure of the Green ring for finite-dimensional pointed rank one Hopf algebras of nilpotent type. The authors also provided a classification of all indecomposable modules and constructed their tensor product decomposition formulas. Subsequently, they studied the Green rings of finite-dimensional pointed rank one Hopf algebras of non-nilpotent type in [6]. Guo and Yang in [7] gave a complete list of all indecomposable modules and described all components of Auslander-Reiten quivers of a class of non-pointed tame Hopf algebras. The authors also established their projective class rings as well as the projective class rings of the category of their Yetter-Drinfeld modules. Su and Yang gave the Green rings of Δ-associative algebras (see [8]). Lin and Yang constructed and classified all string representations of a class of tame type Hopf algebras and gave the explicit description of projective class rings (see [9]).
Yang and Zhang in [10] classified the bialgebra Ore extensions of automorphism type for the 4-dimensional Sweedler Hopf algebra and obtained a series of non-isomorphic bialgebras and Hopf algebras. The most interesting one of them, denoted by A24m, which is noncocommutative and non-pointed with dimension 24m, arouses us to study their representations and describe its representation ring. The results may help us to understand the more general case of non-pointed bialgebras. We first classify all simple and all indecomposable modules of A24m explicitly. Then, the decomposition formulas of the tensor products of them are established, and the description of the representation ring is given.
The outline of this paper is as follows: In Section 1, the definition of the 24m-dimensional bialgebras A24m is reviewed. It is shown that A24m is a basic algebra, and its corresponding bound quiver is described. In Section 2, all indecomposable modules of A24m are classified, and decomposition formulas of their tensor products are established. In Section 3, we focus on characterizing its representation ring of A24m.
Throughout, K is assumed to be an algebraically closed field containing a non-trivial 6m-th primitive root q of unity, and ω=qm.
First of all, let us review the family of bialgebras A24m discussed in the paper.
By definition, the algebra A24m is generated by x,g,z subjecting to the following relations:
g2=1,x2=0,z6m=1,xg=−gx,zx=ωgxz,zg=gz+2xz. |
If we define two K-maps Δ:A24m→A24m⊗A24m and ϵ:A24m→K on generators by
Δ(g)=g⊗g,Δ(x)=x⊗1+g⊗x,Δ(z)=z⊗z−2e1z⊗e1z+2e1z⊗xz,ϵ(x)=0,ϵ(g)=ϵ(z)=1, |
where e1=1−g2, and extend them to the whole space A24m in the natural way. Then, A24m becomes a 24m-dimensional bialgebra with a basis
{xigjzk|k∈Z6m;i,j=0,1}. |
It is easy to see that A24m is non-commutative, non-cocommutative and non-pointed. For the detailed construction, the readers are referred to [10, Page 20, Case 2: (a)(ⅰ)].
Obviously, xA24m of A24m is a bi-ideal; the bialgebra A24m/xA24m deduced from A24m is the group algebra K(C2×C6m), which comultiplication is given by
Δ(g)=g⊗g,Δ(z)=z⊗z−2e1z⊗e1z=12(1⊗1+1⊗g+g⊗1−g⊗g)(z⊗z),ϵ(g)=ϵ(z)=1. |
Indeed, it is straightforward to see that K(C2×C6m) is a bialgebra in this setting.
For i∈Z6m, we set
ϵi=16m6m−1∑j=0q−ijzj,e0=1+g2. |
Lemma 2.1. The set
{e0ϵie0,e1ϵie1|i∈Z6m} |
is a complete set of primitive orthogonal idempotents of A24m.
Proof. For n=0,1,…,6m−1, we have
e0xznϵie0=xe1znϵie0=x(zne1+n∑k=1 oddωk−1xzn+n∑l=1 evenωl−1gxzn)e0=xzne1ϵie0=xzn(ϵie1+16m6m−1∑j=0j∑k=1 oddq−ijωk−1xzj+16m6m−1∑j=0j∑l=1 evenq−ijωl−1gxzj)e0=xznϵie1e0+16m6m−1∑j=0j∑k=1 oddq−ijωk−1xznxzje0+16m6m−1∑j=0j∑l=1 evenq−ijωl−1xzngxzje0=0. |
Similarly, e1xznϵie1=0. It follows that
(e0ϵie0)2=e0ϵie0ϵie0=e0(e0ϵi+16m6m−1∑j=0j∑k=1 oddq−ijωk−1xzj+16m6m−1∑j=0j∑l=1 evenq−ijωl−1gxzj)ϵie0=e0ϵie0+16m6m−1∑j=0j∑k=1 oddq−ijωk−1e0xzjϵie0+16m6m−1∑j=0j∑l=1 evenq−ijωl−1e0gxzjϵie0=e0ϵie0. |
(e1ϵie1)2=e1ϵie1ϵie1=e1(e1ϵi−16m6m−1∑i=0j∑k=1 oddq−ijωk−1xzj−16m6m−1∑i=0j∑l=1 evenq−ijωl−1gxzj)ϵie1=e1ϵie1−16m6m−1∑i=0j∑k=1 oddq−ijωk−1e1xzjϵie1−16m6m−1∑i=0j∑l=1 evenq−ijωl−1e1gxzjϵie1=e1ϵie1. |
Hence, e0ϵie0, e1ϵie1 are idempotents for all i∈Z6m.
Furthermore, for any i≠j, we have
(e0ϵie0)(e0ϵje0)=e0ϵie0ϵje0=e0(e0ϵj+16m6m−1∑j=0j∑k=1 oddq−ijωk−1xzj+16m6m−1∑j=0j∑l=1 evenq−ijωl−1gxzj)ϵje0=e0ϵiϵje0+16m6m−1∑j=0j∑k=1 oddq−ijωk−1e0xzjϵje0+16m6m−1∑j=0j∑l=1 evenq−ijωl−1e0gxzjϵje0=0. |
In a similar way, for any i≠j,
(e1ϵie1)(e1ϵje1)=0,(e0ϵie0)(e1ϵje1)=0, |
and
eiϵjeixeiϵjei=0 |
for all i=0,1 and j∈Z6m. In other words, dim eiϵjeiA24meiϵjei=1.
Also, we have
6m−1∑i=0e0ϵie0+6m−1∑i=0e1ϵie1=e0(6m−1∑i=0ϵi)e0+e1(6m−1∑i=0ϵi)e1=e20+e21=1. |
Therefore, the set
{e0ϵie0,e1ϵie1|i∈Z6m} |
is indeed a complete set of primitive orthogonal idempotents of A24m.
Lemma 2.2. The algebra A24m is a basic algebra.
Proof. Recall that an algebra A is basic if and only if the algebra A/J is isomorphic to K×K×⋯×K, where J=radA is the radical of A. Therefore, the statement is obvious since A24m/J=A24m/xA24m≅K(C2×C6m), which is basic, by the previous remark.
Here, we give the detailed proof by constructing the explicit correspondence between primitive idempotents fij=eiϵjei(i∈Z2,j∈Z6m) of A24m and those of the 12m-dimensional diagonal matrix algebra. It is convenient in describing the bound quiver of A24m and indecomposable projective modules of A24m in the later. It is remarked that fij=eiϵjei, the liftings of those of K(C2×C6m), are not in K(C2×C6m).
Now, for i∈Z2 and j∈Z6m, we set
fij=eiϵjei. |
It is easy to see that
radA24m=∑i∈Z2j∈Z6mKxfij |
of dimension 12m, and
rad2A24m=0. |
The 12m-dimensional algebra
K×K×⋯×K |
can be identified with the 12m-dimensional diagonal matrix algebra
diag(K,K,⋯,K). |
Let
ζp=diag(0,…,0,1p position,0,…,0), |
where p=1,⋯,12m, and only the p-th position is 1 and others are 0. Let
ˉfij=fij+radA24m, where i∈Z2,j∈Z6m. |
Now, define the mapping
φ:A24m/radA24m⟶diag(K,K,⋯,K) |
by
φ(ˉf0j)=ζj+1, and φ(ˉf1j)=ζ6m+1+j, |
where j∈Z6m.
We must prove that φ is an algebra isomorphism. Indeed, we have
φ(ˉf0i)φ(ˉf0j)=ζi+1ζj+1=δijζi+1, |
φ(ˉf0iˉf0j)=φ(δijˉf0i)=δijζi+1, |
φ(ˉf0i)φ(ˉf0j)=φ(ˉf0iˉf0j). |
In the same way, we can obtain
φ(ˉf1k)φ(ˉf1l)=φ(ˉf1kˉf1l). |
Because
φ(ˉf0i)φ(ˉf1k)=ζi+1ζ6m+1+k=0, |
φ(ˉf0iˉf1k)=φ(ˉf0iˉf1k)=0, |
we have
φ(ˉf0i)φ(ˉf1k)=φ(ˉf0iˉf1k). |
Furthermore,
φ(1∑i=06m−1∑j=0ˉfij)=φ(ˉ1)=E, |
1∑i=06m−1∑j=0φ(ˉfij)=6m−1∑j=0φ(ˉf0j)+6m−1∑j=0φ(ˉf1j)=E, |
where E is the identity matrix of 12m×12m.
Hence, φ is an algebra isomorphism, and the algebra A24m is a basic algebra.
Furthermore, we see that
radA24m/rad2A24m=∑i∈Z2j∈Z6mKxfij. |
Hence,
f0i(radA24m/rad2A24m)f0j=0,f1i(radA24m/rad2A24m)f1j=0, |
and
f0i(radA24m/rad2A24m)f1j={K{xf1j}, if i=j+m,0, otherwise, |
f1i(radA24m/rad2A24m)f0j={K{xf1j}, if i=j+4m,0,otherwise, |
where i,j∈Z6m.
We can describe the bound quiver (Q,I) of A24m. Now, let ai,j be the points corresponding to the primitive idempotents fki,lj, where
ki≡(j+1)(mod2), lj≡(i+j+1∑k=0(5+3(−1)k)m2)mod6m, | (2.1) |
and the arrows are endowed with αi,j+1:ai,j+1→ai,j, for all (i,j)∈Zm×Z12.
For example, for m=1, the bound quiver (QA24,I) is
![]() |
with the admissible ideal
I={α0,(j+1)α0,j|j∈Z12}. |
In general, for m≥1, it is observed that the bound quiver of (QA24m,I) is
![]() |
with the admissible ideal
I={αi,(j+1)αi,j|i∈Zm,j∈Z12}. |
In this section, we mainly construct all mutually non-isomorphic indecomposable modules of A24m. Then, we consider their tensor products, respectively. Finally, we conclude that the decomposition formula of their tensor products.
By the bound quiver of algebra (A24m,I), we see that algebra A24m can be decomposed into a direct sum of m blocks Ci,i∈Zm: A24m=⨁i∈ZmCi, where Ci=∑j∈Z12A24mfki,lj, and ki,lj are as in (2.1). Consequently, by [11, Chap. 5, Theorem 3.5], any indecomposable A24m-module is either a 1-dimensional simple module or a 2-dimensional projective A24m-module.
Let Sij be a 1-dimensional vector space spanned by vij,i∈Z2,j∈Z6m, we define the action of A24m on Sij as follows:
x⋅vij=0,g⋅vij=(−1)ivij,z⋅vij=qj⋅vij. |
It is straightforward to see that Sij is a simple A24m-module, and Sij≅Skl if and only if i=k,j=l.
Lemma 3.1. Any simple A24m-module S is isomorphic to Sij, where i∈Z2,j∈Z6m.
Proof. Assume that S is a simple A24m-module. By Lemma 2.2, the dimension of the simple A24m-module S is 1. Let v be the basis of S. We set
x⋅v=av,g⋅v=bv,z⋅v=cv, |
where a,b,c∈K. We have
x2v=a2v=0,g2v=b2v=v,z6mv=c6mv=v. |
Hence, we have a=0,b=±1, and c6m=1. Consequently, we can set c=qj for some j∈Z6m and
x⋅v=0,g⋅v=v,z⋅v=qjv, |
or
x⋅v=0,g⋅v=−v,z⋅v=qjv. |
Therefore, there exist i∈Z2, j∈Z6m such that S=Sij.
Furthermore, the module action of the submodule A24mfij of the left regular A24m-module is as follows:
x⋅fij=xfij,x⋅xfij=0,g⋅fij=(−1)ifij,g⋅xfij=(−1)i+1xfij,z⋅fij=qjfij+(−1)iqjxfij,z⋅xfij=(−1)i+1ωqjxfij. |
Let Pij be the projective cover of simple A24m-module Sij for (i,j)∈Z2×Z6m.
Lemma 3.2. We have Pij≅A24mfij.
Proof. Note that {fij|i∈Z2,j∈Z6m} is the complete set of orthogonal idempotent elements of the algebra A24m. So we have
A24m=(6m−1⨁j=0A24mf0j)⊕(6m−1⨁j=0A24mf1j). |
Also, A24mfij is an indecomposable projective module.
We define the map φ as follows:
φ: A24mfij→Sijfij⟼(−1)ivijxfij⟼0, |
for each (i,j)∈Z2×Z6m.
It is easy to see that φ is a surjective linear map. By Lemma 3.1, we have
φ(xfij)=0=x⋅vij=x⋅φ(fij),φ(x2fij)=0=x⋅φ(xfij),φ(gfij)=φ((−1)ifij)=(−1)iφ(fij)=vij=g⋅φ(fij)=g⋅(−1)ivij,φ(gxfij)=φ((−1)i+1xfij)=0=g⋅φ(xfij),φ(zfij)=φ(qjfij+(−1)iqjxfij)=qj(−1)ivij=z⋅φ(fij)=z⋅(−1)ivij,φ(zxfij)=φ((−1)i+1ωqjxfij)=0=z⋅φ(xfij). |
So the map φ:A24mfij→Sij can define a module epimorphism, and
A24mfij/kerφ≅Sij. |
It is easy to see that kerφ=rad(A24mfij)=K{xfij}, and
A24mfij/rad(A24mfij)≅Sij. |
However,
A24mfij/rad(A24mfij)≅Pij/rad(Pij)≅Sij |
since Pij is a projective cover of simple A24m-module Sij. By the uniqueness of the projective cover of a module, we have
Pij≅A24mfij,i∈Z2,j∈Z6m. |
Now, we identify Pij with A24mfij for (i,j)∈Z2×Z6m.
So far, we have provided a classification of all indecomposable A24m-modules. In the following, we consider their tensor products.
Proposition 3.3. Sij⊗Skl≅S(i+k)(mod2),(j+l)(mod6m)≅Skl⊗Sij, for i∈Z2,j∈Z6m.
Proof. For simple modules Sij,Skl, the tensor product Sij⊗Skl is also an A24m-module with a basis vij⊗vkl. By Lemma 3.1, we have
x⋅(vij⊗vkl)=0,g⋅(vij⊗vkl)=(−1)i+k(vij⊗vkl),z⋅(vij⊗vkl)=qj+l(vij⊗vkl). |
Hence,
Sij⊗Skl≅S(i+k)(mod2),(j+l)(mod6m)≅Skl⊗Sij,(i∈Z2,j∈Z6m). |
The result follows.
Proposition 3.4. Sij⊗Pkl≅P(i+k)(mod2),(j+l)(mod6m)≅Pkl⊗Sij, for i∈Z2,j∈Z6m.
Proof. We consider Pkl=A24mfkl,(k∈Z2,l∈Z6m), which is of basis {fkl,xfkl}. The action of A24m on Pkl is the following:
x⋅fkl=xfkl,x⋅xfij=0,g⋅fkl=(−1)kfkl,g⋅xfkl=(−1)k+1xfkl,z⋅fkl=qlfkl+(−1)kqlxfkl,z⋅xfkl=(−1)k+1ωqlxfkl. |
The corresponding matrix representations of x,g,z are
Ax=(0010),Ag=((−1)k00(−1)k+1),Az=(ql0(−1)kql(−1)k+1ωql), |
respectively.
On the other hand, {vij⊗fkl,vij⊗xfkl} form a basis of the tensor product Sij⊗Pkl. The action of A24m on Sij⊗Pkl is the following:
x⋅(vij⊗fkl)=vij⊗xfkl,x⋅(vij⊗xfij)=0,g⋅(vij⊗fkl)=(−1)k+ivij⊗fkl,g⋅(vij⊗xfkl)=(−1)k+i+1vij⊗xfkl,z⋅(vij⊗fkl)=qjvij⊗(qlfkl+(−1)kqlxfkl)=qj+lvij⊗fkl+(−1)kqj+lvij⊗xfkl,z⋅(vij⊗xfkl)=qjvij⊗(−1)k+1ωqlxfkl=(−1)k+1ωqj+lvij⊗xfkl. |
The corresponding matrix representations of x,g,z are
Bx=(0010),Bg=((−1)k+i00(−1)k+i+1),Bz=(qj+l0(−1)kqj+l(−1)k+1ωqj+l), |
respectively.
Hence, we have
Sij⊗Pkl≅P(i+k)(mod2),(j+l)(mod6m). |
In a similar way, we can also prove that
Pkl⊗Sij≅P(i+k)(mod2),(j+l)(mod6m). |
Proposition 3.5.
Pij⊗Pkl≅P(i+k+1)(mod2),(j+4m+3mi+l)(mod6m)⊕P(i+k)(mod2),(j+l)(mod6m), |
for i∈Z2,j∈Z6m.
Proof. Considering the exact sequence
Pijψ⟶Sij⟶0, |
where ψ is defined by ψ(fij)=(−1)ivij, ψ(xfij)=0. We have Kerψ=K{xfij}=Ω(Sij), and
0⟶Ω(Sij)⟶Pij⟶Sij⟶0. |
On the other hand, for a Hopf algebra H, and any H-module M, if P is a projective H-module, then P⊗M or M⊗P is also a projective H-module (see, for example, [12]). Therefore, we obtain the following splitting short exact sequence:
0⟶Ω(Sij)⊗Pkl⟶Pij⊗Pkl⟶Sij⊗Pkl⟶0. |
Hence, we obtain
Pij⊗Pkl=Ω(Sij)⊗Pkl⊕Sij⊗Pkl. |
Note that q is the 6m-th primitive root of the unit, and ω is the 6-th primitive root of the unit; we have q3m=−1. Also, without loss of generality, we assume that qm=ω. Accordingly, the action of the algebra A24m on Ω(Sij) is as follows:
x⋅xfij=0,g⋅xfij=(−1)i+1xfij,z⋅xfij=(−1)i+1qjωxfij=q3m(i+1)qjqmxfij=q3mi+j+4mxfij. |
Reviewing the construction of simple A24m-module Sij, one sees that
Ω(Sij)≅S(i+1)mod2,(j+4m+3mi)mod6m. |
Therefore,
Pij⊗Pkl≅(S(i+1)mod2,(j+4m+3mi)mod6m⊗Pkl)⊕(Sij⊗Pkl). |
By Proposition 3.4, we obtain
Pij⊗Pkl≅P(i+k+1)(mod2),(j+4m+3mi+l)(mod6m)⊕P(i+k)(mod2),(j+l)(mod6m). |
In this section, we mainly focus on establishing the representation ring of the algebra A24m. Let us first review the definition of the projective class ring.
Assuming that H is a Hopf algebra, we denote by [V], the isomorphic class of a finite-dimensional H-module V. Let F(H) be the free abelian group generated by [V], we define a multiplication on F(H) as [M][N]=[M⊗N]. Then, F(H) is an associative ring with the identity element [K]. The quotient ring r(H):=F(H)/I is called the Green ring of H, where I is the ideal generated by the relation [M⊗N]−[M][N] and [M⊕N]−[M]−[N]. It is clear that {[V]|V∈ind(H)} forms a Z-basis for r(H), where ind(H) is the set consisting of isomorphism classes of all finite-dimensional indecomposable H-modules.
The subring P(H) of r(H) generated by the projective modules and simple modules of H is called the projective class ring of H. It is noted that the representation ring of A24m is the same as its projective class ring of A24m.
Now, let us determine the representation ring r(A24m) of the algebra A24m.
Lemma 4.1. Let t=[S01],f=[P01],u=[S11],h=[P11], then the following equations hold:
(1) t6m=1,u6m=1.
(2) tf=ft,tu=ut,th=ht,uh=hu,uf=fu.
(3) t2=u2,fh=u4m+1h+th,hf=tm+1f+uf.
(4) f2=u4m+1f+tf,h2=tm+1h+uh.
Proof. (1) By Proposition 3.3, we obtain
[S01]6m=[S01⊗S01⊗⋯⊗S01]=[S0,6m(mod6m)]=[S00]=1. |
Therefore, t6m=1. Similarly, u6m=1.
(2) By Proposition 3.4, Sij⊗Pkl=Pkl⊗Sij, so, tf=[S01][P01]=[S01⊗P01]=[P01⊗S01]=[P01][S01]=ft, thus, we have tf=ft. Similarly, we obtain
tu=ut,th=ht,uh=hu,uf=fu. |
(3) By Proposition 3.3, we obtain
[S01]2=[S01⊗S01]=[S02];[S11]2=[S11⊗S11]=[S02]. |
Thus, t2=u2. By Proposition 3.5, we obtain
[P01⊗P11]=[S1,4m+1⊗P11⊕S01⊗P11]=u4m+1h+th. |
[P11⊗P01]=[S0,m+1⊗P01⊕S11⊗P01]=tm+1f+uf. |
Thus, fh=u4m+1h+th,hf=tm+1f+uf.
(4) By Proposition 3.5, we obtain
f2=[P01]2=[P01⊗P01]=[S1,4m+1⊗P01⊕S01⊗P01]=u4m+1f+tf. |
h2=[P11]2=[P11⊗P11]=[S0,m+1⊗P11⊕S11⊗P11]=tm+1h+uh. |
The lemma follows.
Corollary 4.2. {tifj;uti1fj1;ukhl;tuk1hl1} is a Z-basis of r(A24m), where i,i1,k,k1∈Z6m,j,j1,l,l1∈Z2, and i,k are even, i1,k1 are odd.
Proof. Note that r(A24m) has a Z-basis {[Sij],[Pij]∣i∈Z2,j∈Z6m}. Hence, the rank of r(A24m) is 24m. By Lemma 4.1, r(A24m) is spanned by the set {tifj;uti1fj1;ukhl;tuk1hl1}, which consists of 24m elements. Hence, the set is a Z-basis of r(A24m).
Now, we can obtain the main result of the paper.
Theorem 4.3. The representation ring r(A24m) is isomorphic to the quotient ring Z⟨α,β,γ,η⟩/I, where I is the ideal generated by the relations
α6m−1,β2−γ4m+1β−αβ,α2−γ2,η2−αm+1η−γη,βη−γ4m+1η−αη,ηβ−αm+1β−γβ. |
Proof. Let
π:Z⟨α,β,γ,η⟩⟶Z⟨α,β,γ,η⟩/I |
be the natural epimorphism. It is straightforward to check that Z⟨α,β,γ,η⟩/I is spanned by
{αiβj;γαi1βj1;γkηl;αγk1ηl1}, |
where i,i1,k,k1∈Z6m,j,j1,l,l1∈Z2, and i,k are even, i1,k1 are odd. This means that the rank of Z⟨α,β,γ,η⟩/I is equal or less than 24m.
By Corollary 4.2, r(A24m) is generated by t,f,u, and h. So, there is a uniqueness ring epimorphism
Φ:Z⟨α,β,γ,η⟩⟶r(A24m) |
satisfying
Φ(α)=t,Φ(β)=f,Φ(γ)=u,Φ(η)=h. |
By Lemma 4.1, we see that
t6m=1,tf=ft,tu=ut,th=ht,uh=hu,uf=fu, |
t2=u2,f2=u4m+1f+tf,h2=tm+1h+uh, |
fh=u4m+1h+th,hf=tm+1f+uf. |
It follows that
Φ(α6m−1)=0,Φ(αβ−βα)=0, |
Φ(αγ−γα)=0,Φ(αη−ηα)=0,Φ(γη−ηγ)=0, |
Φ(γβ−βγ)=0,Φ(α2−β2)=0, |
Φ(η2−αm+1η−γη)=0, Φ(β2−γ4m+1β−αβ)=0, |
Φ(βη−γ4m+1η−αη)=0, Φ(ηβ−αm+1β−γβ)=0. |
Hence, Φ(I)=0, and Φ induces an epimorphism
¯Φ:Z⟨α,β,γ,η⟩/I⟶r(A24m) |
satisfying ¯Φ(¯v)=Φ(v), for any v∈Z⟨α,β,γ,η⟩, where ¯v=π(v)=v+I.
Comparing the ranks of Z⟨α,β,γ,η⟩/I with r(A24m), we get that ¯Φ is indeed a ring isomorphism.
The result is obtained.
In this paper, all indecomposable modules of A24m are constructed and classified. Furthermore, we obtain the decomposition formulas for the tensor products of indecomposable modules as well as the representation ring of the bialgebra. These results may help us to understand the general representation theory of a non-pointed bialgebra.
Shilin Yang contributed the creative ideas and proof techniques for this paper, including the construction of the article structure and the revision of content; Huaqing Gong consulted the relevant background of the paper and wrote the article. All authors have read and agreed to the published version of the manuscript.
All authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
All authors declare no conflicts of interest in this paper.
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