Cotorsion dimensions and FP-projective dimensions of weak Hopf actions

Xiaoyuan CHEN

Front. Math. China ›› 2025, Vol. 20 ›› Issue (1) : 17 -24.

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Front. Math. China ›› 2025, Vol. 20 ›› Issue (1) : 17 -24. DOI: 10.3868/s140-DDD-025-0001-x
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Cotorsion dimensions and FP-projective dimensions of weak Hopf actions

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Abstract

Let H be a finite dimensional weak Hopf algebra and A be an H-module algebra. In this paper, we mainly discuss the relations of cotorsion dimension and FP-projective dimension between AH and A. As applications, sufficient conditions are given for LCD(AH)=LCD(A) and LFPD(AH)=LFPD(A).

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Cotorsion dimension / FP-projective dimension / weak Hopf algebra

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Xiaoyuan CHEN. Cotorsion dimensions and FP-projective dimensions of weak Hopf actions. Front. Math. China, 2025, 20(1): 17-24 DOI:10.3868/s140-DDD-025-0001-x

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In this paper, k is a field on which all algebras and coalgebras are discussed. H is a k-weak Hopf algebra. Hopf algebra turns out an important structure obtained by weakening the unitality of Hopf algebra comultiplication and multiplicativity of counit in [1]. As a result, many Hopf algebra conclusions can be generalized in weak Hopf algebra theories.
Let R be a ring and M be a left R-module. According to [2, 7], the cotorsion dimension CDR(M)=n of the module M refers to the smallest positive integer such that ExtRn+1(F,M)=0 for any flat R-module F. The left cotorsion dimension LCD(R) of R is the supremum of the cotorsion dimensions of all left R-modules. In particular, if CD(M)=0, then M is called a cotorsion module. The left FP-projective dimension LFPD(M)=n of the module M means the smallest positive integer such that ExtRn+1(M,N)=0 for any FP-injective left R-module N. This index implies how far the ring R is from being Noetherian. If H is a Hopf algebra, then LFPD(AM)=LFPD(AHM) (see [2, 7]). This paper mainly discusses whether this equation holds under the condition of weak Hopf algebra as well as the relations between CD(A) and CD(AH) or not. Please see references [3, 6, 7, 9] for details of group ring and cotorsion dimensions of modules. Then, we will review some basic concepts of weak Hopf algebra. Please see references [1, 5, 8] for details.
Definition 1 A weak bialgebra (WBA) is a quintet (A,μ,η,Δ,ε) satisfying the following axioms 1‒3. If (A,μ,η,Δ,ε,S) further satisfies Axiom 4, then it is called a weak Hopf algebra (WHA).
Axiom 1  A is a finite dimensional associative algebra over the field k with multiplication μ and unit η.
Axiom 2  A is a finite dimensional coalgebra over the field k with comultiplication Δ and counit ε.
Axiom 3 The comultiplication satisfies multiplicativity:
⟨1⟩ Δ(xy)=Δ(x)Δ(y);
the weak multiplicativity of counit:
⟨2⟩ ε(xyz)=ε(xy1)ε(y2z),ε(xyz)=ε(xy2)ε(y1z);
⟨3⟩ Δ2(1)=(Δ(1)1)(1Δ(1)),Δ2(1)=(1Δ(1))(Δ(1)1).
Axiom 4 S:AA is a k-linear map, called an antipode, satisfying
⟨4⟩x1S(x2)=ε(11x)12,
  S(x1)(x2)=ε(x12)11,
  S(x1)x2S(x3)=S(x).
Let A be a weak bialgebra. Then the maps Πl and Πr:AA, Πl(x)=xl=ε(11x)12,Πr(x)=xr=ε(x12)11. Let Hl={hl|hH},Hr={hr|hH}.
Definition 2 Let H be a weak bialgebra over the field k. If an algebra A is a left H-module and satisfies the following conditions, then A is called a left H-module algebra:
⟨5⟩ hab=(h1a)(h2b),
⟨6⟩ h1=hl1.
Lemma 1  Let H be a weak Hopf algebra over the field k and A be a left H-module algebra. For any hH and aA, we have
⟨7⟩ hla=(hl1)a,hra=a(hr1).
Let H be a weak Hopf algebra over the field k and A be a left H-module algebra. The Smash product AH is AHlH as a space, and its multiplication is defined as (ah)(bg)=a(h1b)h2g. It has been proved that AH is an associative algebra with unit AH, and the maps iA:AAH,aa1 and iH:HAH,h1h are all algebraic maps. For more details, see reference [5].
Let AH be defined as mentioned above, M and N be two AH-modules, then HomA(M,N) is a rightH-module, and its module action can be defined as:
(fh)(m)=S(h1)f(h2m),fHomA(M,N),hH,mM.
In fact, for any h,gH,fHomA(M,N) and mM, there are
((fh)g)(m)=S(g1)S(h1)f(h2g2m)=(f(hg))(m);
(f1)(m)=S(11)f(12m)=(1S(11))(S1(12)11)f(m)=((S(11)1S1(12))1S(11)2)f(m)=(11S(12)1311)f(m)=f(m).
Lemma 2  Let AH be defined as mentioned above, M and N be AH-modules. Then there exists a natural isomorphism from functor HomAH(,) to functor HomH(Hl,HomA(,)).
Proof For ιM,N:HomH(Hl,HomA(M,N))HomA(M,N)H,ff(1), andHomA(M,N)H={fHomA(M,N)fh=fhr} is isomorphic. Therefore, we only need to prove
HomA(M,N)H=HomAH(M,N).
For any fHomA(M,N)H,hH,mM, there is
(fh)(m)=(fhr)(m)S(h1)f(h2m)=S(11)f(12hrm)=f(hrm).
Hence,
hf(m)=h2f(S1(h1)lm)=h3f(S1(h2)h1m)=f(S1(h2)rh1m)=f(hm).
ah=(a1)(1h), so fHomAH(M,N).
Conversely, if fHomAH(M,N), then
(fh)(m)=S(h1)f(h2m)=f(hrm)=S(11)f(12hrm)=(fhr)(m).
Next, we need to prove that the naturality establishes. For any fHomAH(M,M), assume
f¯=HomH(Hl,HomA(f,N)),
i.e., f¯:HomH(Hl,HomA(M,N))HomH(Hl,HomA(M,N)),f¯(α)(z)(m)=α(z)f(m), then on one hand (HomAH(f,N)ιM,N)(α)= HomAH(f,N)(ιM,N(α))=α(1)f, and on the other (ιM,Nf¯)(α)= ιM,N(f¯(α))=α(1)f. In the same way, the naturality on N is established. Therefore, the proof is completed. □
In the following part, we assume that H is a semisimple weak Hopf algebra, that is, every H-module is completely reducible.
Lemma 3  Let H be a semisimple weak Hopf algebra, A be a left H-module algebra, M and N be AH-modules satisfying ExtAp(M,N)= 0,p>0. Then ExtAHn(M,N)=0,n>0.
Proof Let 0CPM0 be an AH-module exact sequence, whereP is a projective AH-module. Since ExtAp(M,N)=0,p>0, there are exact sequences
0HomA(M,N)HomA(P,N)HomA(C,N)0,
and ExtAp(C,N)ExtAp+1(M,N)=0,p>0. Therefore, the commutative diagram is obtained:
HomH(Hl,HomA(P,N))HomH(Hl,HomA(C,N))ExtH1(Hl,HomA(M,N))0,
HomAH(P,N)HomAH(C,N)ExtAH1(M,N)0.
From Lemma 2, there exists the commutative diagram:
0ExtHp(Hl,HomA(C,N))ExtHp+1(Hl,HomA(M,N))0,
0ExtAHp(C,N)ExtAHp+1(M,N)0.
It is obvious that ExtAHn(M,N)ExtHn(Hl,HomA(M,N))=0,n>0.
Lemma 4  Let AH be defined as mentioned above, and AH is a flat right A-module. Then for any AH-module M, CDA(M) CDAH(M).
Proof Assume CDAH(M)=n<. Let F be a flat A-module, then AHAF is a flat AH-module. Then ExtAHn+1(AHAF,M)=0 and ExtAn+1(F,M)ExtAHn+1(AHAF,M) (see reference [9]). Hence, ExtAn+1(F,M)=0, which indicates CDA(M)n.
Theorem 1  Let AH be defined as mentioned above and AH is a flat right A-module, then for any AH-module M, CDA(M)= CDAH(M).
Proof According to Lemma 4, we need to prove CDA(M)CDAH(M). Assume CDA(M)=n<. According to reference [3], there exists a AH-module exact sequence 0MX0Xn1X0, where X0,,Xn1 are all cotorsion AH-modules. According to reference [6] and Lemma 6, X0,,Xn1,X are all cotorsion A-modules. Then for any flat A-module F and p>0, ExtAp(F,X)=0. Based on Lemma 3, for any flat AH-module F, ExtAH1(F,X)=0, so CDAH(X)=0. According to reference [6], there exists an exact sequence
0MX0Xn1Xn0,
where X0,,Xn1,Xn are all cotorsion modules, so CDAH(M)n. □
Corollary 1  Let AH be defined as mentioned above and AH be a flat right A-module, then LCD(AH)LCD(A).
It is known that the ring R is left perfect if and only if LCD(R)=0. As an application, we will give several conditions such that LCD(AH) =LCD(A).
Theorem 2  Let H be a semisimple weak Hopf algebra which is free on Hl, and A be a left H-module algebra. If one of the following conditions is met, then LCD(AH)=LCD(A):
(1) H is semisimple;
(2) A is commutative.
Proof According to Corollary 1, we need to prove LCD(AH)LCD(A).
Proof of (1).
From reference [8], AHHEnd(AH)AMn(A). If H is semisimple, we can obtain LCD(AH)LCD(A) directly from Corollary 1.
Proof of (2).
First, we assert that AH is an (A,A)-bimodule, with actions defined as b(ah)=abh and (ah)b=abh,a,bA,hH. In fact, for any a,b,cA,zHl,hH, there are
b(azhazh)=abzh(S1(z)a)bh=7abzh(S1(z)1)abh=7abzhabzh=0;
(azhazh)b=abzh(S1(z)a)bh=7abzhabzh=0;
b(c(ah))=acbh=abch=(bc)(ah),1(ah)=ah,
((ah)c)b=acbh=(ah)(cb),(ah)1=ah.
Next, we prove α:AHA,ahhla is an (A,A)-bimodule epimorphism. In reality, for any a,b,c A,zHl,hH, there are
α(azhazh)=(zh)lahlS1(z)a=7(zh)la(hla)(S1(z)1)=(zh)la(hla)(z1)=7ε(11zh)12azlhla=ε(z1h)z2azlhla=zlhlazlhla=0;
bα(ah)=b(hla)=7(hl1)ab=hl(ab)=α(b(ah));
α(ah)b=(hla)b=7(hl1)ab=hl(ab)=α((ah)b).
Finally, we prove that α is a split (A,A)-bimodule epimorphism. Define β:AAH,β(a)=a1. It is proved that β is an (A,A)-bimodule map and αβ=idA. Therefore, as an (A,A)-bimodule, A, is the direct summand of AH, and for anyA-module M, there is
AHAMM(XAM).
Hence, CDA(M)CDA(AHAM). From Lemma 4,CDA(AHAM)CDAH(AHAM), so we can get LCD(AH)LCD(A). □
Corollary 2  Let H be a semisimple weak Hopf algebra which is free on Hl, and A be a left H-module algebra. If one of the following conditions is met, then A is left perfect if and only if AH is left perfect:
(1) H is semisimple;
(2) A is commutative.
Theorem 3 Let H be a semisimple weak Hopf algebra which is free on Hl. Assume A is a left H-module algebra and M is a AH-module, then M is a finitely presented AH-module if and only if M is a finitely presented A-module.
Proof If M is a finitely presented left AH-module, then there exists an exact sequence 0CPM0, where C is a finitely generated AH-module, and P is a finitely generated projective AH-module. Based on reference [7], C is a finitely generated A-module, and P is a finitely generated projective A-module, so M is a finitely presented A-module.
On the contrary, if M is a finitely presented A-module, then there exists an exact sequence 0CPM0, where C is a finitely generated A-module and Pis finitely generated projective A-module. Since H is free on Hl, there exists an AH-module exact sequence 0AHACAHAPAHAM0. Obviously, AHAC is finitely generated, and AHAP is finitely generated projective. This means that AHAM is finitely presented, and M is the direct summand of AHAM as an AH-module, so M is a finitely presented AH-module.
Theorem 4  Let H be a semisimple weak Hopf algebra which is free on Hl. Assume A is a left H-module algebra, and M is a AH-module. Then M is a FP-projective left AH-module if and only if M is a FP-projective left A-module.
Proof If M is an FP-projective left AH-module, then M is the direct summand of N as an AH-module and N is the union of {Nαα<λ} based on reference [4] , where for λ,N0=0,Nα+1/Nα,α<λ is a finitely presented AH-module. From Theorem 3, Nα+1/Nα,α<λ is a finitely presented A-module, so M is an FP-projective A-module.
Conversely, if M is an FP-projective left A-module, we first assert that for any FP-injective AH-module N, it is also an FP-injective A-module. LetPbe a finitely presented A-module, based on reference [9], ExtA1(P,N)ExtAH1(AHAP,N). According to Theorem 3, AHAP is a finitely presented AH-module, and
ExtA1(P,N)ExtAH1(AHAP,N)=0,
M is an FP-injective A-module. For any FP-injective AH-module N, there are ExtA1(M,N)=0 and
ExtAH1(AHAM,N)ExtA1(M,N)=0,
M is the direct summand of AHAM, so ExtAH1(M,N)=0, that is, M is an FP-projectie AH-module. □
Theorem 5  Let H be a semisimple weak Hopf algebra which is free on Hl. Assume A is a coherent H-module algebra, and M is an AH-module, then FPDA(M)=FPDAH(M)=FPDAH(AHAM).
Proof Since M is the direct summand of AHAM as an AH-module, there is
FPDAH(M)FPDAH(AHAM).
Next, we prove FPDA(M)FPDAH(M). If FPDAH(M)=n<, there exists an AH-module exact sequence 0PnPn1P0M0, where Pn,Pn1,,P0 are all FP-projective AH-modules. From Theorem 4, they are all FP-projective A-modules, so FPDA(M)n.
At last, we must prove FPDAH(AHAM)FPDA(M). Assuming FPDA(M)=n<, there exists an A-module exact sequence 0PnPn1P0M0, where Pn,Pn1,,P0 are all FP-projective A-modules. From reference [7], we can get the exact sequence
0AHAPnAHAPn1AHAP0AHAM0,
where
AHAPn,AHAPn1,,AHAP0
are all FP-projective AH-modules, so FPDAH(AHAM)n. □
Corollary 3  Let H be a semisimple weak Hopf algebra which is free on Hl. AssumeA is a left coherent H-module algebra, then LFPD(AH)LFPD(A).
Corollary 4  Let H be a semisimple weak Hopf algebra which is free on Hl. Assume A is a left coherent H-module algebra and H is semisimple, then LFPD(A)=LFPD(AH).
Proof From Corollary 3, we only need to prove LFPD(A)LFPD(AH). According to reference [8], AHH is isomorphic to Mn(A) and Mn(A) is isomorphic to A. Based on Corollary 3, we can obtain LFPD(A)LFPD(AH). □

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