On Quandle Homology Groups of Alexander Quandles of Prime Order

Home , Domain (ring theory)

TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 365, Number 7, July 2013, Pages 3413–3436 S 0002-9947(2013)05754-6 Article electronically published on January 30, 2013

ON QUANDLE HOMOLOGY GROUPS OF ALEXANDER QUANDLES OF PRIME ORDER

TAKEFUMI NOSAKA

Abstract. In this paper we determine the integral quandle homology groups of Alexander quandles of prime order. As a special case, this settles the delayed Fibonacci conjecture by M. Niebrzydowski and J. H. Przytycki from their 2009 paper. Further, we determine the cohomology group of the Alexander quandle and obtain relatively simple presentations of all higher degree cocycles which generate the cohomology group. Finally, we prove that the integral quandle homology of a ﬁnite connected Alexander quandle is annihilated by the order of the quandle.

1. Introduction A quandle is a set with a certain binary operation: the operation is, roughly speaking, defined under some weaker conditions than groups. In analogy to geometric realizations of group (co)homology, classifying spaces of quandles were first introduced by R. Fenn, C. Rourke and B. Sanderson in [FRS]. As a modification, the quandle (co)homology of a quandle X is defined by J. S. Carter, D. Jelsovsky, S. Kamada, L. Langford and M. Saito in [CJKLS]. The 2, 3or4-cocyclesofthe quandle cohomology give rise to the quandle cocycle invariants for 1-knots and 2-knots (see [CJKLS], [CKS] or [O] for details). Further, concerning higher dimensional knots, J. S. Carter, S. Kamada and M. Saito in [CKS] investigated the correspondence between some higher dimensional link-diagrams colored by X and some cycles of quandle homology; the pairings between these higher degree cycles and cocycles are expected to be invariants of higher dimensional links of codimen- sion two. Therefore, for the study of the invariants of links or of the pairings, it is important to find explicit formulae of quandle cocycles and to determine quandle cohomology groups. However, it is not so easy to calculate quandle (co)homology groups, since there are fewer methods to study them than group (co)homology theory. But some results on quandle (co)homologies are known: T. Mochizuki listed all 2-cocycles for finite Alexander quandles over a finite field in [M1] and all 3-cocycles for Alexander quandles on a finite field in [M2]. A presentation of a 3-cocycle found in [M1] gives rise to some applications for 2-knots (see [AS] for example). R. A. Litherland and S. Nelson analyzed the free and torsion subgroup of the quandle homology group of a finite quandle in [LN]. Regarding quandle homologies of higher degree, M. Niebrzy- dowski and J. H. Przytycki constructed some quandle homological operations and

Received by the editors November 17, 2009 and, in revised form, April 1, 2011. 2010 Mathematics Subject Classiﬁcation. Primary 20G10, 55N35, 58H10; Secondary 57Q45, 57M25, 55S20. Key words and phrases. Rack, quandle, homology, cohomology, knot.

c 2013 American Mathematical Society Reverts to public domain 28 years from publication 3413

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estimated the torsion subgroups of the integral quandle homology groups of some quandles in [NP]. In this paper we determine the integral quandle homology groups of all Alexander quandles of prime order p. Here this quandle is deﬁned to be Zp with a binary operation given by x ∗ y = ωx +(1− ω)y,whereZp means a cyclic group of order 1 p and ω ∈ Zp is neither 0 nor 1. The class of the quandles on Zp is known to be the simplest non-trivial class among quandles. To be speciﬁc, it is known [EGS] that any “connected” quandle of order p is isomorphic to one of the Alexander Q Z quandles. We denote the integral quandle homology by Hn (X; ). Our main Q Z result is to determine Hn (X; ) of all quandles of the simplest class as follows: Theorem 1.1. Let X be the Alexander quandle of order p with ω =0 , 1. Then the Q Z ∼ Z ⊕ Zb1 Q Z ∼ Zbn integral quandle homology groups are H1 (X; ) = p and Hn (X; ) = p for n ≥ 2,wherebn is determined by

bn+2e = bn + bn+1 + bn+2,b1 = b2 = ···= b2e−2 =0, and b2e−1 = b2e =1, and e is the order of ω, that is, e>0 is the minimal number satisfying ωe =1. Q Z ∼ We discuss some corollaries. Theorem 1.1 shows that Hn (X; ) = 0forthe Alexander quandle X with ω = −1, 0, 1and2≤ n ≤ 4 (Corollary 2.3), since e>2 and 2e − 1 > 4. Therefore, this tells us that only the case ω = −1isusefulin Alexander quandles of prime order for the study of quandle cocycle invariants of 1-knots and 2-knots. Furthermore, when ω = −1, the Alexander quandle X is said to be a dihedral quandle. As a corollary of Theorem 1.1, we settle the delayed F ibonacci conjecture by M. Niebrzydowski and J. H. Przytycki [NP] (see Corollary Q Z 2.2): they conjectured that the quandle homologies Hn (X; ) of dihedral quandles are characterized by a “delayed Fibonacci sequence”. Q Z Besides, we discuss the torsion subgroup of Hn (M; ). Here we deal with, more generally, connected Alexander quandles M of finite order: the quandle is defined by a Z[T ±]-module M satisfying (1 − T )M = M with a binary operation given by ∗ − Q Z | | x y = Tx+(1 T )y. A remarkable result is that Hn (M; ) is annihilated by M for n ≥ 2 (Corollary 6.2). As a special case, if X is the Alexander quandle of order ∈ Z Q Z p with ω p,thenHn (X; ) is annihilated by p (Corollary 6.4, which settles Q Z [NP, Conjecture 16]). It is known [LN, Theorem 1] that Hn (X; ) is annihilated by |X|n for a connected quandle X with a certain condition. Namely, Corollary 6.2 is a stronger estimate for Alexander quandles. Corollary 6.2 also plays a role in the proof of Theorem 1.1. Let X be the Q Z Alexander quandle of order p.SinceHn (X; ) is annihilated by p as above, it is a Zp-vector space; hence, to show Theorem 1.1, it suffices to determine the dimension of the space. To determine the dimension, we will compute that of the quandle cohomologies, since it is easier to deal with cohomologies than homologies in general. We give a more detailed outline of the computations in Section 2.3, and note that our proof follows from a generalization of the Mochizuki approach to the third cohomologies in [M1, M2].

1We ignore the case of ω = 0 and 1 throughout this paper. If ω = 0, then the binary operation is forbidden by quandle axioms. When ω = 1, the Alexander quandle is a trivial quandle. Hence − Q ∼ p(p−1)n 1 we can easily obtain the quandle homology: Hn (X; Z) = Z .Wethusomitthecase p =2.

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Our approach to proving our theorem provides beneﬁts on quandle cocycles. More concisely, we determine the quandle cohomolgies (Theorem 3.3), and obtain simple presentations of all higher degree cocycles (Corollary 5.8): we write down explicitly the presentations in Examples 5.1, 5.2 and 5.3. A key to obtaining the cocycles is an operation ⊕ n Z ⊕ n+1 Z ⊕ n+2 Z −→ n+2e Z Ωn Ωn−1 : HQ(X; p) HQ (X; p) HQ (X; p) HQ (X; p), where e is the order of ω. The operation is shown to be isomorphic (Corollary 5.6 and Remark 5.7); hence, presentations of all higher degree cocycles are obtained from those of lower degree via the operation by induction on n as desired. We present concretely the 4-cocycles for use in Example 5.9. These presentations of all quandle cocycles are expected to apply to invariants of higher dimensional links. In a future paper [No], the author will discuss the 4-cocycles obtained above. Unfortunately, the quandle cocycle invariants using the 4-cocycles of linked surfaces are equal to those using the 3-cocycle [No, Corollary 3.8]. On the other hand, recently, Clauwens [Cla] calculated the quandle homology of the dihedral quandle by algebraic topology. This paper is organized as follows. Section 2 reviews the quandle homology and reformulates Theorem 1.1. Section 3 contains preliminaries for the proof. Section 4 n Z ≤ − introduces some isomorphisms and calculates HQ(X; p)forn 2e 1. Section 5 determines all the quandle cohomology, leading to Theorem 1.1. Section 6 discusses the torsion parts of the quandle homologies for connected Alexander quandles.

2. Results Throughout this paper we fix an odd prime number p. We review quandle homologies of Alexander quandles in Section 2.1. We state the main result (Theorem 2.1) in Section 2.2 and outline the proof in Section 2.3. 2.1. Preliminaries: Rack and quandle homology groups. We begin by re- viewing Alexander quandles, which are an interesting class of quandles in this paper. An Alexander quandle is defined to be a Z[T ±]-module with a binary operation given by x ∗ y = Tx+(1− T )y. It is known [EGS] that any “connected” quandle of prime order is isomorphic to an Alexander quandle of type Z[T ±]/(p, T − ω)for some ω ∈ Zp,whereω is neither 0 nor 1 (see also [O, Section 5.1]). As it were, this type is the simplest non-trivial quandle among quandles. If ω = −1, the Alexander quandle is called a dihedral quandle. We next review the rack and quandle homology groups introduced in [CJKLS]. ∗ R Given an Alexander quandle (X, ) and a commutative ring A,letCn (X; A)be n the free A-module generated by n-tuples (x1,x2,...,xn)ofX ;inotherwords R n R R → Cn (X; A)=A X . Define its boundary homomorphism ∂n : Cn (X; A) R ≥ Cn−1(X; A)forn 2tobe R − i ∗ ∗ ∂n (x1,...,xn)= ( 1) (x1 xi,...,xi−1 xi,xi+1,...,xn) ≤ ≤ 2 i n − (x1,...,xi−1,xi+1,...,xn) , R R ◦ R and ∂1 to be the zero map. The composite ∂n−1 ∂n is known to be zero. The R R rack complex of X is the pair (C∗ (X; A),∂∗ ). Furthermore, since x ∗ x = x ∈ D ⊂ R for any x X,wehaveasubcomplexCn (X; A) Cn (X; A) generated by n- tuples (x1,...,xn) with xi = xi−1 for some i. Then we call the quotient complex

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Q R D Cn (X; A)=Cn (X; A)/Cn (X; A)thequandle complex of X. We denote the ho- R D Q mologies of these complexes by Hn (X; A), Hn (X; A), and Hn (X; A), respectively. R Q This Hn (X; A) is called the rack homology,andHn (X; A) is called the quandle homology. We recall known results on the quandle homologies of Alexander quandles X R Z ∼ Z with ω of order p. By direct calculation it can be veriﬁed that H1 (X; ) = Q Z ∼ Z and H1 (X; ) = (see also [LN]). T. Mochizuki showed [M1, Corollary 2.2] Q Z ∼ Q Z ∼ − H2 (X; ) = 0 and proved [M1, Theorem 3.1] that H3 (X; ) = 0ifω = 1, 0, 1. Also, in the case ω = −1, M. Niebrzydowski and J. Przytycki showed in [NP] that Q Z ∼ Z Z ⊂ Q Z H3 (X; ) = p and p H4 (X; ). It is also known [NP, Corollary 10] that Q Z n−2 Hn (X; ) is annihilated by p .

2.2. Main results. We will state our main theorem and give some corollaries and examples. We determine the integral quandle homology group of every Alexander quandle of order p as follows.

Theorem 2.1. Let ω ∈ Zp be ω =0 , 1.LetX be an Alexander quandle of type Z − Q Z ∼ X = [T ]/(p, T ω). Then the integral quandle homology groups are Hn (X; ) = Q Zbn ≥ Z ∼ Z ⊕ Zb1 p for n 2 and H1 (X; ) = p ,wherebn is determined by bn = bn−2e + bn−2e+1 + bn−2e+2, b1 = b2 = ··· = b2e−2 =0, and b2e−1 = b2e =1,ande is the order of ω.Inotherwords,e is the minimal number satisfying ωe =1.

As a special case, we obtain the homology in the case ω = −1. This settles the delayed F ibonacci conjecture by M. Niebrzydowski and J. H. Przytycki [NP, Conjecture 5]:

Corollary 2.2 ([NP, Conjecture 5]). Let X be the dihedral quandle of order p. Q Q Z ∼ Zbn ≥ Z ∼ Z ⊕ Zb1 Then Hn (X; ) = p for n 2 and H1 (X; ) = p ,wherebn is determined by bn+3 = bn+2 + bn,b1 = b2 =0, and b3 =1. i ∈ Z Proof. Note that e =2.PutFb(x)= i≥1 bix [[x]]. Theorem 2.1 says

b x1 + b x2 + b x3 + b x4 x3 + x4 x3 F (x)= 1 2 3 4 = = . b 1 − x2 − x3 − x4 1 − x2 − x3 − x4 1 − x − x3 Therefore, the generating function leads to the condition as required.

For the study of quandle cocycle invariants of 1-knots and 2-knots, it is important Q Z to determine Hn (X; )forn =2, 3 or 4 (see [CJKLS] and [CKS] for details). It follows from the next corollary that only the dihedral quandle is useful in Alexander quandles of prime order for the invariants.

Corollary 2.3. Let X be an Alexander quandle of order p with ω = −1, 0, 1.Then Q Z ∼ Hn (X; ) = 0 for n =2, 3 or 4. Proof. Since ω = −1, we see 2e − 1 > 4.

Q Z By Corollaries 2.2 and 2.3, the Betti number of the homology Hn (X; )inthe case of n ≤ 4orω = −1 does not depend on the odd prime p. However, the higher degree homology groups with ω = −1 do depend on p and ω in general.

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2.3. Outline of the proof. Here we outline the proof of Theorem 2.1. Let X be Q Z an Alexander quandle of order p. As we will show later on, Hn (X; ) is annihilated Q Z Z by p (Corollary 6.4). This implies that Hn (X; ) is a finite dimensional p-vector Q Z ∼ Zbn space, i.e., Hn (X; ) = p for some bn. Hence, if we know the dimension of Q Z Hn (X; ), then the proof of Theorem 2.1 would be complete. For this we will calculate the quandle “cohomology” group with Zp-coefficient. n Z The outline for studying the cohomology HQ(X; p) is a generalization of Mochi- k Z ≤ zuki’s approach [M2] to HQ(X; p)fork 3. Similarly to [M2, Section 2], we first n Z give a decomposition of HQ(X; p) in Section 3. We will define another cohomology n(0) n Z ∼ n(0) ⊕ n−1(0) H (X) given by (4), and show that HQ(X; p) = H (X) H (X)for ≥ each n 2 (Proposition 3.2). As a corollary, by the universal coefficient theorem, (0) (0) n(0) we will show bn = cn for n ≥ 2 (Lemma 5.5), where cn denotes dim H (X) . Therefore, we shall turn to the study of the dimension of Hn(0)(X). For the study on Hn(0)(X), we will construct an isomorphism φ¯ from Hn(0)(X)toaquotient vector space (Proposition 3.7). This quotient is a slight modification of a quotient vector space introduced in [M2, Section 3.2.4], where the third cohomology was reduced to an analysis of the quotient space. However, we need further steps to determine Hn(0)(X)forn ≥ 4, as contrasted with n ≤ 3. To begin, we set up some other quotient vector spaces in Section 4.1. Using them, a key step is to construct the following composite of some isomorphisms: −1 −1 −1 m m−4(0) m−3(0) m−2(0) φ¯ ◦ (Θ1) ◦···◦(Θe−1) ◦ Ψ : H (X) ⊕ H (X) ⊕ H (X) −→ Hn(0)(X),

where m = n − 2e + 4. In more detail, in Section 4.1 we will construct the map Θi given by (15) and show that the map Θi is isomorphic (Proposition 4.2). Further, m m we will deﬁne Ψ by (29) in Section 5.1, and show that Ψ is an isomorphism in Section 5.2. (0) (0) After the constructions, the above isomorphisms tell us that cn = cn−2e + (0) (0) (0) (0) ··· (0) cn−2e+1 + cn−2e+2. On the other hand, we show that c1 = c2 = = c2e−2 = (0) (0) (0) 0,c2e−1 = c2e = 1 in Section 4.3. Since bn = cn as mentioned above, this completes the proof of Theorem 2.1. Here we remark on presentations of all n-cocycles of Hn(0)(X). The above isomorphisms are constructed by polynomial segments (see Propositions 3.10 and 5.4, and Corollary 4.4). Then we obtain relatively simple presentations of all higher degree cocycles which generate Hn(0)(X) from those of lower degree cocycles (Corol- lary 5.8).

3. Quandle cohomology groups of Alexander quandles As mentioned in Section 2.3, for the proof of Theorem 2.1, we shall investigate the quandle cohomologies. In Section 3.1, we reformulate quandle cochain groups and decompose the groups. In Section 3.2 we state Theorem 3.3 determining the cohomology. In Section 3.3 we prepare a diﬀerential operation and an integral operation on the cochain group. These operations play a fundamental role in this paper. In Section 3.4, for the search of Hn(0)(X), we will construct an isomorphism from Hn(0)(X) to a quotient space (Proposition 3.7).

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3.1. Preliminaries: A reformulation of quandle cochain groups. For calculations of quandle (co)homology groups of Alexander quandles, it is convenient to change another “coordinate system” such as in [M2]. There is not anything new in RU this subsection. For an Alexander quandle M, we deﬁne Cn (M; A)tobethefree n A-module generated by n-tuples (U1,...,Un)ofM and deﬁne the boundary map by i ∂n(U1,...,Un)= (−1) (T · U1,...,T · Ui−1,T · Ui + Ui+1,Ui+2,...,Un) ≤ ≤ − 1 i n 1 (1) − (U1,...,Ui−1,Ui + Ui+1,Ui+2,...,Un) ,

for n ≥ 2, and ∂1 to be the zero map. It can be veriﬁed that ∂n−1 ◦ ∂n =0.Further we have a subcomplex group generated by n-tuples (U1,...,Un) with Ui =0for ≤ ≤ − QU some 1 i n 1. Then we have the quotient complex denoted by Cn (M; A). Dually, we can deﬁne the cochain groups denoted by Cn (M; A)andCn (M; A). RU QU We will give a canonical correspondence between these two complexes of Alexan- der quandles. Let us consider a bijection from M n to M n given by n n M (x1,...,xn) → (x1 − x2,x2 − x3,...,xn−1 − xn,xn) ∈ M .

R Q RU The map induces an isomorphism Cn (M; A)(resp. Cn (M; A)) to Cn (M; A) QU (resp. Cn (M; A)). Moreover, these isomorphisms are easily veriﬁed to be chain RU QU maps. We will mainly deal with the complexes Cn (M; A)andCn (M; A)in Sections 3 to 6. We will review a reformulation of the quandle cochain groups used in [M2]. Let X be an Alexander quandle of order p with ω ∈ Zp in all of the following. For n ≥ 1, we deﬁne n { · i1 ··· in ∈ Z | C (X)= ai1,...,in U1 Un p[U1,...,Un]

1 ≤ ij ≤ p − 1(j ≤ n − 1),in ≤ p − 1}, 0 n and C (X)=Zp. The coboundary is deﬁned as follows: for f ∈ C (X)andn ≥ 1,

δn(f)(U1,...,Un+1) i−1 := (−1) f(ω · U1,...,ω· Ui−1,ω· Ui + Ui+1,Ui+2,...,Un+1) ≤ ≤ 1 i n − f(U1,...,Ui−1,Ui + Ui+1,Ui+2,...,Un+1) ∈ Zp[U1,...,Un+1], n n+1 and δ0 is a zero map. We can check that δn(C (X)) ⊂ C (X)andδn+1 ◦ δn =0 for any n. The cohomology group is denoted by Hn(X). The complex Cn(X)is isomorphic to the cochain group Cn (X; Z ) in Section 2. Hence, the universal QU p coeﬃcient theorem leads to

n ∼ QU Z Z ⊕ 1 QU Z Z (2) H (X) = Hom(Hn (X; ), p) Ext (Hn−1(X; ), p). We will decompose the complex Cn(X) by the homogenous degree, that is, n { · i1 ··· in ∈ n | } Cd (X):= ai1,...,in U1 Un C (X) ik = d . 1≤k≤n n ⊂ n+1 Since δn(Cd (X)) Cd (X), we obtain a direct sum decomposition of the com- n n plexes as (C (X),δn)=( d Cd (X),δn). Next we review the formula in (3). Let ∈ n · a us decompose f Cd (X)asf = 0≤a≤p−1 fa(U1,...,Un−1) Tn , where we denote

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the n-th variable by Tn instead of Un. By an easy calculation, we have the following fundamental formula (see [M2, Lemma 3.2]), which the reader should keep in mind:

(3) · a δn(f)(U1,...,Un,Tn+1)= δn−1(fa)(U1,...,Un) Tn+1 0≤a≤p−1 n−1 d −1 a a +(−1) fa(U1,...,Un−1) · ω · (Un + ω Tn+1) − (Un + Tn+1) . 0≤a≤p−1

By the following lemma shown in [M1, M2], we may consider only the case of ωd =1.

d n Lemma 3.1 ([M1, Lemma 3.1]). If ω =1, then the complex Cd (X) is acyclic. ∈ n Proof. Let f Cd (X)beann-cocycle. Applying Tn+1 = 0 to (3), we can form − n−1 d − · a 0=δn−1(f0)(U1,...,Un)+( 1) (ω 1) fa(U1,...,Un−1) Un . 0≤a≤p−1

n d −1 We thus have f =(−1) (ω −1) ·δn−1(f0), which implies that f is cohomologous to zero.

Next, we consider the following submodule of Cn(X): for n ≥ 1

n(1) n C (X):={f0(U1,...,Un−1) ∈ C (X)| f0 ∈ Zp[U1,...,Un−1]},

0(1) { } n(1) n(1) ∩ n n(1) and C (X):= 0 .Put Cd (X):=C (X) Cd (X). Since δn(Cd (X)) is n+1(1) contained in Cd (X), we deﬁne the cohomologies as follows:

n(1) Zd (X):=Ker(δn), n(1) n−1(1) Bd (X):=δn−1(Cd (X)), n(1) n(1) n(1) Hd (X):=Zd (X)/Bd (X).

n n(1) We further deﬁne a cohomology of the quotient complex Cd (X)/Cd (X)asusual, i.e., n ∩ n Zd (X):=Ker(δn) Cd (X), n n−1 (4) Bd (X):=δn−1(Cd (X)), n(0) n n n(1) Hd (X):=Zd (X)/(Bd (X)+Zd (X)). These notation will be used later. Further, we denote Hn(1)(X)and d d n(0) n(1) n(0) d Hd (X)byH (X)andH (X), respectively. By Lemma 3.1, we have n(0) ∼ n(0) n(1) ∼ n(1) (5) H = Hd (X),H = Hd (X). d: ωd=1 d: ωd=1 (1) (0) n(1) n(0) Further cn and cn denote dim H (X) and dim H (X) , respectively. Note (0) (1) that by deﬁnition c0 = c1 = 1 is clear.

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3.2. Quandle cohomology and a decomposition of quandle complex. In this subsection, we state a decomposition of and the dimension of the quandle cohomology group for an Alexander quandle of order p. We ﬁrst show that the following short exact sequence splits:

in −→ n(1) −→d n −→ n n(1) −→ 0 Cd (X) Cd (X) Cd (X)/Cd (X) 0. Proposition 3.2 (cf. [M2, Theorem 3.12]). Let X be an Alexander quandle of n n → n(1) order p with ω.Letsd be the canonical projection Cd (X) Cd (X) as a linear map. Let n ≥ 2. d n n ∼ n(0) ⊕ (I) If ω =1,thensd is a chain map. In particular, Hd (X) = Hd (X) n(1) Hd (X). d n(1) → n−1 (II) If ω =1, then the canonical inclusion Cd (X) Cd (X) induces n−1(0) ∼ n(1) n ∼ n−1(0) ⊕ an isomorphism Hd (X) = Hd (X).Asaresult,H (X) = H (X) Hn(0)(X). This will be shown in the next section. In conclusion, for the study on Hn(X), it is enough to search Hn(0)(X). Furthermore, the following theorem determines Hn(0)(X).

(0) Theorem 3.3. Let X be as above, and let e be the order of ω,andcn := dim(Hn(0)(X)).Then (0) (0) (0) ≤ ≤ − (I) c0 = c2e−1 =1,andcn =0 for 1 n 2e 2. (0) (0) (0) (0) ≥ (II) cn = cn−2e+2 + cn−2e+1 + cn−2e for n 2e. Our purpose in Sections 4 to 6 is to show this theorem, which is the key to proving Theorem 2.2 as mentioned in Subsection 2.3:

n Remark 3.4. Consequently, we can determine H (X) as follows. Put the generat- i i ∈ Z ing function Fc(x):= i≥0 dim(H (X))x [[x]]. Since Proposition 3.2 follows n (0) (0) − 2e−2 2e · dim(H (X)) = cn−1 + cn , Theorem 3.3 indicates Fc(x)=(1 x + x ) (1 − x2e−2 − x2e−1 − x2e)−1. Furthermore, in Section 5.4 we will give concrete presentations of all n-cocycles n(0) n which span H (X) (Corollary 5.8). Therefore, we can determine dim(Hd (X)), n(0) and dim(Hd (X)) for any d and n, although we omit the explicit formulae. 3.3. Calculus on quandle cochain groups. We will set up a differential operation and an integral operation on the quandle cochain group. Using them, we will show Proposition 3.2 at the end of this subsection. Dn n → n − We first define a homomorphism d : Cd (X) Cd−1(X)ofdegree 1by Dn · a · · a−1 d ( fa(U1,...,Un−1) Tn )= a fa(U1,...,Un−1) Tn . 0≤a≤p−1 0≤a≤p−1

Dn n(1) · Note that Ker( d )=Cd (X) and that any element of the form fp−1(U1,...,Un−1) p−1 ∈ n Dn Dn Tn Cd−1(X) is not contained in the image of d .Further d is shown to be ◦Dn Dn+1 ◦ a chain homomorphism, that is, δn d = d δn. On the other hand, for n, d ∈ Z, we will introduce an integral operation from Dn n · a ∈ Dn Im( d )toCd (X). Let f = 0≤a≤p−2 fa(U1,...,Un−1) Un Im( d ). We then

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deﬁne the integration by the formula −1 · · a+1 (f)(U1,...,Un)= (a +1) fa(U1,...,Un−1) Un . n 0≤a≤p−2 · a ∈ n ◦Dn Note that for g = 0≤a≤p−1 ga(U1,...,Un−1) Un Cd (X), ( n d )(g)= g − g0. Furthermore, by direct calculation we can show the following relation between the integration and δn. ∈ Dn−1 n(1) Lemma 3.5. For f Im( d ), let us regard n−1(f) as an element of Cd (X). Then ◦ ◦ − n−1 − d ∈ n (6) (δn−1 )(f)=( δn−1)(f)+( 1) (1 ω ) (f) Cd (X). n−1 n n−1 In particular, if ωd =1, then the integration commutes with the coboundary map δn.

Proof of Proposition 3.2. To prove the part (I), we formulate any element f ∈ n k n n C (X)byf = f (U ,...,U − )U . Notice that the section s : C (X) → d k≤p−1 k 1 n 1 n d d n(1) n ◦Dn − Cd (X) is defined by sd (f)=f0(U1,...,Un−1). We first note that ( n d )(f − ∈ n n ◦ ◦Dn f0)=f f0 Cd (X)andthatsd n d is the zero map by definitions. Hence, sn+1(δ (f)) = sn+1(δ (f − f )) + sn+1(δ (f )) d n d n 0 d n 0 n+1 ◦Dn − = sd δn(( d )(f f0)) + δn(f0) n n+1 ◦ ◦Dn − n n =(sd ) (δn d )(f f0) + δn(sd (f)) = δn(sd (f)), n n where we use (6) in the third equality. Namely, the section sd is a chain map as desired. n(1) → n−1 To prove (II), reverify the canonical inclusion ι : Cd (X) Cd (X)fromthe n(1) definitions of these complexes. This map induces a bijection between Cd (X)and n−1 n−1(1) d Cd (X)/Cd (X). When ω = 1, formula (3) immediately concludes δn(f0)= δn−1(ι(f0)). Thus, the bijection is also a chain isomorphism, which immediately n(1) ∼ n−1(0) leads to Hd (X) = Hd (X) as required. 3.4. Mochizuki’s techniques in the general case of n. In [M1, M2], T. Mochi- zuki discovered all 2- and 3-cocycles of the quandle cohomology of a certain Alexan- der quandle. He reduced them to a quotient vector space in [M2, Section 3.2.4]. Following his techniques, in the general case of n, we will construct an isomorphism n(0) ∈ n from Hd (X) to a quotient space given by (8) below. Assume that f Cd (X)is an n-cocycle and ωd = 1. Then by (3) we obtain · a − − n 0= δn−1(fa)(U1,...,Un) Tn+1 ( 1) fa(U1,...,Un−1) ≤ ≤ − 0 a p 1 −1 a a · (Un + ω Tn+1) − (Un + Tn+1) . 1 By comparing with the coefficients of Tn+1 on both the left and right sides, we have − n −1 − · · a−1 (7) δn−1(f1)(U1,...,Un)=( 1) (ω 1) fa(U1,...,Un−1) a Un . 0≤a≤p−1

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− n −1 − ·Dn ∈ −1 Dn Namely, δn−1(f1)=( 1) (ω 1) d (f). Hence f1 δn−1(Im( d )). We thus n → −1 Dn obtain a homomorphism φ : Zd (X) δn−1(Im( d )) given by φ(f)=f1. · a ∈ n−1 d Lemma 3.6. Let g = a ga(U1,...,Un−2) Tn−1 Cd (X).Ifω =1,then − n −1 − ·Dn−1 φ(δn−1(g)) = δn−2(g1)+( 1) (ω 1) d (g). Proof. Straightforward. n(1) { } Further, it is evident that φ(Zd (X)) = 0 . Hence the homomorphism φ induces n δ−1 (Im(Dn)) ¯ Zd (X) n−1 d (8) φ : −→ − − . n n(1) Im(Dn 1)+Bn 1(X) Bd (X)+Zd (X) d d−1 T. Mochizuki showed that the homomorphism φ¯ is an isomorphism in the case of n = 3 [M2, Lemma 3.17]. It holds for general n as follows: Proposition 3.7. If ωd =1, then the homomorphism φ¯ is an isomorphism. Fur- · p−1 ∈ −1 Dn thermore, for a representative of the form gp−1(U1,...,Un−2) Un−1 δn−1(Im( d )), the preimage via φ¯ is represented by a polynomial gˆ of the form

gˆ(U1,...,Tn)=gp−1(U1,...,Un−2) −1 p p −1 p (9) (1 − ω )T − (U − + T ) +(U − + ω T ) · n n 1 n n 1 n . (1 − ω−1)p ¯ ∈ n Proof. We ﬁrst show that the homomorphism φ is injective. Let f Cd (X)be ¯ Dn−1 an n-cocycle such that φ(f)=0.Thenwehavef1 = δn−2(h)+ d (g)for ∈ n−2 ∈ n−1 ¯ ¯ some h Cd−1 (X)andg Cd (X). Put an n-cocycle f given by f = f + − n − −1 −1 · ∈ n Dn ¯ Dn ( 1) (1 ω ) δn−1(g) Cd (X). It follows from (7) that d (f)= d (f)+ − n − −1 −1 · Dn−1 Dn n(1) ( 1) (1 ω ) δn−1( d (g)) = 0. Since Ker( d )=Cd (X), it implies ∈ n n(1) f Bd (X)+Zd (X). ∈ n−1 ∈ Next, we will show its surjectivity. For g Cd−1 (X) satisfying δn−1(g) Dn Im( d ), we shall construct an n-cocycleg ˆ so that φ(ˆg)=g. Since we consider g Dn−1 modulo Im( d ), we may deal with only g of the form · p−1 g(U1,...,Un−2,Tn−1)=gp−1(U1,...,Un−2) Tn−1 , n−2 where gp−1 ∈ C − (X). Then we deﬁne a polynomialg ˆ by d p · p−1 ∈ n (10)g ˆ(U1,...,Un−1,Tn)= δn−1 gp−1(U1,...,Un−2) Un−1 Cd (X). n We now check thatg ˆ is an n-cocycle. Actually, since ωd = 1, equality (6) leads to ◦ · p−1 δn(ˆg)= δn δn−1 gp−1(U1,...,Un−2) Un−1 =0. n+1 It remains to show that φ(ˆg)=g and (9). For this, Lemma 3.9 below immediately indicates (11) − n − −1 · ∈ n−1(1) δn−2(gp−1)(U1,...,Un−1)=( 1) (1 ω ) gp−1(U1,...,Un−2) Cd−p (X). Using this equality, let us expand the right side in (10). Then we obtain the required formula (9). Further, by the formula (9), we easily see φ(ˆg)=g.

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Remark 3.8. We have easily obtained the simple formula (9) above. However, T. Mochizuki originally showed a slightly complicated formula to produce n-cocycles of Alexander quandles on R (see [M1, Proposition 3.1] for details). By pulling back the formula to Zp, he found a non-trivial 3-cocycle of the dihedral quandle of order p [M1, Theorem 3.1].

The following lemma will be used several times later on.

∈ n−1 n−1(1) Lemma 3.9. Let g Cd−1 (X) with no term of Cd−1 (X). Let us decompose g · a ∈ Dn as g = 1≤a≤p−1 ga(U1,...,Un−2) Un−1.Ifδn−1(g) Im( d ), then it satisﬁes (12) − n − d−1 · ∈ n−1(1) δn−2(gp−1)(U1,...,Un−1)=( 1) (1 ω ) gp−1(U1,...,Un−2) Cd−p (X). Proof. By (3) we thus have · a δn−1(g)(U1,...,Un−1,Tn)= δn−2(ga)(U1,...,Un−1) Tn ≤ ≤ − 1 a p 1 n d−1 −1 a a +(−1) ga(U1,...,Un−2) ω · (Un−1 + ω Tn) − (Un−1 + Tn) .

∈ Dn p−1 Since δn−1(g) Im( d ), by comparing with the coeﬃcients of Tn on the right side, we obtain equality (12).

Before going into the next section, we show Proposition 3.10 on the forms of the ω ∈ Z cocycles. For this, we now introduce the following polynomial En−1 p[Un−1,Un]: ω p −1 p −1 p E − (U − ,U )= (U − + U ) − (1 − ω )U − (U − + ω U ) /p n 1 n 1 n n 1 n n n 1 n ≡ − j−1 −1 · − −j · p−j · j (13) ( 1) j (1 ω ) Un−1 Un (mod p). 1≤j≤p−2

Proposition 3.10. Let X be an Alexander quandle of order p.ThenHn(0)(X) is · ω generated by some elements of the forms gp−1 En−1,wheregp−1 is contained in n−2 − n − −1 · ∈ n−1(1) Cd−p (X) for some d and satisﬁes δn−2(gp−1)=( 1) (1 ω ) gp−1 Cd−p (X) and ωd =1.

· ω Proof. It follows from Proposition 3.7 that n-cocycles of the forms gp−1 En−1 n(0) generate H (X). Moreover, by (11) in the previous proof, gp−1 have to satisfy − n − −1 · ∈ n−1(1) d δn−2(gp−1)=( 1) (1 ω ) gp−1 Cd−p (X)andω =1.

4. Proof of Theorem 3.3(I) Throughout this section we ﬁx the order e of ω,inotherwords,e is the minimal number of satisfying ωe = 1. Also, we assume that ωd = 1. In Section 4.1, we will search the quotient vector space in (8) in the case ω = −1. We will construct a homomorphism Θi and show that the map is an isomorphism (Proposition 4.2). Roughly speaking, the isomorphism makes the degree of the quotient space a lower degree. In Section 4.3, as a corollary, we will prove Theorem 3.3(I), i.e., (0) ··· (0) (0) (0) c1 = = c2e−2 =0andc0 = c2e−1 =1.

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4.1. Isomorphisms Θi. We will construct a map given below in (15). In this subsection we assume that ω = −1, and we ﬁx 1 ≤ i ≤ e − 2 such that n ≥ · j 2i + 2. To begin with, we put g = j gj (U1,...,Un−2i) Un−2i+1 contained in −1 Dn−2i+2 δn−2i+1(Im( d−ip+p )). By Lemma 3.9 we have

n−2i+2 d−ip+p−1 n −i δn−2i(gp−1)=(−1) (1 − ω ) · gp−1 =(−1) (1 − ω ) · gp−1. Moreover, we decompose · k gp−1(U1,...,Un−2i+1)= gk,p−1(U1,...,Un−2i) Un−2i+1. 1≤k≤p−1 Then the above equalities mean

n −i (−1) (1 − ω ) · g − (U ,...,U − ) p 1 1 n 2i · k = δn−2i−1(gk,p−1(U1,...,Un−2i) Tn−2i+1 ≤ ≤ − 1 k p 1 n−1 +(−1) gk,p−1(U1,...,Un−2i−1) ≤ ≤ − 1 k p 1 d−ip −1 k k · ω (Un−2i + ω Tn−2i+1) − (Un−2i − Tn−2i+1) .

1 Comparing the coeﬃcients of Tn−2i+1 in both the left and right sides shows an equality − n −i−1 − ·Dn−2i (14) δn−2i−1(g1,p−1)=( 1) (ω 1) d−ip (gp−1). Therefore we obtain a homomorphism −1 Dn−2i+2 −→ −1 Dn−2i (15) Θi : δn−2i+1(Im( d−ip+p )) δn−2i−1(Im( d−ip )),

given by Θi(g)=g1,p−1. We can verify the following lemma, similar to Lemma 3.6.

n−2i Lemma 4.1. Let h be an element of Cd−ip+p−1(X). We decompose · a · b h = ha,b(U1,...,Un−2i−2) Un−2i−1 Tn−2i. a,b If ωd =1,then

n −i−1 Θ (δ − (h)) = δ − − (h − )+(−1) (ω − 1) i n 2i n 2i 2 1,p 1 · · · a−1 ( a ha,p−1(U1,...,Un−2i−2) Un−2i−1). 1≤a≤p−1 Dn−2i+1 { } Further, it is clear that Θi Im( d−ip+p ) = 0 . Hence, the homomorphism Θi induces −1 n−2i+2 −1 n−2i δ − (Im(D − )) δ − − (Im(D − )) Θ : n 2i+1 d ip+p −→ n 2i 1 d ip . i Dn−2i+1 n−2i+1 Dn−2i−1 n−2i−1 Im( d−ip+p )+Bd−ip+p−1(X) Im( d−ip )+Bd−ip−1 (X) We remark that if i = 1, then the left side is exactly the right side in (8). In order to state Proposition 4.2, we now introduce the polynomial ω,i ∈ Z En−2i−1(Un−2i−1,Un−2i) p[Un−2i−1,Un−2i]

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given by the formula −i−1 p −i −1 p (1 − ω )U + ω (U − − + ω U − ) n−2i n 2i 1 n 2i p −i p − (U − − + U − ) +(1− ω )U /p (16) n 2i 1 n 2i n−2i−1 ≡ − j−1 −1 · −i−j − · p−j · j ( 1) j (ω 1) Un−2i−1 Un−2i (mod p). 1≤j≤p−1

d Proposition 4.2. If ω =1, n − 2i − 2 ≥ 0,and0

Before going into the proof, we make some remarks. Since ωd =1,wehave d−ip−p−1 −i −i −i−1 ω = ω . Also, note that 1 − ω , 1 − ω ∈ Zp are non-zero from the deﬁnition of e.

Proof. First, we study the injectivity. Assume that Θi(g) = 0, that is, g1,p−1 ∈ Dn−2i−1 n−2i−1 Im( d−ip )+Bd−ip−1 (X). We will show that such g can be killed below. We have Dn−2i−1 ∈ n−2i−1 ∈ n−2i−2 g1,p−1 = d−ip (g )+δn−2i−2(h)forsomeg Cd−ip (X)and h Cd−ip−1 (X). 0 We may assume that the coeﬃcient of Un−2i−1 in g is zero. Then we integrate (14) by Tn−2i,andobtain − n −i−1 − · ◦ ◦Dn−2i−1 ( 1) (ω 1) gp−1 =( δn−2i−1 d−ip )(g ) n−2i (18) ◦Dn−2i − − n −i − · =( d−ip ) δn−2i−1(g ) = δn−2i−1(g ) ( 1) (ω 1) g , n−2i

0 where the last equality is obtained from that the coeﬃcient of Un−2i−1 in δn−2i−1(g ) n−2i−2 d−ip is (−1) (ω −1)·g by (3). We putg ˇ(U1,...,Un−2i):=g (U1,...,Un−2i−1)· p−1 ∈ n−2i Un−2i Cd−ip+p−1(X). Then we have

δn−2i(ˇg)(U1,...,Un−2i,Tn−2i+1) · p−1 = δn−2i−1(g )(U1,...,Un−2i) Tn−2i+1 n+1 +(−1) g (U ,...,U − − ) 1 n 2i 1 −i −1 p−1 p−1 · ω (U − + ω T − ) − (U − + T − ) n 2i n 2i+1 n 2i n 2i+1 ≡ − n −i−1 − · −i − · · p−1 ( 1) (ω 1) gp−1 +(ω 1) g (U1,...,Un−2i) Tn−2i+1 − n+1 −i − · · p−1 +( 1) (ω 1) g (U1,...,Un−2i−1) Tn−2i+1 − n −i−1 − · · p−1 =( 1) (ω 1) gp−1(U1,...,Un−2i) Tn−2i+1 ≡ − n −i−1 − · Dn−2i+1 ( 1) (ω 1) g(U1,...,Un−2i,Tn−2i+1)mod(Im(d−ip+p )),

where the second equality is derived from (18). We remark that ωi+1 =1,since − ∈ n−2i+1 0

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∈ −1 Dn−2i Next, we will show the surjectivity. We put g δn−2i−1(Im( d−ip )). Consid- Dn−2i−1 ering it modulo Im( d−ip ), we may assume that g is of the form · p−1 g (U1,...,Un−2i−1)=gp−1(U1,...,Un−2i−2) Un−2i−1. ∈ Dn−2i+2 Here we claim that the following polynomialg ˘ satisfies δn−2i+1(˘g ) Im( d−ip+p ): ◦ · p−1 (19)g ˘ (U1,...,Un−2i,Tn−2i+1):=( δn−2i−1)(g )(U1,...,Un−2i) Tn−2i+1. n−2i p−1 − n Indeed, the coefficient of Tn−2i+2 in δn−2i+1(˘g ) vanishes, since ( 1) δn−2i+1(˘g )is reduced to be − n ◦ · p−1 ( 1) δn−2i+1 ( δn−2i−1)(g ) Un−2i+1 n−2i − n ◦ · p−1 ◦ =( 1) δn−2i ( δn−2i−1)(g ) Tn−2i+2 + ( δn−2i−1)(g ) n−2i n−2i d−ip+p−1 −1 p−1 p−1 · (ω (U − + ω T − ) − (U − + T − ) n 2i+1 n 2i+2 n 2i+1 n 2i+2 n =(−1) ( ◦δn−2i ◦ δn−2i−1)(g ) n−2i+1 − n − d−ip+p−1 ◦ · p−1 +( 1) (1 ω )( δn−2i−1)(g ) Tn−2i+2 + δn−2i−1(g ) n−2i n−2i −i p−1 p−1 · ω (ωU − + T − ) − (U − + T − ) n 2i+1 n 2i+2 n 2i+1 n 2i+2 · − −i p−1 = δn−2i−1(g ) (1 ω )Tn−2i+2 n−2i −i p−1 p−1 + ω (ωUn−2i+1 + Tn−2i+2) − (Un−2i+1 + Tn−2i+2) , where the first and second equalities are obtained from (3) and (6), respectively. Finally we will show that the above (−1)ng˘ is equal to (17), immediately leading to Θi(˘g )=g . For this, note that by Lemma 3.9 we obtain · p−1 − n − −i δn−2i−2(gp−1)(U1,...,Un−2i−1) Un−2i =( 1) (1 ω )gp−1(U1,...,Un−2i−2). n Therefore, by using (3), (−1) δn−2i−1(g )(U1,...,Un−2i)equals

n p−1 (−1) δ − − (g )(U ,...,U − − ) · U + g (U ,...,U − − ) n 2i 2 p−1 1 n 2i 1 n−2i p−1 1 n 2i 2 −i −1 p−1 p−1 · ω (Un−2i−1 + ω Un−2i) − (Un−2i−1 + Un−2i) = g − (U ,...,U − − ) p 1 1 n 2i 2 · − −i p−1 −i −1 p−1 − p−1 (1 ω )Un−2i +ω (Un−2i−1 +ω Un−2i) (Un−2i−1 + Un−2i) . Hence, by integrating this, (−1)ng˘ in (19) turns out to be the required formula in (17).

4.2. Composition of Θi’s. We will consider the composition of Θi’s in Corollary ω,i − 4.3. For this, we prepare the following product of the polynomials En−2i:ifω = 1, we deﬁne ω ω,i (20) En−2e+3≤n−2(Un−2e+3,...,Un−2):= En−2k−1(Un−2i−1,Un−2i). 1≤i≤e−2

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− ω ∈ Z Further, when ω = 1, we deﬁne En−2e+3≤n−2 to be 1 p.Wethenreacha conclusion on the composition of the isomorphisms presented in Proposition 4.2 as follows.

Corollary 4.3. If n ≥ 2e − 1 and ω = −1, then the composite map Θe−2 ◦···◦Θ1 gives rise to an isomorphism −1 n −1 Dn−2e+4 δ − (Im(D )) δ − (Im( − )) (21) n 1 d −→ n 2e+3 d ep+2p . Dn−1 n−1 Dn−2e+3 n−2e+3 Im( d )+Bd (X) Im( d−ep+2p)+Bd−ep+2p−1(X) Further, for a representative of the form · p−1 ∈ −1 Dn−2e+4 g(U1,...,Un−2e+2) Un−2e+3 δn−2e+3(Im( d−ep+2p)), ◦···◦ −1 · p−1 the preimage (Θe−2 Θ1) (g Un−2e+3) is represented by · ω · p−1 g(U1,...,Un−2e+2) En−2e+3≤n−2(Un−2e+3,...,Un−2) Tn−1 . Proof. This is proven according to Proposition 4.2 and the presentation of (19). In Section 5, we will deal with the right side of (21). ¯ ¯ We now consider the composition of Θi and φ,whereφ is given by (8), and we present n-cocycles. Note that the following statement also holds for ω = −1. ≥ − d ∈ n−2e+2 Corollary 4.4. Assume n 2e 1 and ω =1.Letg Cd−ep+p (X) be divisible − n − · ∈ n−2e+3(1) by Un−2e+2.Ifω =0, 1 and δn−2e+2(g)=( 1) (1 ω) g Cd−ep+p (X),then · p−1 −1 Dn−2e+4 g(U1,...,Un−2e+2) Un−2e+3 is contained in δn−2e+3(Im( d−ep+2p)). Moreover, for such g, the following polynomial is an n-cocycle of degree d: · ω · ω (22) g(U1,...,Un−2e+2) En−2e+3≤n−2(Un−2e+3,...,Un−2) En−1(Un−1,Un). Proof. The former is obtained by direct calculation similar to the proof in Lemma 3.9. We will show the latter part. Note that if ω = −1, i.e., e = 2, then the above statement is the same as Proposition 3.7. For ω = −1, since g(U1,...,Un−2e+2) · p−1 ∈ −1 Dn−2e+4 · Un−2e+3 δn−2e+3(Im( d−ep+2p)), then Corollary 4.3 yields g(U1,...,Un−2e+2) ω · p−1 ∈ −1 Dn En−2e+3≤n−2 Tn−1 δn−1(Im( d−1)). Moreover, according to Proposition 3.7 we conclude that the polynomial presented by (22) is an n-cocycle. 4.3. The proof of Theorem 3.3(I). Proof. First, the proof in the cases n ≤ 3orω = −1 immediately results from the (0) following facts. As mentioned in Section 3.1, c0 = 1. Further, we easily check 1(0) (0) 2 Z ∼ Z (X) = 0, which implies c1 =0.Forn =2,HQ(X; p) = 0isshown[M1, 3(0) ∼ Corollary 2.2]. It is known [M1, Theorem 3.1] that H (X) = Zp if ω = −1, and ∼ that H3(0)(X) = 0ifω = −1. ≥ − (0) ··· Next, we may assume that n 4andω =1, 0, 1. For the proof of c1 = = (0) c2e−2 = 0, by Proposition 3.7 it suﬃces to show that the right side of (8) vanishes for n<2e − 1. For this, we consider the two cases where n is odd or even. First we consider the case where n is even. Proposition 4.2 ensures an isomorphism − −1 n 1 D2 δ − (Im(D )) δ1 (Im( − )) n 1 d −→ d+p np/2 , Dn−1 n−1 D1 1 Im( d )+Bd−1 (X) Im( d+p−np/2)+Bd−1+p−np/2(X)

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given by the composite Θn/2−1 ◦···◦Θ1. Any element of the right side is represented · p−1 ∈ Z by a1 U1 for some a1 p. Assume a1 = 0. By the homogenous degree we verify p − 1=d + p − np/2 − 1, i.e., d = pn/2. Since n/2 ≤ e − 1 by assumption, we have ωd = ωn/2 = 1, which contradicts ωd = 1. Hence a =0,thatis,theabove 1 ∼ right side has no non-trivial elements. Therefore, Hn(0)(X) = 0. On the other hand, in the odd case of n, Proposition 4.2 gives rise to an isomorphism − −1 n 1 D3 δ − (Im(D )) δ2 (Im( − − )) (23) n 1 d −→ d p(n 3)/2 , Dn−1 n−1 D2 2 Im( d )+Bd−1 (X) Im( d−p(n−3)/2)+Bd−1−p(n−3)/2(X)

obtained from the composite Θ(n−3)/2 ◦···◦Θ1. Any element of the right side is · b · p−1 ∈ 2 ∈ Z represented by a1 U1 U2 Cd−1−p(n−3)/2(X)forsomea1 p and b =0.We will show a = 0 as follows. By Lemma 3.9, it satisﬁes that 1 b − · b · b b − b (24) a1(ω 1) U1 = δ1(a1 U1 )=a1 (ωU1 + U2) (U1 + U2) .

By elementary calculation we have b =1ora1 =0.Ifa1 = 0, then by comparing with the homogenous degrees, we have 1 + (p − 1) = d − 1 − p(n − 3)/2, i.e., d (n+1)/2 d = p(n +1)/2+1− p, which implies ω = ω = 1. Hence a1 = 0, i.e., the right side in (23) vanishes. (0) Finally, we will show c2e−1 = 1. Note that it follows from Proposition 4.2 that the isomorphism given by (23) holds for the case of n =2e − 1 as well. Hence, it suﬃces to show that the right side in (23) is one dimensional. Note that if d = ep − p +1,thenωd = 1. By equality (24) and the above discussion, the right · p−1 ∈ 2 (0) ≤ side of (23) is generated by U1 U2 Cp (X), which implies c2e−1 1. We · p−1 ∈ 2 claim that U1 U2 Cp (X) cannot be killed. Indeed the homogenous degree of 2 any element in Bd−1−p(n−1)/2(X)issmallerthanp by deﬁnition. In the sequel we (0) obtain c2e−1 =1.

Remark 4.5. Since δ1(U1)=(ω − 1) · U1 by deﬁnition, it follows from Corollary 4.4 2e−1(0) ∼ that H (X) = Zp is spanned by · ω · ω ∈ 2e−1 U1 E2≤2e−3(U2,...,U2e−3) E2e−2(U2e−2,U2e−1) Cep−p+1(X). − · ω In the case of ω = 1andn = 3, the resulting 3-cocycle U1 E2 (U2,U3) is found in [M1, Theorem 0.3] and [M2, Example 2.4.1].

5. Quandle cocycles and the proof of Theorem 2.1 In this section we deal with Hn(0)(X)forn ≥ 2e. For this, recall from Corollary 4.3 that Hn(0)(X) is isomorphic to the quotient space on the right side of (21). In Section 5.1 we will present some examples of the cocycles and construct a homomorphism from Hn−2e(0)(X) ⊕ Hn−2e+1(0)(X) ⊕ Hn−2e+2(0)(X)tothequotient space. In Section 5.2 we will show that the map is isomorphic (Proposition 5.4). In Section 5.3 we will prove Theorem 3.3 and the main theorem 2.2. As a result, we will formulate presentations of all cocycles which span Hn(X) (Corollary 5.8). Throughout this section we ﬁx n ≥ 2e and d satisfying ωd =1.Further,we denote by e the order of ω. To simplify the exposition, we deﬁne d = d +(2− e)p and m = n − 2e +4.Notethatm ≥ 4, ωd −p−1 = ωd −2p =1and(−1)m =(−1)n. Also notice that, if ω = −1, then d = d and m = n.

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5.1. Examples of n-cocycles of the Alexander quandle of prime order. ∈ m−3 − Example 5.1. Let fm−3 Cd−p−1(X)bean(m 3)-cocycle which is divisible by Um−3.Putgp−1(U1,...,Um−2):=fm−3(U1,...,Um−3) · Um−2.Thenwecan m check δm−2(gp−1)=(−1) (1 − ω) · gp−1 using formula (3). According to Corollary 4.4, by applying such gp−1 to (22), we have an n-cocycle given by (25) · · ω · ω ∈ n fm−3(U1,...,Um−3) Um−2 Em−1≤n−2(Um−1,...,Un−2) En−1(Un−1,Tn) Cd (X), ω where En−1(Un−1,Tn) is the polynomial given by (13). ω ∈ Example 5.2. We deﬁne the following polynomial Fm−3(Um−3,Um−2) Zp[Um−3,Um−2]: (26) ω p p p F − (U − ,U − )= (U − + U − ) − (ωU − + U − ) +(ω − 1)U /p m 3 m 3 m 2 m 3 m 2 m 3 m 2 m−3 ≡ − j−1 · −1 · − 1−j · p−j · j ( 1) j (1 ω ) Um−3 Um−2 (mod p). 2≤j≤p−1 ∈ m−4 − m−4(1) If fm−4 Cd−2p(X)isan(m 4)-cocycle with no term of Cd−2p (X), then we put · ω gp−1(U1,...,Um−2):=fm−4(U1,...,Um−4) Fm−3. Hence by direct calculation we m have δm−2(gp−1)=(−1) (1 − ω) · gp−1, which satisﬁes the condition in Corollary 4.4. Thus, by applying such gp−1 to (22), we obtain an n-cocycle given by · ω fm−4(U1,...,Um−4) Fm−3(Um−3,Um−2) · ω · ω ∈ n Em−1≤n−2(Um−1,...,Un−2) En−1(Un−1,Tn) Cd (X).

For example, when ω = −1, n =4,andf0 = 1, we have a 4-cocycle of the form ψ (U ,U ,U ,T ) 4,0 1 2 3 4 p − p − p · p − − p − p 2 = (U1 + U2) +(U1 U2) 2U1 (U3 + T4) (U3 T4) 2T4 /p . ω ∈ Example 5.3. We introduce the following polynomial Gm−3(Um−3,Um−2) Z [U − ,U − ]: p m 3 m 2 ω p+1 −1 p+1 G − = (U − + U − ) − ω(U − + ω U − ) m 3 m 3 m 2 m 3 m 2 − (1 − ω)U p+1 − (1 − ω−1)U p+1 /p (27) m−3 m−2 ≡ − j−1 · −1 −1 · − −j · p−j · j+1 ( 1) j (j +1) (1 ω ) Um−3 Um−2 (mod p). 1≤j≤p−2

Let fm−2 be an (m − 2)-cocycle of the form · ω fm−2(U1,...,Um−2)=fm−4(U1,...,Um−4) Em−3(Um−3,Um−2). · ω Then we put gp−1 := fm−4 Gm−3. By direct calculations we get δm−2(gp−1)= n (−1) (1 − ω) · gp−1. Therefore, because of (22) we ﬁnd an n-cocycle given by · ω fm−4(U1,...,Um−4) Gm−3(Um−3,Um−2) · ω · ω ∈ n Em−1≤n−2(Um−1,...,Un−2) En−1(Un−1,Tn) Cd (X). − · ω For example, when ω = 1, n =5,andf3 = U1 E2 (U2,U3)istheabove3-cocycle presented in Remark 4.5, we have a 5-cocycle given by · p+1 − p+1− p+1− p+1 · p− − p− p 2 U1 (U2+U3) +(U2 U3) 2U2 2U3 (U4+T5) (U4 T5) 2T5 /p .

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We will construct a map from Hm−2(0)(X) ⊕ Hm−3(0)(X) ⊕ Hm−4(0)(X)given below in (29). Examples 5.1, 5.2, and 5.3 enable us to deﬁne a homomorphism. m m−4 m−4(1) m−3 m−3(1) Ψ : Z (X)/Z (X) ⊕ Z (X)/Z (X) d d −2p d −2p d −p−1 d −p−1 m−2 m−4 ω ⊕ Z (X) ∩ C (X) · E (28) d −p−1 m−3 −1 m δ − (Im(D )) −→ m 1 d , Dm−1 m−1 Im( d )+Bd−1 (X) m · ω Ψd (fm−4,fm−3,fm−4 Em−3) · ω · p−1 · · p−1 · ω · p−1 := fm−4 Fm−3 Tm−1 + fm−3 Um−2 Tm−1 + fm−4 Gm−3 Tm−1. Furthermore, we extend the map on the third direct summand to a map on m−2 m−4 · ω Zd−p−1(X) deﬁned by zero outside C (X) Em−3. By elementary calculation, we can check that the map does not depend on the coboundaries. We thus obtain the induced map as follows: (29) −1 m δ (Im(D )) m m−4(0) ⊕ m−3(0) ⊕ m−2(0) −→ m−1 d Ψ : H (X) H (X) H (X) m−1 m−1 , Im(D )+B (X) d d d −1 where the direct sums are over all d satisfying ωd = ω2, i.e., ωd = 1. We will show

d m Proposition 5.4. Let ω =1. The above map Ψd is isomorphic. Hence, so is m Ψ . 5.2. The proof of Proposition 5.4. Since the proof is technical, the reader may skip it without loss of continuity. For the proof, we need differential operations p − 1 times and p times: Let us Dn ◦Dn ◦ ··· ◦ Dn n → n consider the composite d−j+1 d−j+2 d : Cd (X) Cd−j(X). We Dn Dn denote it by d−j≤d for short. It goes without saying that d−j≤d is a chain ◦Dn Dn+1 ◦ Dn homomorphism, δn d−j≤d = d−j≤d δn. Also note that d−p≤d(f)=0andthat Dn ∈ n(1) p−1 − n(1) d−p+1≤d(f) Cd−p+1(X)meansthecoefficientofTn in f.SinceCd−p+1(X) n−1 Dn injects Cd−p+1(X) as shown in Proposition 3.2, we often regard d−p+1≤d(f)as n−1 being in Cd−p+1(X). m Proof. We will show the surjectivity of Ψd in Step 1 and the injectivity in Step 2. ∈ m−1 Step 1. Let g be an element of the right side of (28). As usual, g Cd−1 (X) · p−1 ∈ may satisfy g(U1,...,Um−2,Tm−1)=gp−1(U1,...,Um−2) Tm−1 and δm−1(g) Dm Im( d ). Then gp−1 satisfies (12) by Lemma 3.9, which is rewritten in (30) − m − d −1 · ∈ m−1(1) δm−2(gp−1)(U1,...,Um−1)=( 1) (1 ω ) gp−1(U1,...,Um−2) Cd−p (X). Dm−1 We will show equality (31) below. For this, taking d−p to (30), we obtain Dm−1 Dm−2 0= d−p(δm−2(gp−1)) = δm−2( d−p(gp−1)). Dm−2 ∈ m−2 Namely, d−p(gp−1) Cd−p−1 is a cocycle. Recall here the decomposition m−2 ∼ m−2(0) ⊕ m−3(0) Hd−p−1 = Hd−p−1 Hd−p−1 in Proposition 3.2. Hence, there exist two repre- m−2(0) m−3(0) sentative cocycles fm−2 of Hd−p−1 (X)andfm−3 of Hd−p−1 (X) and a polynomial

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∈ m−3 h Cd−p−1(X) satisfying Dm−2 ∈ m−2 (31) d−p(gp−1)=fm−2 + fm−3 + δm−3(h) Cd−p−1(X). · Here, by Proposition 3.10, we may replace fm−2 by a cocycle of the form fm−4 ω Em−3. Wefocusonthethirdtermδm−3(h)in(31)operatedbythecomposite Dm−2 Dm−2 · ω d−2p≤d−p−1.Notethat d−2p≤d−p−1 annihilates fm−4 Em−3 from the deﬁni- ω m−4 ∈ m−4 Dm−3 tion of Em−3.Lethd−2p Cd−p−1 denote d−2p≤d−p−1(h). Hence, by applying Dm−2 d−2p≤d−p−1 to (31), we have Dm−3 m−4 · 0 m−4 0=δm−3( d−2p≤d−p−1(h)) = δm−3(hd−2p Um−3)=δm−4(hd−2p), d−2p − m−4 where the third equality follows from (3) and ω =1.Thus hd−2p equals some − · p−1 Dm−3 ∈ m−3 (m 4)-cocycle fm−4. Hence h = fm−4 Um−3 + d−p(k )forsomek Cd−p (X). To summarize, equality (31) is rewritten as Dm−2 · ω · p−1 Dm−2 (32) d−p(gp−1)=fm−4 Em−3 + fm−3 + δm−3(fm−4 Um−3)+ d−p(δm−3(k )). Here we present the following two formulas obtained by direct calculations: Dm−2 · ω · p−1 − (fm−4 Fm−3)=δm−3(fm−4 Um−3), (33) d p+1 Dm−2 · ω · ω d−p+1(fm−4 Gm−3)=fm−4 Em−3.

Therefore, integrating (32) by Um−2, we conclude · ω · · ω (34) gp−1 = fm−4 Gm−3 + fm−3 Um−3 + fm−4 Fm−3 + δm−3(k ). Here, we claim δm−3(k ) = 0. This is because that gp−1 satisﬁes equality (30) above, and that, by the discussions in Examples 5.1, 5.2, and 5.3, we have known a similar equality − − m − d −1 − ∈ m−1(1) δm−2 gp−1 δm−3(k ) =( 1) (1 ω ) gp−1 δm−3(k ) Cd−p (X). p−1 m Since g = gp−1Um−1 by deﬁnition, (34) precisely means the surjectivity of Φd . · ω Step 2. To prove the injectivity, for cocycles fm−2,fm−3,and fm−4 Em−3,we ∈ m−2 ∈ m−1 let h Cd−1 and k Cd satisfy − · ω · · ω · p−1 Dm−1 fm−4 Fm−3 + fm−3 Um−2 + fm−4 Gm−3 Tm−1 = δm−2(h)+ d (k). Our purpose is to check that each of the cocycles are cohomologous to zero. By Dm−1 operating d−p≤d−1 to both the left and right sides, we have (35) · ω · · ω Dm−2 ∈ m−1(1) fm−4 Fm−3 + fm−3 Um−2 + fm−4 Gm−3 = δm−2( d−p≤d−1(h)) Cd−p (X). Dm−2 ∈ m−2(1) m−3 The polynomial d−p≤d−1(h) Cd−p (X) may be regarded as that in Cd−p (X). ∈ m−3 Letusdenoteitbyh Cd−p (X). Then, because of (3), we notice Dm−2 · 0 δm−2( d−p≤d−1(h)) = δm−2(h Um−2) − m − −1 · ∈ m−2(1) = δm−3(h )+( 1) (1 ω ) h Cd−p (X). Dm−2 · 0 Dm−2 Also notice d−p(h Um−2) = 0. Thereby, by applying d−p to (35), Dm−2 · ω · ω Dm−3 ∈ m−2 (36) d−p(fm−4 Fm−3)+fm−3 +fm−4 Em−3 = δm−3( d−p(h )) Cd−p−1(X).

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Dm−2 · ω · p−1 Recall from (33) that d−p+1(fm−4 Fm−3)=δm−3(fm−4 Um−3). Therefore,

· ω Dm−3 − · p−1 (37) fm−3 + fm−4 Em−3 = δm−3( d−p(h ) fm−4 Um−3). · ω Namely, it follows from Proposition 3.2(I) that fm−3 and fm−4 Em−3 are null- cohomologous.

Finally, it suﬃces to show that fm−4 is cohomologous to zero. To begin, by Dm−2 ω ω applying d−2p+1≤d−p−1 to (36), from the deﬁnitions of Em−3 and Fm−3,we have (38) − 2 1 ω 2 m−3 m−2(1) (1−ω)f − ·U − + f ·U = δ − (D (h )) ∈ C (X). m 4 m 3 2 m−4 m−3 m 3 d −2p+1≤d −p d −2p+1 Dm−2 Dm−3 ∈ m−3(1) Similarly to d−p≤d−1(h)above, d−2p+1≤d−p(h ) Cd−2p+1 may be contained m−4 in Cd−2p+1.Wedenoteitbyh . Then the right side of (38) is expanded to

· 0 − m − −1 · ∈ m−2(1) = δm−3(h Um−3)=δm−4(h )+( 1) (1 ω ) h Cd−2p+1(X). Dm−3 Hence, equality (38) sent by the operation d−2p+1 is reduced to − − 2 · Dm−4 ∈ m−3 (1 ω)fm−4 +(1 ω )fm−4 Um−3 = δm−4( d−2p+1(h )) Cd−2p(X).

d−2p m−3 ∼ m−3(0) ⊕ Therefore, since ω = 1, it follows from the decomposition Hd−2p = Hd−2p m−4(0) Hd−2p in Proposition 3.2 that the cocycle fm−4 is null-cohomologous.

5.3. The proofs of Theorems 3.3(II) and 2.1. We now prove Theorem 3.3(II), and, using it, we show Theorem 2.1.

m Proof of Theorem 3.3(II). Recall the isomorphisms Θi and Ψd in Propositions 4.2 and 5.4. By composing them as mentioned in Section 2.3, we have an isomorphism from Hn(0)(X) ⊕ Hn+1(0)(X) ⊕ Hn+2(0)(X)toHn+2e(0)(X), which immediately implies Theorem 3.3(II).

Q Z Proof of Theorem 2.1. By Corollary 6.4 (which we will show later) Hn (X; )isan- Q Z Z QU Z ∼ nihilated by p;thatis,Hn (X; )isa p-vector space. We thus have Hn (X; ) = (0) Zbn n(0) p for some bn. Recall Theorem 3.3(II): the dimension cn = dim(H ) is deter- (0) (0) (0) (0) (0) ··· (0) (0) (0) mined by cn+2e = cn+2 +cn+1 +cn ,c1 = = c2e−2 =0, and c2e−1 = c2e =1. (0) By Lemma 5.5 below, bn = cn for n ≥ 2. Hence, we obtain the recurrence formula in Theorem 2.2.

(0) Lemma 5.5. The above bn satisﬁes bn = cn for n ≥ 2. 2 Z ∼ Proof. To begin, if n = 2, it is shown [M1, Corollary 2.2] that HQ(X; p) = 0. By QU Z ∼ (0) (2) we obtain H2 (X; ) = 0. Hence b2 = c2 =0.Furthermore,fork>2, we k (0) (0) have bk + bk−1 = dim(H (X)) = ck + ck−1 by the universal coeﬃcient theorem (0) and Proposition 3.2. Hence, the induction on k concludes bn = cn for n ≥ 2.

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5.4. Cohomological operations and presentations of n-cocycles. As a result of Theorem 3.3, we will present all cocycles which span the cohomology Hn(X). To ω ω ω ω see this, we use the four polynomials En−1, En−2e+3≤n−2, Fn−2e+1,andGn−2e+1; see the formulas (13), (20), (26), and (27) for the deﬁnitions, respectively. For this purpose, we now set up an operation on the quandle cohomology group (cf. [NP, Sections 2 and 6] for a study on homological operations for certain quandles). m−4 m−4(1) m−3 m−3(1) Ω : Z (X)/Z (X) ⊕ Z (X)/Z (X) m d −2p d −2p d −p−1 d −p−1 ⊕ m−2 ∩ m−4 · ω Zd−p−1(X) C (X) Em−3 −→ n(0) Hd (X), · ω Ωm(fm−4,fm−3,fm−4 Em−3) · ω · · ω · ω · ω := (fm−4 Fm−3 + fm−3 Um−2 + fm−4 Gm−3) Em+1≤n−2 En−1, where m = n − 2e + 4. We notice that this does not depend on the coboundaries by Propositions 3.7, 4.2, and 5.4. Therefore, the Ωm induces n−2e(0) n−2e+1(0) n−2e+2(0) n(0) Ωn−2e+4 : H (X) ⊕ H (X) ⊕ H (X) −→ H (X). From the construction, we can check that by Proposition 3.7 and Corollary 4.4 the −1 −1 −1 m above Ωn−2e+4 is equal to the composite φ¯ ◦ (Θ1) ◦···◦(Θe−1) ◦ Ψ .Since each homomorphism is an isomorphism, we thus obtain

Corollary 5.6. Ωn−2e+4 is an isomorphism for n ≥ 2e. ∼ − Remark 5.7. Recall Hn(X) = Hn(0)(X)⊕Hn 1(0)(X) from Proposition 3.2. Then, the direct sum Ωn−2e+4 ⊕ Ωn−2e+3 induces an isomorphism on the quandle cohomology Hn(X). Next, we will present all cocycles which span the cohomology Hn(X). For this ω,(0) purpose we will define sets of some n-cocycles denoted by Cocn as follows. We ω,(0) ∈ Z ω,(0) first define Coc0 to be 1 p. Define Coci by the empty sets for 0

ω ∈ n−2e ω,(0) noting that fn−2e+2/En−2e+1 C (X) from the deﬁnition of Cocn−2e+2.Also, ω,(0) n(1) by deﬁnition, notice that any element of Cocn has no term of C (X). Corollary 5.8. Let X be the Alexander quandle of order p with ω =0 , 1.Then n(0) ω,(0) H (X) is independently generated by Cocn . Further, the quandle cohomology n ω,(0)∪ ω,(0) − H (X) is independently generated by Cocn Cocn−1 , where we regard the (n 1)- ω,(0) n(1) cocycles in Cocn−1 as n-cocycles of C (X).

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0 ∼ Proof. By deﬁnition H (X) = Zp is generated by 1 ∈ Zp. By Theorem 3.3 we have i(0) ∼ 2e−1(0) ∼ H (X) = 0for0

Finally, we end this section by taking examples of 4-cocycles. − 4 Z ∼ Z 2 Example 5.9. Let ω = 1andn =4.ThenHQ(X; p) = ( p) . Further, the two bases are represented by ψ (x, y, z, w):=(x − y) · 2(z − w)p − (2z − w − y)p − (y − w)p /p, 4,0 ψ (x, y, z, w):= (x − 2y + z)p +(x − z)p − 2(x − y)p 4,1 · 2wp − (2z − w)p − zp /p2.

4 Z Here, we use the usual coordinate system of CQ(X; p), such as in [CJKLS, CKS].

6. Torsion subgroup of integral quandle homology The goal in this section is to show Theorem 6.1. As a corollary, for an Alexander quandle X of order p, the quandle homology is annihilated by p (Corollary 6.4). We will review the finite connected Alexander quandles and the torsion subgroups of the quandle homology groups. Let M be an Alexander quandle, that is, M is a Z[T,T−1]-module with a binary operation given by x ∗ y = Tx+(1− T )y.If the quotient module M/(1 − T )M is zero, the quandle is said to be connected.Itis known [LN, Proposition 1] that this definition is equivalent to the standard definition of the connectivity. R. A. Litherland and S. Nelson showed in [LN, Theorem 1] R Z Z that if M is finite and connected, then the free subgroup of Hn (M; )is and the ∈ R Z generator is represented by (0, 0,...,0) Cn (M; ) and that the torsion subgroup is annihilated by |M|n for each n ≥ 1. It is shown in [NP, Theorem 17] that for the R Z dihedral quandle X of order 3, the torsion part of Hn (X; ) is annihilated by 3. More generally, in this paper we will show a stronger estimate of the integral quandle homology group for finite connected Alexander quandles. Theorem 6.1. Let M be a finite connected Alexander quandle. Then, for each n, R Z | | the torsion part of Hn (M; ) is annihilated by M . Before the proof, we will present some corollaries and remarks. Q Z | | Corollary 6.2. Let M be as above. Then Hn (M; ) is annihilated by M for n ≥ 2. Q Z Proof. According to [LN, Theorem 4], Hn (M; )isshowntobeadirectsummand R Z Q Z Q Z of Hn (M; ). Since (0,...,0) is zero in Cn (M; ), Hn (M; ) has no free part.

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Q Z Remark 6.3. It is false that in general, for a ﬁnite connected quandle, Hn (X; ) is annihilated by |X|. For example, let QS(6) be the connected quandle of order 6 presented by [CKS, Example 2.2]. QS(6) is not isomorphic to any Alexan- der quandle (see [O, Section 5.1]). Then it is known [CKS, Example 2.5] that Q Z ∼ Z Z ∼ Z 3Z ⊕ Z Z H3 (QS(6); ) = /24 = /2 /3 . We immediately obtain the following corollaries, which we state without proof. Corollary 6.4. Let X be an Alexander quandle of order p, that is, X = Z − Q Z p[T ]/(T ω) for some ω =0, 1.ThenHn (X; ) is annihilated by p. 2 Corollary 6.5. If X = Z2[T ]/(T + T +1) with the Alexander quandle operation, Q Z then Hn (X; ) is annihilated by 4. Remark 6.6. Corollary 6.4 is conjectured in [NP, Conjecture 16]. Proof of Theorem 6.1. We will work with our coordinate system of the rack com- j j RU Z → RU Z plex in (1). We now introduce two chain maps f+,f0 : Cn (M; ) Cn (M; ) given by ⎧ ⎪ ⎨⎪ 0,...,0,y,Uj+1,...,Un for 0

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 3436 TAKEFUMI NOSAKA j j − j j−1 − j By direct calculation we obtain ∂n+1Dn,+ + Dn−1,+∂n =( 1) f+ f0 for j j j+1 j>1. Therefore Dn,+ are chain homotopies between f+ and f0 . This completes the proof. Acknowledgments The author thanks Toshiyuki Akita, Maciej Niebrzydowski, J´ozef Przytycki, Masahico Saito, Shin Satoh, Kokoro Tanaka, Michihisa Wakui, and Tadayuki Watanabe for valuable comments and discussions. He is sincerely grateful to Takuro Mochizuki for carefully reading the paper and making suggestions for its improve- ment. He also expresses his gratitude to Tomotada Ohtsuki for encouragement and useful advice. References

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Research Institute for Mathematical Sciences, Kyoto University, Sakyo-ku, Kyoto, 606-8502, Japan E-mail address: [email protected]

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