Toric rings, inseparability and rigidity

arXiv:1508.01290v1 [math.AC] 6 Aug 2015

TORIC RINGS, INSEPARABILITY AND RIGIDITY ¨ MINA BIGDELI, JURGEN HERZOG AND DANCHENG LU Abstract. Let K be a field, H an affine semigroup and R = K[H] its toric ring over K. In this paper we give an explicit description of the ZH-graded components of the cotangent module T 1 (R) which classifies the infinitesimal deformations of R. In particular, we are interested in unobstructed deformations which preserve the toric structure. Such deformations we call separations. Toric rings which do not admit any separation are called inseparable. We apply the theory to the edge ring of a finite graph. The coordinate ring of a convex polyomino may be viewed as the edge ring of a special class of bipartite graphs. It is shown that the coordinate ring of any convex polyomino is inseparable. We call a bipartite graph G semi-rigid if T 1 (R)a = 0 for all a with −a ∈ H. Here R is the edge ring of G. A combinatorial description of semi-rigid graphs is given. The results are applied to show that for n = 3, Gn is semi-rigid but not rigid, while Gn is rigid for all n ≥ 4. Here Gn is the complete bipartite graph Kn,n with one edge removed.

Introduction In this paper we study infinitesimal deformations and unobstructed deformations of toric rings which preserve the toric structure, and apply this theory to edge ideals of bipartite graphs. Already in [1] and [2], infinitesimal and homogeneous deformations of toric varieties have been considered from a geometric point of view. The view point of this paper is more algebraic and does not exclude non-normal toric rings, having in mind toric rings which naturally appear in combinatorial contexts. This aspect of deformation theory has also been pursued in the papers [4],[5] and [3] where deformations of Stanley-Reisner rings attached to simplicial complexes were studied. Let K be a field. The infinitesimal deformations of a finitely generated K-algebra R are parameterized by the elements of the cotangent module T 1 (R) which in the case that R is a domain is isomorphic to Ext1R (ΩR/K , R), where ΩR/K denotes the module of differentials of R over K. The ring R is called rigid if T 1 (R) = 0. We refer the reader to [12] for the theory of deformation. Let H be an affine semigroup and K[H] its affine semigroup ring. We are interested in the module T 1 (K[H]). This module is naturally ZH-graded. Here ZH denotes the associated group of H which for an affine semigroup is a free group of 2010 Mathematics Subject Classification. Primary 13D10, 05E40; Secondary 13C13. Key words and phrases. Deformation, Toric ring, Rigid, Inseparable, Bipartite graph, Convex Polyomino. The paper was written while the third author was visiting the Department of Mathematics of University Duisburg-Essen. He wants to express his thanks for the hospitality. 1

finite rank. For each a ∈ ZH, the a-graded component T 1 (K[H])a of T 1 (K[H]) is a finite dimensional K-vector space. In Section 1 we describe the vector space T 1 (K[H])a and provide a method how to compute its dimension. Let H ⊂ Zm with generators h1 , . . . , hn . Then the associated group ZH of H is a subgroup of Zm , and K[H] is the K-subalgebra of the ±1 h1 hn ring K[t±1 1 , . . . , tm ] of Laurent polynomials generated by the monomials t , . . . , t . a(1) a(m) Here ta = t1 · · · tm for a = (a(1), . . . , a(m)) ∈ Zm . Let S = K[x1 , . . . , xn ] be the polynomial ring over K in the indeterminates x1 , . . . , xn . Then S may be viewed as a ZH-graded ring with deg xi = hi , and the K-algebra homomorphism S → K[H] with xi 7→ thi is a homomorphism of ZH-graded K-algebras. We denote by IH the kernel of this homomorphism. The ideal IH is called the toric ideal associated with H. It is generated by homogeneous binomials. To describe these binomials, consider the group homomorphism Zn → Zm with εi 7→ hi , where ε1 , . . . , εn is the canonical basis of Zn . The kernel L of this group homomorphism is a lattice of Zn and is called the relation lattice of H. For v = (v(1), . . . , v(n)) ∈ Zn we define the Q Q v(i) −v(i) binomial fv = fv+ − fv− with fv+ = i, v(i)≥0 xi and fv− = i, v(i)≤0 xi , and let IL be the ideal generated by the binomials fv with v ∈ L. P It is well known that IH = IL . Each fv ∈ IH is homogeneous of degree h(v) = i, v(i)≥0 v(i)hi . Let fv1 , . . . , fvs be a system of generators of IH . We consider the (s × n)-matrix   v1 (1) v1 (2) . . . v1 (n)  v2 (1) v2 (2) . . . v2 (n)  AH =  . .. ..   ... . .  vs (1) vs (2) . . . vs (n)

Summarizing the results of Section 1, dimK T 1 (K[H])a can be computed as follows: let l = rank AH , la be the rank of the submatrix of AH whose rows are the ith rows of AH for which a + h(vi ) 6∈ H, and let da be the rank of the submatrix of AH whose columns are the jth columns of AH for which a + hj ∈ H. Then dimK T 1 (K[H])a = l − la − da . In Section 2 we introduce the concept of separation for a torsionfree lattice L ⊂ Zn . Note that a lattice L ⊂ Zn is torsionfree if and only if it is the relation lattice of some affine semigroup. Given an integer i ∈ [n] = {1, 2, . . . , n}, we say that L admits an i-separation if there exists a torsionfree lattice L′ ⊂ Zn+1 of the same rank as L such that πi (IL′ ) = IL , where πi : S[xn+1 ] → S is the K-algebra homomorphism which identifies xn+1 with xi . An additional condition makes sure that this deformation which induces an element in T 1 (K(H))−hi is non-trivial, see 2.1 for the precise definition. We say that L is inseparable, if for all i, the lattice L admits no i-separation, and we call H and its toric ring inseparable if its relation lattice is inseparable. In particular, if the generators of H belong to a hyperplane of Zm , so that K[H] also admits a natural standard grading, then H is inseparable if T 1 (K[H])−1 = 0. In general, the converse is not true since the infinitesimal deformation given by nonzero element of T 1 (K[H])−1 may be obstructed. We demonstrate this theory and 2

show that a numerical semigroup generated by three elements which is not a complete intersection is i-separable for i = 1, 2, 3, while if it is a complete intersection it is i-separable for at least two i ∈ {1, 2, 3}. For the proof of this fact we use the structure theorem of such semigroups given in [8]. The last Section 3 is devoted to the study of T 1 (R) when R is the edge ring of a bipartite graph. This class of rings has been well studied in combinatorial commutative algebra, see e.g. [11] and [13]. For a given simple graph G of the vertex set [n] one considers the edge ring R = K[G] which is the toric ring generated over K by the monomials ti tj for which {i, j} is an edge of G. Viewing the edge ring a semigroup ring K[H], the edges ei of G correspond the generators hi of the semigroup H. We say that G is inseparable if the corresponding semigroup is inseparable. The main result of the first part of this section is a combinatorial criterion for G being inseparable. Let C be a cycle of G and e a chord of G. Then e splits C into two disjoint connected components C1 and C2 which are obtained by restricting C to the complement of e. A path P of G is called a crossing path of C with respect to e if one end of P belongs to C1 and the other end to C2 . Now the criterion (Corollary 3.5) says that a bipartite graph G is inseparable if and only if for any cycle C which has a unique chord e, there exists a crossing path of C with respect to e. In particular, if no cycle has a chord, then G is inseparable. By using this criterion we show in Theorem 3.6 that the coordinate ring of any convex polyomino, which may be interpreted as a special class of edge rings, is inseparable. For the rest of this section we consider the semi-rigidity and rigidity of bipartite graphs. We call H semi-rigid if T 1 (K(H))−a = 0 for all a ∈ H, and characterize in Theorem 3.10 semi-rigidity of bipartite graphs in terms of the non-existence of certain constellations of edges and cycles of the graph. To classify rigidity of bipartite graphs is much more complicated, and we do not have a general combinatorial criterion for when a bipartite graph is rigid. Instead we consider for each n ≥ 3 the graph Gn which is obtained by removing an edge from the complete bipartite graph Kn,n . It is shown in Proposition 3.12 that for n = 3, Gn is not rigid while for n ≥ 4, Gn is rigid. It remains a challenging problem to classify all rigid bipartite graphs.

1. T 1 for toric rings Let H be an affine semigroup, that is, a finitely generated subsemigroup of Zm for some m > 0. Let h1 , . . . , hn be the minimal generators of H, and fix a field K. The ±1 toric ring K[H] associated with H is the K-subalgebra of the ring K[t±1 1 , . . . , tm ] of a(1) a(m) Laurent polynomials generated by the monomials th1 , . . . , thn . Here ta = t1 · · · tm for a = (a(1), . . . , a(m)) ∈ Zm . Let S = K[x1 , . . . , xn ] be the polynomial ring over K in the variables x1 , . . . , xn . The K-algebra R = K[H] has a presentation S → R with xi 7→ thi for i = 1, . . . , n. The kernel IH ⊂ S of this map is called the toric ideal attached to H. Corresponding to this presentation of K[H] there is a presentation Nn → H of H which can be extended to the group homomorphism Zn → Zm with εi 7→ hi for i = 1, . . . , n, where ε1 , . . . , εn denotes the canonical basis of Zn . Let L ⊂ Zn be the kernel of this 3

group homomorphism. The lattice L is called the relation lattice of H. Note that L is a free abelian group and Zn /L is torsion-free. For a vector v ∈ Zn with v = (v(1), . . . , v(n)), we set X X v+ = v(i)εi and v− = −v(i)εi . i, v(i)≥0

i, v(i)≤0

Then v = v+ −v− . It is a basic fact and well-known (see e.g. [7]) that IH is generated by the binomials fv with v ∈ L, where fv = xv+ − xv− . We define an H-grading on S by setting deg xi = hi . Then IH is a graded ideal with deg fv = h(v), where X X (1) h(v) = v(i)hi (= −v(i)hi ). i, v(i)≥0

i, v(i)≤0

Let v1 , . . . , vr be a basis of L. Since IH is a prime ideal we may localize S with respect to this prime ideal and obtain IH SIH = (fv1 , . . . , fvr )SIH . In particular, we see that (2)

height IH = rank L.

Let ΩR/K be the module of differentials of R over K. Since R is a domain, the cotangent module T 1 (R) is isomorphic to Ext1R (ΩR/K , R), and since R is H-graded it follows that ΩR/K is H-graded as well, and hence Ext1R (ΩR/K , R) and T 1 (R) are ZHgraded. Here ZH denotes the associated group of H, that is, the smallest subgroup of Zm containing H. It is our goal to compute the graded components T 1 (R)a of T 1 (R) for a ∈ ZH. The module of differentials has a presentation ΩR/K = (

n M

Rdxi )/U,

i=1

L where U is the submodule of the free R-module ni=1 Rdxi generated by the elements dfv with v ∈ L, where n X dfv = (∂fv /∂xi )dxi . i=1

Here ∂fv /∂xi stands for partial derivative of fv with respect to xi , evaluated modulo IH . One verifies at once that n X dfv = v(i)th(v)−hi dxi . (3) i=1

L For i ∈ [n], the basis element dxi of ΩS/K ⊗S R = ni=1 Rdxi is given the degree hi . Then U is an H-graded submodule of ΩS/K ⊗S R, and deg dfv = deg fv = h(v). 4

For any ZH-graded R-module M we denote by M ∗ the graded R-dual HomR (M, R). Then the exact sequence of H-graded R-modules 0 → U → ΩS/K ⊗S R → ΩR/K → 0 gives rise to the exact sequence (ΩS/K ⊗S R)∗ → U ∗ → T 1 (R) → 0 of ZH-graded modules. This exact sequence may serve as the definition of T 1 (R), namely, to be the cokernel of (ΩS/K ⊗S R)∗ → U ∗ . Let fv1 , . . . , fvs be a system of generators of IH , where we may assume that for r ≤ s, the elements v1 , . . . , vr form a basis of L. In general s is much larger than r. Observe that the elements dfv1 , . . . , dfvs form a system of generators of U. We let F be a free graded R-module with basis g1 , . . . , gs such that deg gi = deg dfvi for i = 1, . . . , s, and define the R-module epimorphism F → U by gi 7→ dfvi for i = 1, . . . , s. The kernel of F → U we denote by C. The composition F → ΩS/K ⊗S R of the epimorphism F → U with the inclusion map U → ΩS/K ⊗S R will be denoted by δ. We identify U ∗ ⊂ F ∗ with its image in F ∗ . Then T 1 (R) = U ∗ / Im δ ∗ and U ∗ is the submodule of F ∗ consisting of all ϕ ∈ F ∗ with ϕ(C) = 0. We first describe the ZH-graded components of U ∗ . Let a ∈ ZH. We denote by KL the K-subspace of K n spanned by v1 , . . . , vs and by KLa the K-subspace of KL spanned by the vectors vi with i ∈ / Fa . Here the set Fa is defined to be Fa = {i ∈ [s] : a + h(vi ) ∈ H}. Then we have Theorem 1.1. For all a ∈ ZH, we have dimK (U ∗ )a = dimK KL − dimK KLa . Proof. Let σ1 , . . . , σs be the canonical basis of K s and W ⊂ K s be the kernel of the K-linear map K s → KL with σi 7→ vi for i = 1, . . . , s. We will show that (4)

(U ∗ )a ∼ = {µ ∈ K s : µ(i) = 0 for i ∈ [s] \ Fa and hµ, λi = 0 for all λ ∈ W },

as K-vector space. Assuming this isomorphism has been proved, let Xa be the image of W ⊂ K s L under the canonical projection K s → Va = i∈Fa Kσi . Then (4) implies that (U ∗ )a is isomorphic to the orthogonal complement of Xa in Va . Thus, (5)

dimK (U ∗ )a = |Fa | − dimK Xa . 5

L Let Za = i6∈Fa Kσi and Ya the cokernel of Xa → Va . Then we obtain a commutative diagram with exact rows and columns 0   y

0 −−−→ W ∩ Za −−−→   y

0 −−−→

0 −−−→

W   y

Xa   y

0   y

0   y

Za −−−→ KLa −−−→ 0     y y

−−−→ K s −−−→ KL −−−→ 0     y y −−−→ Va −−−→ Ya     y y

−−−→ 0

0 0 0 ∗ Now (5) implies that dimK (U )a = dimK Ya , and the diagram shows that dimK Ya = dimK KL − dimK KLa . It remains to prove the isomorphism (4). Observe that (U ∗ )a =P {ϕ ∈ (F ∗ )a : ϕ(C) = ∗ 0}, where C is the kernel of F → U. Let ϕ ∈ (F )a . Then ϕ = si=1 ϕ(gi )gi∗ , where g1∗ , . . . , gs∗ is the basis of F ∗ dual to g1 , . . . , gs . Since deg gi∗ = − deg dfvi = −h(vi ), it follows that ϕ ∈ (F ∗ )a if and only if ϕ(gi ) = µ(i)ta+h(vi ) with µ(i) ∈ K and µ(i) = 0 if a + h(vi ) 6∈ H. Hence X µ(i)ta+h(vi ) gi∗ )(C) = 0}. (U ∗ )a ∼ = {µ ∈ K s : µ(i) = 0 for i ∈ [s] \ Fa and ( i∈Fa

In order to complete the proof of (4) we only need to prove the following statement: (6)

(

s X

µ(i)ta+h(vi ) gi∗ )(C) = 0 if and only if hµ, λi = 0 for all λ ∈ W.

i=1

P Let z ∈ Cb for some b ∈ H. Then z = i∈[s] λ(i)tb−h(vi ) gi with λ(i) ∈ K for i = 1, . . . , s and λ(i) = 0 if b−h(vi ) ∈ / H since z ∈ Fa . Moreover, since z ∈ Ker(F → U) b−h(v1 ) it follows that λ(1)t dfv1 + · · · + λ(s)tb−h(vs ) dfvs = 0. This implies that X X X X λ(i)vi (j)tb−hj )dxj = 0. λ(i)tb−h(vi ) vi (j)th(vi )−hj dxj = ( i∈[s] b−h(vi )∈H

j∈[n]

j∈[n]

i∈[s] b−h(vi )∈H

Note that if b − hj ∈ / H, then for all i ∈ [s] with b − h(vi ) ∈ H, one has h(vi ) − h ∈ / H and so v (j) = 0. Here we use the definition of P h(vi ), see (1). Therefore, i Pj i∈[s],b−h(vi )∈H λ(i)vi (j) = 0 for j = 1, . . . , n. This implies i∈[s],b−h(vi )∈H λ(i)vi = 0. In conclusion we see that X X λ(i)tb−h(vi ) gi ∈ Cb if and only if λ(i)vi = 0. i∈[s],b−h(vi )∈H

i∈[s],b−h(vi )∈H

6

P This particularly implies that if z = i∈[s] λ(i)tb−h(vi ) gi ∈ Cb , then λ = (λ(1), . . . , λ(s)) ∈ W. Since s s X X X a+h(vi ) ∗ b−h(vi ) ( µ(i)t gi )( λ(i)t gi ) = ( µ(i)λ(i))ta+b , i=1

i=1

i∈[s],b−h(vi )∈H

Ps

/ H or hµ, λi = it that ( i=1 µ(i)ta+h(vi ) gi∗)(Cb ) = 0 if and only if either a+b ∈ Pfollows s all i with b − h(vi ) ∈ / H. In i=1 µ(i)λ(i) = 0 for all λ ∈ W satisfying λ(i) = 0 forP particular, we have if hµ, λi = 0 for all λ ∈ W , then ( si=1 µ(i)ta+h(vi ) gi∗ )(C) = 0. P a+h(vi ) ∗ gi )(C) = 0. Write a = a+ −a− For the converse, we assume that ( si=1 µ(i)t Ps with a+ ∈ H and a− ∈ H, and set b0 = Pi=1 h(vi ) + a− . Since a + b0 ∈ H and b0 − h(vi ) ∈ H for all i ∈ [s], and since ( si=1 µ(i)ta+h(vi ) gi∗ )(Cb0 ) = 0, it follows that (µ, λ) = 0 for all λ ∈ W . Therefore the statement (6) has been proved and this completes the proof. Now for any a ∈ ZH we want to determine the dimension of (Im δ ∗ )a . We observe that the ZH-graded R-module Im δ ∗ is generated by the elements s s X X ∗ ∗ ∗ δ ((dxi ) ) = (∂fvj /∂xi )gj = vj (i)th(vj )−hi gj∗ . j=1

∗

j=1

∗

Note that deg δ ((dxi ) ) = −hi for i = 1, . . . , n. For i = 1, . . . , n we set wi = (v1 (i), . . . , vs (i)), and for a ∈ ZH we let KDa be the K-subspace of K s spanned by the vectors wi for which i ∈ Ga . Here the set Ga is defined to be Ga = {i ∈ [n] : a + hi ∈ H}. Proposition 1.2. Let a ∈ ZH. Then dimK (Im δ ∗ )a = dimK KDa . Proof. The K-subspace (Im δ ∗ )a ⊂ (F ∗ )a is spanned by the vectors s X a+hi ∗ ∗ t δ ((dxi ) ) = vj (i)ta+h(vj ) gj∗ j=1

with i ∈ Ga . The desired formula dimK (Im δ ∗ )a follows once we have shown that X X µ(i)wi = 0. µ(i)ta+hi δ ∗ ((dxi )∗ ) = 0 if and only if i∈Ga

i∈Ga

Here µ(i) ∈ K for any i ∈ Ga . To prove it we notice that s s X X X X X a+h(vj ) ∗ a+hi ∗ ∗ µ(i)vj (i)ta+h(vj ) gj∗). µ(i)( vj (i)t gj ) = ( µ(i)t δ ((dxi ) ) = i∈Ga

P

i∈Ga

j=1

j=1 i∈Ga

P Thus i∈Ga µ(i)ta+hi δ ∗ ((dxi )∗ ) = 0 if and only if P i∈Ga µ(i)vj (i) = 0 for j = 1, . . . , s. Since vj (i) = wi (j), this is the case if and only if i∈Ga µ(i)wi = 0. 7

Corollary 1.3. Let a ∈ ZH. Then dimK KDa + dimK KLa ≤ dimK KL. Equality holds if and only if T 1 (R)a = 0. Summarizing our discussions of this section we observe that all information which is needed to compute dimK T 1 (R)a can be obtained from the (s × n)-matrix   v1 (1) v1 (2) . . . v1 (n)  v2 (1) v2 (2) . . . v2 (n)  . AH =  ..  ..  ... .  . vs (1) vs (2) . . . vs (n)

Indeed, dimK T 1 (K[H])a can be computed as follows: let l = rank AH , ra the rank of the submatrix of AH whose rows are the ith rows of AH for which a + h(vi ) 6∈ H, and let ca be the rank of the submatrix of AH whose columns are the jth columns of AH for which a + hj ∈ H. Then dimK T 1 (K[H])a = l − la − da .

(7)

Corollary 1.4. Suppose a ∈ H. Then T 1 (R)a = 0. Proof. Since a ∈ H, it follows that G(a) = [n] and dimK Da = dimK KL = rank AH . Thus the assertion follows from Corollary 1.3. The inequality of Corollary 1.3 can also be deduced from the following lemma. Lemma 1.5. Fix a ∈ ZH. Then vi (j) = 0 for every pair i, j with i ∈ / Fa and j ∈ Ga . Proof. Assume on the contrary that vi (j) 6= 0, say vi (j) < 0, for some i ∈ / Fa and j ∈ Ga . Then X X −vi (k)hk + (−vi (j) − 1)hj ∈ H. vi (k)hk = hj + b, where b = h(vi ) = − k6=j vi (k) 4 because C has a chord. Since every induced cycle of G(P) is a 4-cycle and since C has only one chord, C must be a 6-cycle. Assume that the vertices of C are hi , vk1 , hℓ1 , vj , hℓ2 , vk2 , listed counterclockwise, and the chord of C is e = (hi , vj ) as above. With the notation introduced, it follows that (i, j), (i, k2 ), (ℓ2 , k2 ), (ℓ2 , j), (ℓ1, j), (ℓ1 , k1 , ), (i, k1) are vertices of P. We consider the following cases. Suppose first that (ℓ1 − i)(ℓ2 − i) > 0. Without loss of generality, we may assume ℓ2 > ℓ1 > i. Then, since P is convex and (i, k2 ) and (ℓ2 , k2 ) are both vertices of P, we have (ℓ1 , k2 ) is a vertex of P. It follows that {hℓ1 , vk2 } is an edge of G(P) which 19

is a chord of C, contradicting our assumption that C has a unique chord. Similarly the case that (k1 − j)(k2 − j) > 0 is also not possible. It remains to consider the case when (ℓ1 − i)(ℓ2 − i) < 0 and (k1 − j)(k2 − j) < 0. Without loss of generality we may assume that ℓ1 < i < ℓ2 and k1 < j < k2 . Then either (i−1, j +1) or (i+1, j −1) is a vertex of P by the connectedness and convexity of P. We may assume that (i − 1, j + 1) ∈ V (P). Note that (i − 1, k1 ) and (ℓ2 , j + 1) belong to V (P). Thus we obtain the path vk1 , hi−1 , vj+1 , hℓ2 in G(P) which is a crossing path chord of C with respect to e. Semi-rigidity. We say that R is semi-rigid if T 1 (R)a = 0 for all a ∈ ZH with −a ∈ H. In this subsection we consider this weak form of rigidity which however is stronger than inseparability. We again let G be a finite bipartite graph on the vertex set [m] with edge set E(G) = {e1 , e2 , . . . , enP }. The edge ring of G is the toric ring K[H] whose generators are the elements hi = j∈V (ei ) δj , i = 1, . . . , n. Here δ1 , . . . , δm is the canonical basis of Zm . As above we may assume that each edge of G belongs to a cycle and that C1 , C2 , . . . Cs is the set of cycles of G and where C1 , . . . , Cs1 is the set of induced cycles of G. Let Ci be one of these cycles with edges ei1 , ei2 , . . . , ei2t labeled counterclockwise. Two distinct edges e and e′ of Ci are said to be of the same parity in Ci if e = eij and e′ = eik with j − k an even number. Lemma 3.7. Let a = −hj − hk , and let i ∈ [s1 ]. Then i ∈ Fa′ , if and only if ej and ek have the same parity in Ci . Moreover, if Fa′ 6= ∅, then KLa = KL−hj + KL−hk . Proof. Since i ∈ [s1 ], the cycle Ci is an induced cycle. , . . . , ei2t be the P Let ei1 , ei2P edges of Ci labeled counterclockwise. Then h(vi ) = tk=1 hi2k−1 = tk=1 hi2k . Thus if ej and ek have the same parity in Ci , it follows that hj and hk belong to either one of the above summands, so that a + h(vi ) ∈ H. This shows that i ∈ Fa′ . Conversely, suppose that i ∈ Fa′ . Let hj = δj1 + δj2 and hk = δk1 + δk2 . For simplicity, we may assume that δ1 , . . . , δ2t correspond to the vertices of Ci and that the edges of Ci correspond to the elements δ2t + δ1 and δi + δi+1 for i = 1, . . . , 2t − 1. Then h(vi ) = δ1 + · · · + δ2t and a + h(vi ) = −δj1 − δj2 − δk1 − δk2 + δ1 + . . . + δ2t ∈ H. P In general, let h ∈ H, h = m i=1 zi δi with zi ∈ Z. Then it follows that zi ≥ 0 for all i. Hence it follows from (14) that ej and ek are edges of Ci with V (ej ) ∩ V (ek ) = ∅ (that is, the vertices j1 , j2 , k1 , k2 are pairwise different), and that a + h(vi ) is the sum of all δi , i = 1, . . . , 2t with i 6= j1 , j2 , k1, k2 . Suppose the edges ej and ek do not have the same parity in Ci . Then a + h(vi ) is the sum of S1 and S2 , where each of S1 and S2 consists of an odd sum of δi . Hence none of these summands belongs to H. Since S1 + S2 ∈ H, there exists a summand δr1 in S1 and a summand δr2 in S2 such that δr1 + δr2 ∈ H. This implies that {r1 , r2 } ∈ E(Ci ) because Ci has no chord. However this is not possible. Indeed, if {r1 , r2 } ∈ E(Ci ), then r2 ≡ r1 + 1 mod 2t. But this is not the case. (14)

20

′ ′ , Next we show that KLa = KL−hj +KL−hk if Fa′ 6= ∅. Note that Fa′ ⊆ F−h ∩F−h j k we have KL−hj + KL−hk ⊆ KLa by Lemma 3.1. In order to obtain the desired ′ equality, we only need to show that vi ∈ KL−hj + KL−hk for each i ∈ (F−h ∩ j ′ ′ F−hk ) \ Fa . ′ ′ ) \ Fa′ . Since Fa′ 6= ∅, there exists an induced cycle, say C, Let i ∈ (F−h ∩ F−h j k such that ej and ek have the same parity in C. We may assume that V (C) = [2t] and E(C) = {{1, 2}, {2, 3}, . . . , {2t − 1, 2t}, {2t, 1}}, and that ej = {1, 2} and ek = {2k − 1, 2k} with 1 < k ≤ t. Since ej , ek do not have the same parity in Ci , we can assume without loss of generality that E(Ci ) is

{{1, 2}, {2, i1}, {i1 , i2 }, . . . , {i2h , i2h+1 }, {i2h+1 , 2k}, {2k, 2k − 1}} ∪{{2k − 1, i2h+2 }, . . . , {i2ℓ , i2ℓ+1 }, {i2ℓ+1 , 1}}. Then we have even closed walks W1 : 2, 3, . . . , (2k − 1), 2k, i2h+1 , i2h . . . , i1 , 2 and W2 : 1, 2, 3, . . . , (2k − 1), i2h+2 , . . . , i2ℓ+1 , 1. Let w1 = v(W1 ) and w2 = v(W2 ). Since the vertex 1 belongs to ej but is not a vertex of W1 , Lemma 3.2 implies that w1 ∈ KL−hj . Similarly it follows that w2 ∈ KL−hk . Since vi differs at most by a sign from either w1 − w2 or w1 + w2 , it follows that vi ∈ KL−hj + KL−hk , as required. ′ ′ . Then KL−hj 6= KL−hk . 6= F−h Lemma 3.8. Suppose that F−h j k ′ ′ Proof. Let i ∈ F−h \ F−h . Then vi ∈ KL−hk and vi (j) 6= 0, since ej is an edge of j k Ci . However the vectors v which belong to KL−hj have the property that v(j) = 0. Hence vi ∈ KL−hk \ KL−hj , and this implies KL−hj 6= KL−hk .

Corollary 3.9. Assume that K[G] is inseparable. Let a = −hj − hk . Then ′ ′ ′ dimK KLa = dimK KL − 1 if F−a 6= ∅ and F−h = F−h . j k

Otherwise, dimK KLa = dimK KL. Proof. Since we assume that G is inseparable, it follows from Corollary 1.1 and Proposition 1.2 that dimK KL − dimK KL−hj = dimK (Im δ ∗ )−hj . Since by assumption each edge of G belongs to a cycle, it follows that dimK (Im δ ∗ )−hj = 1. Thus dimK KL−hj = dimK KL − 1. Similarly, dimK KL−hk = dimK KL − 1. If F ′ −hj = F ′ −hk , then KL−hj = KL−hk , and if moreover, F ′−a 6= ∅, then together with Lemma 3.7 we have dimK KLa = dimK KL − 1, as desired. Otherwise, there are two cases to consider. If F ′ −a = ∅, then KLa = KL, by the definition of KLa and by Lemma 3.1. If F ′ −a 6= ∅ and F ′ −hj 6= F ′ −hk , then KLa = KL−hj + KL−hk = KL, using Lemma 3.7 together with Lemma 3.8. Theorem 3.10. Let G be a bipartite graph such that R = K[G] is inseparable. Then the following statements are equivalent: (a) K[G] is not semi-rigid; 21

(b) there exist edges e, f and an induced cycle C such that e, f have the same parity in C and for any other induced cycle C ′ , e ∈ E(C ′ ) if and only if f ∈ E(C ′ ). Proof. (b) ⇒ (a): Let a = −g − h, where g and h are vectors in H corresponding to the edges e and f respectively. Then dimK KLa = dimK KL − 1 by Corollary 3.9. Note that Ga = ∅, we have (Im δ ∗ )a = 0. Therefore T 1 (R)a 6= 0 by Corollary 1.3, and in particular, R is not semirigid. P (a) ⇒ (b): By assumption, there exists a = i∈[n] −ai hi ∈ ZH with ai ≥ 0 for i = 1, · · · , n such that T 1 (R)a 6= 0. Note that ai ∈ {0, 1}, for otherwise, Fa′ = ∅ and so KLa = KL. In particular T 1 (R)a = 0, a contradiction. Since R is inseparable, it follows that |{i : ai 6= 0}| ≥ 2. If |{i : ai 6= 0}| = 2, then a = −hk − hj for ′ ′ = F−h by Corollary 1.1 and some 1 ≤ i 6= j ≤ n. Therefore, Fa′ 6= ∅ and F−h j k Corollary 3.9. Let e and f be the edges corresponding to the vectors hj and hk , respectively. Then, since Fa′ 6= ∅, there exists an induced cycle C of G such that e and f have ′ ′ implies that for any the same parity in C, by Lemma 3.7. Moreover, F−h = F−h j k ′ ′ ′ induced cycle C of G, e ∈ E(C ) if and only if f ∈ E(C ). Now suppose that |{i : ai 6= 0}| ≥ 3. Then there exists j and k with aj 6= 0 and ak 6= 0, and we set b = −hj −hk . Note that Fa′ ⊆ Fb′ . This implies that KLb ⊆ KLa . Therefore, since (Im δ ∗ )a = (Im δ ∗ )b = 0, we have T 1 (R)b 6= 0, and we are in the previous case. Corollary 3.11. Let P be a convex polyomino. Then K[P] is semi-rigid if and only if P contains more than one cell. Proof. Assume that P contains a unique cell. Then G(P) is a square and it is not semi-rigid by Theorem 3.10. Conversely, assume that K[P] is not semi-rigid. Then there exist two edges e, f and an induced cycle C of G(P) satisfying the condition (b) in Theorem 3.10. Let (i, j) and (k, ℓ) be vertices of P corresponding to the edge e and f , respectively. Then the two edges of C other than e and f correspond to the vertices (i, ℓ) and (k, j) of P. It follows that k 6= i and ℓ 6= j. Without loss of generality, we may assume that k > i and ℓ > j. Then (i + 1, j + 1) ∈ V (P). Let C ′ be the induced cycle of G(P) corresponding to the cell [(i, j), (i + 1, j + 1)] of P. Since C ′ contains the edge e, C ′ must contain f by the condition (b) and thus k = i + 1 and ℓ = j + 1. We claim that [(i, j), (i + 1, j + 1)] is the only cell of P. Suppose that this is not the case. Then we let Ct , t = 1, 2, 3, 4 be four cells which share a common edge with the cell [(i, j), (i + 1, j + 1)]. Note that P contains at least one of the Ct . Indeed, since P is connected and since by assumption P contains a cell C different from [(i, j), (i + 1, j + 1)], there exists a path in P between the cell [(i, j), (i + 1, j + 1)] and C. This path must contain one of the Ct . However V (Ct ) contains exactly one of the two vertices (i, j) and (i + 1, j + 1) for t = 1, . . . , 4. In other words, there exists an induced cycle of G(P) which contains exactly one of the edges e and f . This is contradicted to the condition (b) and thus our claim has been proved. 22

Classes of bipartite graphs which are semi-rigid or rigid. Let Gn be the bipartite graph on vertex set [2n] with edge set E(Gn ) = {{i, j} : i, j ∈ [2n], i − j is odd and {i, j} = 6 {1, 2n}}. Thus Gn is obtained from the complete bipartite graph Kn,n by deleting one of its edges. We observe that for n ≥ 3, Gn = G(Pn ) where Pn is the polyomino with V (Pn ) = {(i, j) : 1 ≤ i, j ≤ n, (i, j) 6= (n, n)}. For any edge e = {i, j} ∈ E(G) we use h(e) to denote the vector δi + δj ∈ Z2n . Let C be a cycle. Then, as before, v(C) stands for the vector corresponding to C and V (C) denotes the set of vertices of C. Our main result of this subsection is the following: Proposition 3.12. Let R be the edge ring of Gn . (a) If n = 3, then R is semirigid, but not rigid. (b) If n ≥ 4, then R is rigid. We need some preparations. First, we introduce some notation. P Let a = i∈[2n] ai δi ∈ Z2n with ai ∈ Z for i = 1, . . . , 2n. We set X X ai and ao = ai . ae = i is even i is odd We also set X ℓ(a) = a1 + a2n and r(a) = ai . i∈{1,2n} /

Lemma 3.13. Let a =

P

i∈[2n]

ai δi and H = H(Gn ). Then

(a) a ∈ ZH if and only if ae = ao . (b) The following conditions are equivalent: (i) a ∈ H; (ii) ae = ao , ℓ(a) ≤ r(a) and ai ≥ 0 for all i = 1, . . . , 2n. Proof. (a) First note that if i is even and j is odd, then δi + δj ∈ ZH. Indeed, if {i, j} = 6 {1, 2n}, then δi + δj ∈ H; if {i, j} = {1, 2n}, then δ1 + δ2n = (δ1 + δ2 ) + (δ3 + δ2n ) − (δ2 + δ3 ) ∈ ZH. Suppose that ae = ao . We prove that a ∈ ZH by induction on |ae |. If ae = 0 then a = 0, and the assertion is trivial. Suppose that |ae | > 0. We only consider the case when ae > 0 since the other case is similar. In this case there exist i, j ∈ [2n] such that i is even, j is odd, ai > 0 and aj > 0. Let b = a − (δi + δj ). Then be = ae − 1 = ao − 1 = bo , and so b ∈ ZH by induction. This implies that a = b + (δi + δj ) ∈ ZH. Conversely, it is obvious that ae = ao if a ∈ ZH. (b) (i) ⇒ (ii): Note that ℓ(h(e)) / E(G). P ≤ r(h(e)) for any e ∈ E(G) since {1, 2n} ∈ Now given a ∈ H. Then a = e∈E(G) ce h(e), where ce is a non-negative integer for P P each e ∈ E(G). It follows that ℓ(a) = e∈E(G) ce ℓ(h(e)) ≤ e∈E(G) ce r(h(e)) = r(a), as required. 23

(ii) ⇒ (i): We use induction on ℓ(a). If ℓ(a) = 0, we see that a ∈ H by induction on ae as in the proof of (a). Assume that ℓ(a) > 0. Without restriction we may further assume that a1 ≥ a2n . Then X ae − a2n = ai > 0, i∈even, i6=2n

for otherwise ae = a2n ≤ a1 < ao , a contradiction. (For the inequality a1 < ao we used that 0 < ℓ(a) ≤ r(a) = a3 + a5 + · · · + a2n−1 .) Hence there exists an even number j 6= 2n with aj > 0. Set b = a − (δ1 + δj ). Then b ∈ H by induction and so a = b + (δ1 + δj ) ∈ H. Corollary 3.14. Let a ∈ ZH with ai ≥ 0 for all i ∈ [2n]. Then either a ∈ H or a = b + k(δ1 + δ2n ), where k ≥ 1 and b ∈ H with ℓ(b) = r(b). Proof. Suppose that a ∈ / H. Then ℓ(a) > r(a), by Lemma 3.13. Note that by Lemma 3.13 we have ae = ao , and hence ℓ(a) −r(a) = a1 + a2nP −(ae −a2n ) −(ao −a1 ) is an even number, say 2k. Set b = (a1 − k)δ1 + (a2n − k)δ2n + i∈{1,2n} ai δi . Suppose / that a1 < k. Then X X ai > a1 + ai ≥ ao , a2n = 2k − a1 + i∈{1,2n} /

i∈{1,2n} /

a contradiction, because ae = ao . Therefore a1 ≥ k. Similarly, a2n ≥ k. Since ℓ(b) = r(b), Lemma 3.13 implies that b ∈ H. Moreover, a = b + k(δ1 + δ2n ) by the choices of k and b. Lemma 3.15. Let a ∈ ZH such that E(Gn ) \ {e : a + h(e) ∈ H} contains no cycle. Then dimK KL = dimK Da . In particular, T 1 (R)a = 0. Proof. Let Γ′ be the graph with E(Γ′ ) = E(Gn ) \ {e : a + h(e) ∈ H}, and let Γ be the graph obtained from Γ′ by adding all vertices of Gn which do not belong to V (Γ′ ). Then Γ is a graph with no cycle and V (Γ) = [2n]. If Γ is a tree, then Γ is a spanning tree of Gn . If Γ is not a tree, we choose a connected component Γ0 of Γ and let Γ1 be the induced graph of Γ on the set [2n] \ V (Γ0 ). Since Gn is connected there exists an edge e of Gn with one end in V (Γ0 ) and the other end in V (Γ1 ). We let Γ′′ be the graph obtained from Γ by adding the edge e. Since neither Γ0 nor Γ1 contains a cycle, it follows that Γ′′ has also no cycle, but one more edge than Γ. Proceeding in this way we obtain after finitely many steps a spanning tree T of Gn which contains E(Gn ) \ {e : a + h(e) ∈ H}. In particular, there exists a subset of {e : a + h(e) ∈ H}, say {e1 , . . . , ek } such that E(T ) = E(Gn ) \ {e1 , e2 , . . . , ek }. Since all spanning trees have the same number of edges, namely, 2n − 1, it follows that k = n2 − 2n. For each i = 1, . . . , k, T + ei contains a unique induced cycle, say Ci . Then for all i = 1, . . . , k we have vi (i) ∈ {±1} and vi (j) = 0 if j 6= i and 1 ≤ j ≤ k. Here vi = v(Ci ), the vector corresponding to the cycle Ci , for i = 1, . . . , k. It follows that dimK Da ≥ k since (v1 (i), . . . , vk (i), . . . , vs1 (i)) ∈ Da for i = 1, . . . , k. Here s1 is the number of induced cycles of Gn . On the other hand, dimK Da ≤ dimK KL and 24

dimk KL = |E(Gn )| − |V (Gn )| + 1 which is equal to k = n2 − 2n. It follows that dimK KL = dimK Da , completing the proof by Proposition 1.2. Proof of Proposition 3.12. (a) Since G3 is associated with a polyomino containing 8 vertices, we have R(= K[G3 ]) is semi-rigid by Corollary 3.11. Let a = δ6 −δ1 −δ2 −δ4 . Then KLa is spanned by the vectors corresponding to the cycles C1 : 3, 2, 5, 4, C2 : 3, 4, 5, 6 and C3 : 3, 2, 5, 6. This implies that dimK KLa = 2. Since dimK Da = 0 and dimK KL = 3, we have T 1 (R)a 6= 0. In particular, R is not rigid, as required. P (b) Let a = i∈[2n] ai δi ∈ ZH. We want to prove that T 1 (R)a = 0. There are following cases to consider. Case 1 : ai ≥ 0 for all i. By Corollary 3.14, either a ∈ H or a = b+k(δ1 +δ2n ), where k ≥ 1 and b ∈ H with ℓ(b) = r(b). If a ∈ H, then T 1 (R)a = 0, see Corollary 1.4. If a = b + k(δ1 + δ2n ) with k = 1, then for any edge e = {i, j} with e ∩ {1, 2n} = ∅, we have a + δi + δj ∈ H by Lemma 3.13. It follows that E(G) \ {e : a + h(e) ∈ H} contains no cycle, and so T 1 (R)a = 0 by Lemma 3.15. If k = 2, then for any induced cycle C, a + h(v(C)) ∈ H if and only if V (C) ∩ {1, 2n} = ∅. This follows from Lemma 3.13 and the fact that any induced cycle of Gn is a 4-cycle. To prove KL = KLa , we have to show if V (C) ∩ {1, 2n} = ∅, then v(C) ∈ KLa . Given an induced cycle C : i1 , i2 , i3 , i4 with V (C) ∩ {1, 2n} = ∅, where i1 , i3 are even and i2 , i4 are odd. Then we obtain two cycles C1 : i1 , i2 , i3 , 1 and C2 : i3 , i4 , i1 , 1. Note that v(C1 ), v(C2 ) ∈ KLa and v(C) is a linear combination of v(C1 ), v(C2), we have KL = KLa and so T 1 (R)a = 0. If k ≥ 3, then for any induced cycle C, one has a + h(v(C)) ∈ / H by Lemma 3.13 and so KLa = KL. In particular, T 1 (R)a = 0. Case 2: There exists a unique i ∈ [2n] with ai < 0. If ai ≤ −2, then Fa′ = ∅ and so KLa = KL. In particular T 1 (R)a = 0. Hence we assume ai = −1. By symmetry, we only need to consider the cases when i = 1 and when i = 3. We first assume that i = 1. Since ae = ao , there exists an odd integer j 6= 1 such that aj > 0, and so a = b + δj − δ1 , where be = bo and bℓ ≥ 0 for each ℓ ∈ [2n]. By Corollary 3.14, either b ∈ H or b = c + k(δ1 + δ2n ) with c ∈ H and k > 0. The second case cannot happen because a1 = −1. Hence for any e ∈ E(G), a + h(e) ∈ H if and only if 1 ∈ e. In other words, a + h(e) ∈ H if and only if e ∈ {{1, 2}, {1, 4}, . . . , {1, 2n − 2}}. Denote {1, 2i} by ei for i = 1, . . . , n − 1. Let Ci be the cycle 1, 2i, 3, 2n − 2 and let vi = v(Ci ) for i = 1, . . . , n − 2. Then for i = 1, . . . , n − 2, we have vi (i) ∈ {±1} and vi (j) = 0 for j 6= i and j = 1, . . . , n − 2. This implies that dimK Da ≥ n − 2. To compute dimK KLa , we notice that if C is an induced cycle with 1 ∈ / V (C), then a + h(v(C)) ∈ / H and thus KLa contains the cycle space of the complete bipartite graph with bipartition {3, 5, . . . , 2n − 1} and {2, . . . , 2n}, which has the dimension (n − 1)n − n − (n − 1) + 1 = n2 − 3n + 2, see (10). Thus T 1 (R)a = 0 because dimK Da = n2 − 2n. Next we assume that i = 3. Then a = b + δj − δ3 , where bℓ ≥ 0 for all ℓ ∈ [2n] and be = bo , j 6= 3 and j is odd. Moreover by Corollary 3.14, we have either b ∈ H or b = c + k(δ1 + δ2n ) for some k ≥ 1 and with c ∈ H and ℓ(c) = r(c). Suppose first that b ∈ / H and k ≥ 2. Then for any cycle C, a + h(v(C)) ∈ H 25

implies V (C) ∩ {1, 2n} = ∅. Thus, similarly as in Case 1 we see that KLa = KL and T 1 (R)a = 0. Suppose next that j 6= 1 and that b ∈ H or b ∈ / H and k = 1. Then a + h(e) ∈ H for any e ∈ {{3, 2}, . . . , {3, 2n − 2}}. Denote {3, 2t} by et for t = 1, . . . , n−1. For t = 1, . . . , n−1, let Ct be the cycle 3, 2t, 5, 2n and let vt = v(Ct ), the vector corresponding to Ct . Then vt (t) ∈ {±1} for t = 1, . . . , n − 1 and vt (k) = 0 for k 6= t. This implies that dimK Da = n − 1. On the other hand, KLa contains the cycle space of the subgraph of Gn induced on {1, 5, . . . , 2n − 1} ∪ {2, 4, . . . , 2n}, which has the dimension n2 − 3n + 1. Thus T 1 (R)a = 0. Finally suppose that j = 1 and that and also k = 1 if b ∈ / H. If b ∈ H, then we check that a + h(e) ∈ H for any e ∈ {{3, 2}, . . . , {3, 2n − 2}} and deduce that T 1 (R)a = 0, in the same process as in the last case. If b ∈ / H and k = 1, then for any induced cycle C, we have a + h(v(C)) ∈ H if and only if 3 ∈ V (C) and {1, 2n} ∩ V (C) = ∅. We claim that KLa = KL. Given an induced cycle C : 3, i1 , i2 , i3 with a + h(v(C)) ∈ H. Here i1 and i3 are even and i2 is odd. We let C1 : 3, i1 , i2 , 2n and C2 : i2 , i3 , 3, 2n. Then v(C1) and v(C2 ) belong to KLa and v(C) is a linear combination of v(C1 ) and v(C2 ). Thus KLa = KL, as claimed. In particular, T 1 (R)a = 0. Case 3: |{k : ak < 0}| = 2. Without restriction we may assume ai = aj = −1 for some i 6= j. Indeed, if some ak ≤ −2, then Fa′ = ∅ by Lemma 3.13 and so T 1 (R)a = 0. Assume first that both i and j are even. Then for any induced cycle C such that {i, j} * V (C), we have v(C) ∈ KLa . Let C : k, i, ℓ, j be a cycle with {i, j} ⊆ V (C). We choose an even number d ∈ [2n] \ {i, j, 2n}. Then we obtain two cycles C1 :k, i, ℓ, d and C2 :ℓ, j, k, d. Since v(C) is a linear combination of v(C1) and v(C2 ) and since v(Ct ) ∈ KLa for t = 1, 2, we have v(C) ∈ KLa and thus KLa = KL. In particular T 1 (R)a = 0. Next assume that i is even and j is odd and {i, j} = 6 {1, 2n}. Notice that we can write a as a = b + k(δ1 + δ2n ) − (δi + δj ), where b ∈ H and k ≥ 0. Moreover, if k > 0 then ℓ(b) = r(b). If k = 0, then dimK Da = 1, and KLa contains the cycle space of the graph which is obtained from Gn by deleting the edge {i, j}. Hence dimK KLa ≥ dimK KL − 1, and so T 1 (R)a = 0. If k = 1, then for any induced cycle C, we have a + h(v(C)) ∈ H if and only if {i, j} ⊆ V (C) and V (C) ∩ {1, 2n} = ∅. Let C : i, j, k, ℓ be an induced cycle such that a + h(v(C)) ∈ H. Then the vectors v1 , v2 which correspond to cycles j, k, ℓ, 2n and ℓ, i, j, 2n respectively belong to KLa and v(C) is a linear combination of v1 , v2 . It follows that KL = KLa and T 1 (R)a = 0. If k ≥ 2, we first note that {i, j} ∩ {1, 2n} = ∅. Indeed, if {i, j} ∩ {1, 2n} = 6 ∅, ′ then either ai ≥ 1 or aj ≥ 1, a contradiction. Thus Fa = ∅ by Lemma 3.13 and it follows that KL = KLa . In particular T 1 (R)a = 0. Finally assume that {i, j} = {1, 2n}. Then a = b − δ1 − δ2n with b ∈ H and so Fa′ = ∅. It follows that KLa = KL and T 1 (R)a = 0. Case 4: |{k : ak < 0}| = 3. We may assume that ai = aj = ak = −1. We only need to consider the case when Fa′ 6= ∅. So we may assume i, k are even and j is odd, and {1, 2n} * {i, j, k}. Let C : i, j, k, ℓ be an induced cycle such that 26

a+h(v(C)) ∈ H. We choose an even number d ∈ [2n]\{i, k, 2n} and let C1 : j, k, ℓ, d and C2 : ℓ, i, j, d be two cycles in Gn . Then v(C1 ) and v(C2 ) belong to KLa and v(C) is a linear combination of v(C1 ) and v(C2 ). This implies KLa = KL, and in particular, T 1 (R)a = 0. Case 5: |{k : ak < 0}| ≥ 4. If |{k : ak < 0}| = 4, we may assume that ai = aj = ak = aℓ = −1. Then for any induced cycle C, a + h(v(C)) ∈ H implies that V (C) = {i, j, k, ℓ}. We may assume that i and k are even numbers. Let t be an odd number in [2n] \ {j, ℓ}, and let C1 : i, j, k, t and C2 : k, l, i, t be 4-cycles of Gn . Since v(C) is a linear combination of v(C1 ) and v(C2 ), we have KLa = KL, and consequently, T 1 (R)a = 0. If |{k : ak < 0}| > 4, then Fa′ = ∅ and so T 1 (R)a = 0. Thus we have shown that T 1 (R)a = for all a, and this shows that R is rigid, as desired. References [1] K. Altmann, Computation of the vector space T 1 for affine toric varieties, J. Pure Appl. Algebra 95 (1994) 239–259 [2] K. Altmann, Minkowski sums and homogeneous deformations of toric varieties, T´ohoku Math. J. 47 (1995) 151–184 [3] K. Altmann, M. Bigdeli, J. Herzog, D. Lu, Algebraically rigid simplicial complexes and graphs, arXiv:1503.08080 [4] K. Altmann, J. A. Christophersen, Deforming Stanley-Reisner schemes, Math. Ann. 348 (2010), 513–537. [5] K. Altmann, J. A. Christophersen, Cotangent cohomology of Stanley-Reisner rings. Manuscripta Math. 115 (2004), 361–378. [6] W. Bruns, J. Herzog, Cohen-Macaulay rings, Cambridges Studies in Advanced Mathematics, 39, 1993. [7] V. Ene, J. Herzog, Gr¨ obner bases in Commutative Algebra, Graduate Studies in Mathematics 130, AMS 2012. [8] J. Herzog, Generators and relations of abelian semigroups and semigroup rings, Manuscripta Math. 3 (1970), 175–193. [9] A. A. Qureshi, Ideals generated by 2-minors, collections of cells and stack polyominoes, J. Algebra, 357 (2012), 279–303. [10] H. Ohsugi, T. Hibi, Koszul bipartite graphs, Advances in Applied Mathematics, 22 (1999) 25–28. [11] H. Ohsugi, T. Hibi, Toric ideals generated by quadratic binomials, J. Algebra. 218 (1999), 509–527. [12] J. Stevens, Deformations of singularities, Lecture Notes in Mathematics (1811), Berlin: Springer, 2003. [13] R. Villarreal, Monomial Algebras, Pure and Applied Mathematics, Marcel Dekker, 2001. Mina Bigdeli, Faculty of Mathematics, Institute for Advanced Studies in Basic Sciences (IASBS), 45195-1159 Zanjan, Iran E-mail address: [email protected] ¨rgen Herzog, Fakulta ¨t fu ¨r Mathematik, Universita ¨t Duisburg-Essen, 45117 Ju Essen, Germany E-mail address: [email protected] 27

Dancheng Lu, Department of Mathematics, Soochow University, 215006 Suzhou, P.R.China E-mail address: [email protected]

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