On isomorphism of simplicial complexes and their related rings

In this paper, we provide a simple proof for the fact that two simplicial complexes are isomorphic if and only if their associated Stanley-Reisner rings, or their associated facet rings are isomorphic as K-algebras. As a consequence, we show that two graphs are isomorphic if and only if their associated edge rings are isomorphic as K-algebras. Based on an explicit K-algebra isomorphism of two Stanley-Reisner rings, or facet rings or edge rings, we present a fast algorithm to find explicitly the isomorphism of the associated simplicial complexes, or graphs. Introduction Let X be a finite nonempty set. A simplicial complex ∆ on X is a set of subsets of X such that, for any x ∈ X, {x} ∈ ∆, and if E ∈ ∆ and F ⊆ E, then F ∈ ∆. A set in ∆ is called a face and a maximal face in ∆ is called a facet. Let X = {x1, . . . , xn}, and ∆ be a simplicial complex on X. Let R = K[x1, . . . , xn] be the polynomial ring in n indeterminates and with coefficients in a field K. Let I(∆) be the ideal in R generated by all square-free monomials xi1 . . . xis , provided that {xi1 , . . . , xis} 6∈ ∆. The quotient ring R/I(∆) is called the Stanley-Reisner ring of the simplicial complex ∆. It is easy to see that any quotient of a polynomial ring over an ideal generated by square-free monomials of degree greater than 1, is the Stanley-Reisner ring of a simplicial complex. A natural question arises: If two Stanley-Reisner rings are isomorphic, are their corresponding simplicial complexes isomorphic? In 1996, W. Bruns and J. Gubeladze proved that if the isomorphism of the rings is a Kalgebra isomorphism, then the corresponding simplicial complexes are isomorphic [1]. In this paper we provide a simple proof for this result. In 2002, S. Faridi in [2] defined the notion of facet ideal for a simplicial complex which is a generalization of the notion of edge ideal for a graph defined by R. Villarreal [6]. 1 Definition 1. Let ∆ be a simplicial complex on the set X = {x1, . . . , xn}. Let F (∆) be the ideal of R = K[x1, . . . , xn] generated by all square-free monomials xi1 . . . xis , provided that {xi1 , . . . , xis} is a facet in ∆. The quotient ring R/F (∆) is called the facet ring of the simplicial complex ∆. Definition 2. Let G be a finite, simple and undirected graph with vertex set {x1, . . . , xn}. The ideal E(G) of R = K[x1, . . . , xn] generated by all square-free monomials xixj , provided that xi is adjacent to xj in G, is called the edge ideal of G. The quotient ring R/E(G) is called the edge ring of the graph G. A question similar to the case of simplicial complexes can be stated for facet rings and edge rings. In 1997, H. Hajiabolhassan and M. L. Mehrabadi in [3] proved that two graphs are isomorphic if and only if their corresponding edge rings are isomorphic as K-algebras. In this paper we will prove the statement for the facet rings and conclude it for the edge rings. Finally, we will present a fast algorithm which admits a K-algebra isomorphism of two Stanley-Reisner rings, or two facet rings, or two edge rings as input and returns explicitly the isomorphism of corresponding simplicial complexes or graphs, as output. The isomorphism First we prove some lemmas. Lemma 1. Let I and J be ideals in R = K[x1, . . . , xn] and S = K[y1, . . . , ym], respectively, both generated by monomials of degree greater than 1. Let φ : R/I → S/J be a K-algebra isomorphism. Let f be a monomial in R and denote the image of f in R/I by f̄ . If f̄ is a zero-divisor in R/I, then the constant term of φ(f̄) in S/J is zero. Proof. Let ḡ be a nonzero monomial in R/I and f̄ ḡ = 0. Let φ(f̄) = h̄ = h+J , φ(ḡ) = l̄ = l + J . Let h̄ = h̄0 + h̄1 + · · · + h̄r and l̄ = l̄i + l̄i+1 + · · · + l̄t be homogeneous decompositions of h̄ and l̄ with l̄i 6= 0. If h̄0 6= 0, then h̄l̄ = h̄0l̄i + monomials of higher degree. But f̄ ḡ = 0 and then, 0 = φ(f̄ ḡ) = h̄l̄. That is, h.l ∈ J , and hence li ∈ J , which is a contradiction. For any i, 1 ≤ i ≤ n, let L(xi) denote the set of yj’s such that ȳj appears in the linear part of φ(x̄i) with nonzero coefficient. Lemma 2. With the above notations, if φ : R/I → S/J is a K-algebra isomorphism, then m = n and L(xi) 6= ∅, for each i, 1 ≤ i ≤ n.