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The group of automorphisms of a zero-divisor graph based on rank one upper triangular matrices
Linear Algebra and its Applications 460 (2014) 242–258
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Linear Algebra and its Applications www.elsevier.com/locate/laa
The group of automorphisms of a zero-divisor graph based on rank one upper triangular matrices Dein Wong ∗,1 , Xiaobin Ma, Jinming Zhou 2 Department of Mathematics, China University of Mining and Technology, Xuzhou 221116, China
a r t i c l e
i n f o
Article history: Received 16 March 2014 Accepted 27 July 2014 Available online 14 August 2014 Submitted by R. Brualdi MSC: 05C25 05C60 05E15 11T99 Keywords: Zero-divisor graph Graph automorphisms Transitivity
a b s t r a c t Let Fq be a finite field with q elements, n(≥ 2) a positive integer, Matn (q) the set of all n × n matrices over Fq , R(n, q) the set of all rank one upper triangular matrices in Mat(n, q). The zero-divisor graph of Matn (q), written as Γ (Matn (q)), is a directed graph with vertex set all nonzero zero-divisors of Matn (q), and there is a directed edge from a vertex A to a vertex B, written as A → B, if and only if AB = 0. In this paper, we determine the automorphisms of an induced subgraph, written as Γ (R(n, q)), of Γ (Matn (q)) with vertex set R(n, q). The main theorem of this article proves that a bijective map σ on R(n, q) with n ≥ 3 is an automorphism of Γ (R(n, q)) if and only if σ(X) = aX P −1 π(xij ) P,
∀X = [xij ] ∈ R(n, q),
where aX ∈ Fq∗ depends on X; P is an invertible upper triangular matrix; π is an automorphism of the field Fq , [π(xij )] denotes the matrix whose (i, j)-entry is π(xij ). © 2014 Elsevier Inc. All rights reserved.
* Corresponding author. 1 2
E-mail address: [email protected] (D. Wong). Supported by the National Natural Science Foundation of China (No. 11171343). Supported by NSF grant BK20140177 of Jiangsu Province.
http://dx.doi.org/10.1016/j.laa.2014.07.041 0024-3795/© 2014 Elsevier Inc. All rights reserved.
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1. Introduction Let R be a commutative ring with identity 1. The zero-divisor graph of R, denoted by Γ (R), is an undirected graph with vertices Z ∗ (R), the set of nonzero zero-divisors of R, and for distinct elements x, y ∈ Z ∗ (R), there is an edge joining x and y if and only if xy = 0. The concept of zero-divisor graph was first defined and studied for commutative rings by Beck in [7], and further studied by many authors (see, e.g., Akbari et al. [1]; Anderson et al. [2–5]; Axtell et al. [6]; Levy et al. [12]; Lucas [17]; Wu [24]). A lot of results about the diameter, the girth of Γ (R) and so on have been obtained. The main idea of these researches is to study the interplay between the ring-theoretic properties of a commutative ring R and the graph-theoretic properties of Γ (R). Redmond [20] extended the concept of zero-divisor graphs of commutative rings to zero-divisor graphs of noncommutative rings. Recently, zero-divisor graphs of matrix rings attracted a lot of attention. Akbari and Mohammadian [1] studied the problem of determining when the zero-divisor graphs of rings are isomorphic, given that the zero-divisor graphs of their matrix rings are isomorphic. I. Bo˘zić and Z. Petrović [8] further investigated the properties of (directed) zero-divisor graphs of matrix rings, studied the relation between the diameter of the zero-divisor graph of a commutative ring R and that of the matrix ring Matn (R). Li [13] proved that when studying Γ (R), one can replace R with its total quotient ring, and they determined the girth of Γ (T ) of T , triangular matrices over a commutative ring. The diameter, the girth and the bounds for the number of edges of Γ (T ) were given in [14]. Generally, determining the full automorphisms of a graph is an important however a difficult problem both in graph theory and in algebraic theory. Searching the literature, we find that little is known for the automorphisms of zero-divisor graphs of rings. In [3], Anderson and Livingston have shown that Aut(Γ (Zn )) is a direct product of symmetric groups for n ≥ 4 a nonprime integer. For the noncommutative case, it was shown in [11] that Aut(Γ (R)) is isomorphic to the symmetric group of degree p+1, when R = Mat2 (Zp ) (p is a prime). S. Park and J. Han [19] proved that Aut(Γ (R)) ∼ = Sq+1 for R = Mat2 (Fq ) with Fq an arbitrary finite field. It is slightly regrettable that the authors of [11] and [19] only concerned with the special case when n = 2. Till now nothing is known about the automorphisms of the zero-divisor graphs of n × n matrices with n ≥ 3. This observation motivates us to do some work on this topic. It seems much difficult to determine the full automorphisms of Γ (Matn (q)), so we focus our attention on the subgraph of Γ (Matn (q)) induced by all rank one upper triangular matrices. Let T (n, q) be the set of all n × n upper triangular matrices over Fq , R(n, q) the set of all rank one matrices in T (n, q). The subgraph of Γ (Matn (q)) induced by all rank one upper triangular matrices is denoted by Γ (R(n, q)). The main theorem of this article proves that any graph automorphism of Γ (R(n, q)) with n ≥ 3 can be decomposed into the product of an inner automorphism, a field automorphism and a local scalar multiplication. For proving the main theorem of this paper we apply a technique from some recent papers which focused on determining the automorphisms of some graphs based on row
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vectors over finite fields. Here we list some papers closely related to our topic. Let Fq2v be the 2v-dimensional row vector space over a finite field Fq , and let K be a non-singular 2v × 2v skew symmetric matrix over Fq . The symplectic graph Sp(2v, q) with respect to K is defined to be the graph with vertices of all 1-dimensional subspace [α] of Fq2v , two vertices [α] and [β] are adjacent if and only if αKβ T = 0. Tang and Wan [21] determined the automorphisms of Sp(2v, q) and proved that this graph is strongly regular. Li and Wang [15] further showed that the subconstituents of the symplectic graph are strictly Deza graphs except the trivial case when υ = 2. Meemark and Prinyasart [18] studied the symplectic graph Sp(2υ, Zpn ) over a commutative ring Zpn and showed that it is strongly regular when υ = 1 and it is arc transitive when p is an odd prime. Li et al. [16] continued the research on Sp(2υ, Zpn ), showing that Sp(2υ, Zpn ) is arc transitive for any prime p, and it is a strictly Deza graph if υ ≥ 2 and n ≥ 2. Gu [9] studied the subconstituents of the symplectic graphs Sp(2υ, Zpn ). Wan and Zhou [22,23] studied the (isotropic) unitary graphs and the (isotropic) orthogonal graphs of characteristic 2. Gu [10] defined the (isotropic) orthogonal graph O(n, q), with respect to a nonsingular symmetric matrix H over Fq , with the vertex set to be all isotropic lines of Fqn , two isotropic lines [α] and [β] are adjacent if and only if αHβ T = 0. Gu determined the automorphisms of O(n, q) and proved that O(n, q) is vertex transitive and edge transitive. 2. Notations and preliminaries Let Fq be a finite field with q elements and Fq∗ = Fq \ {0}, where q is a power of a prime p. For a matrix M , we denote by M T the transpose of M . For n ≥ 2, the symbol T (n, q) denotes the set of all n × n upper triangular matrices over Fq , R(n, q) the set of all rank one matrices in T (n, q). All invertible ones in T (n, q) form a group under multiplication of matrices, which we denote by Tn (q). By I we denote the n × n identity matrix. Denote by Eij (1 ≤ i, j ≤ n) the matrix unit, whose (i, j) position is 1 and all others are 0. Since all Eij , for 1 ≤ i ≤ j ≤ n, form a basis of T (n, q), each A ∈ T (n, q) can be uniquely written as A = 1≤i≤j≤n aij Eij with aij ∈ Fq . Before studying graph automorphisms we give an elementary property about rank one upper triangular matrices. Definition 2.1. Let A = 1≤i≤j≤n aij Eij ∈ R(n, q) be a rank one matrix, if ast = 0, asl = 0 for l = 1, 2, . . . , t − 1 and ak,t = 0 for k = s + 1, . . . , n, then we say that A is supported by (s, t). Lemma 2.2. Let A =
1≤i≤j≤n
aij Eij be a rank one upper triangular matrix over Fq .
(i) There exists a unique pair (s, t) by which A is supported. (ii) akl = 0 if 1 ≤ l ≤ t − 1 or s + 1 ≤ k ≤ n.
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Proof. Since A is of rank ⎛ a1one, ⎞ it can be written as the product of a row vector and a a2
column vector as A = ⎝ .. ⎠ (b1 b2 . . . bn ). Thus aij = ai bj for 1 ≤ i ≤ j ≤ n. Suppose . an
that as = 0, bt = 0, and as+1 = . . . = an = b1 = . . . = bt−1 = 0. Since A is upper triangular, s ≤ t. Then, it is obvious that A is supported by (s, t). Further, the second assertion holds. 2
For example, if n = 6 and if A = 00 M is a 6 × 6 matrix over Fq , where 0 ad bd cd M = 0 0 0 with abcd = 0, then A is supported by (3, 4). a b c
For A ∈ R(n, q), denote by [A] the subspace of T (n, q) spanned by A. As [A] = [aA] for a ∈ Fq∗ , the representation of [A] is not unique. However, if we require the representative A, supported by (s, t), to take 1 at the (s, t) position, then the representation is unique. For the aim of describing all automorphisms of Γ (R(n, q)), we need firstly to study the automorphisms of a related graph. Let Rn (q) be the graph with vertex set {[A] | A ∈ R(n, q)}; there is a directed edge from [A] ∈ V (Rn (q)) to [B] ∈ V (Rn (q)), written as [A] → [B], if and only if AB = 0. The definition is well since AB = 0 ⇔ (aA)(bB) = 0 for a, b ∈ Fq∗ . We also say a vertex [A] ∈ V (Rn (q)) is supported by (s, t), if so is A. The automorphism group of Rn (q) is written as Aut(Rn (q)). In the last section we will show that the problem of determining automorphisms of Γ (R(n, q)) can be reduced to those of Rn (q). 3. Automorphisms of Rn (q) In this section, we determine the automorphisms of Rn (q). We begin with studying the vertex number of Rn (q), the in-degree and the out-degree of vertices. For 1 ≤ s ≤ t ≤ n, denote by Σ(s, t) the set of vertices of Rn (q) which are supported by (s, t). Lemma 3.1. (i) V (Rn (q)) is the disjoint union of all Σ(s, t) for 1 ≤ s ≤ t ≤ n. (ii) The number of vertices in Σ(s, t) is q s+n−t−1 . (iii) The vertex number of Rn (q) is 1≤s≤t≤n q s+n−t−1 . Proof. (i) follows immediately from Lemma 2.2. Let [A1 ], [A2 ] ∈ Σ(s, t). Suppose the (s, t) positions of Ai are taken to be 1. Then Ai (i = 1 or 2) can be uniquely written as ⎞ (i) a1 ⎜ (i) ⎟ ⎜ a2 ⎟ (i) (i)
(i) ⎟ Ai = ⎜ ⎜ .. ⎟ b1 b2 . . . bn , ⎝ . ⎠ (i) an ⎛
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(i)
(i)
(i)
(i)
(i)
(i)
where as − 1 = as+1 = . . . = an = 0 and b1 = . . . = bt−1 = bt − 1 = 0. It is (1) (2) (1) (2) clear that [A1 ] = [A2 ] if and only if ak = ak and bk = bk for all k. On the (1) (1) other hand, ak (1 ≤ k ≤ s − 1) and bl (t + 1 ≤ l ≤ n) can be chosen from Fq arbitrarily. Thus, |Σ(s, t)| = q s−1 · q n−t = q s+n−t−1 , which confirms (ii). By applying (i), |V (Rn (q))| = 1≤s≤t≤n |Σ(s, t)| = 1≤s≤t≤n q s+n−t−1 , which proves (iii). 2 Now, we consider the degree of each vertex. For [A] ∈ V (Rn (q)), we denote by No ([A]) (resp., Ni ([A])) the set of vertices [X] ∈ V (Rn (q)) for which [A] → [X] (resp., [X] → [A]). The out-degree (resp., in-degree) of [A], written as do ([A]) (resp., di ([A])), is the number of vertices contained in No ([A]) (resp., Ni ([A])). Lemma 3.2. (i) If [A] and [B] are contained in the same Σ(s, t), then do ([A]) = do ([B]) and di ([A]) = di ([B]). (ii) For [A] ∈ Σ(s, t) and [B] ∈ Σ(k, l), if t < l then do ([A]) < do ([B]), if s < k then di ([B]) < di ([A]). (iii) For [A] ∈ Σ(s, t) and [B] ∈ Σ(k, l), do ([A]) = do ([B]) and di ([A]) = di ([B]) if and only if s = k and t = l. Proof. To complete the proof of (i), it suffices to prove that do ([A]) = do ([Est ]) and s n di ([A]) = di ([Est ]) for all [A] supported by (s, t). Suppose that A = k=1 l=t akl Ekl with ast = 1. Since A is a rank one matrix we know that akl = akt asl for 1 ≤ k ≤ s −1 and n s−1 t + 1 ≤ l ≤ n. Let P = (I − l=t+1 asl Etl )(I + k=1 akt Eks ). Then P −1 AP = Est . Let ρ be the mapping from No ([A]) to No ([Est ]) defined by [X] → [P −1 X]. Obviously, the mapping is well defined and is bijective. Hence, do ([A]) = do ([Est ]). Similarly, di ([A]) = di ([Est ]). By (i) of this lemma, we only need to prove that do ([Est ]) < do ([Ekl ]) when t < l. Note that a vertex [X] lies in No ([Ekl ]) if and only if the lth row of X is zero. Now we define a mapping θ from No ([Est ]) to No ([Ekl ]) by sending [X] ∈ No ([Est ]) to [Ptl X], where Ptl denote the elementary matrix by permuting the tth and the lth rows of the n × n identity matrix. Since [Ptl X] ∈ No ([Ekl ]) for [X] ∈ No ([Est ]), the mapping is well defined. It is easy to see that θ is injective, since [Ptl X1 ] = [Ptl X2 ] if and only if [X1 ] = [X2 ]. We claim that [Ett ] ∈ No ([Ekl ]) is not the image of any element of No ([Est ]). Suppose on the contrary that θ([X]) = [Ett ] for certain [X] ∈ No ([Est ]). Then [X] = [Ptl Ett ] = [Elt ]. This contradicts the fact that X is upper triangular. Consequently, θ is not surjective, which implies do ([Est ]) < do ([Ekl ]). Similarly, di ([B]) < di ([A]) if s < k. (iii) follows from (i) and (ii) immediately. 2 Next, three standard automorphisms of Rn (q) are constructed. Based on them we can describe any automorphism of Rn (q).
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• Let P ∈ Tn (q) be invertible. We define σP from V (Rn (q)) to itself by [X] → [P −1 XP ]. Then it is easy to check that σP is an automorphism of Rn (q), which is called an inner automorphism of Rn (q). • Let n ≥ 3, and let π be an automorphism of the field Fq . Define σπ from V (Rn (q)) to itself by
ast Est →
1≤s≤t≤n
π(ast )Est .
1≤s≤t≤n
It is not difficult to check that σπ is an automorphism of Rn (q), which is called a field automorphism of Rn (q). • Let ω be a permutation on Fq such that ω(0) = 0. When n = 2, define σω on any vertex [A] ∈ V (R2 (q)) with A = aE11 + bE12 + cE22 as follows: if a = 1, c = 0, [A] is sent to [E11 +ω(b)E12 ]; if c = 1, a = 0, [A] is sent to [−ω(−b)E12 + E22 ]; if a = c = 0, b = 1, [A] is fixed by σω . It is trivial to verify that σω is an automorphism of R2 (q). The main result of this section is as follows. Theorem 3.3. When n ≥ 3, σ is an automorphism of Rn (q) if and only if σ = σP · σπ , where σP is an inner automorphism of Rn (q) and σπ is a field automorphism of Rn (q). When n = 2, σ is an automorphism of Rn (q) if and only if σ = σU · σω , where ω is a permutation on Fq fixing 0, U is a 2 × 2 unit upper triangular matrix (all diagonal elements are 1) over Fq . Proof. The sufficiency is obvious. Now we consider the necessity for the case when n ≥ 3. Suppose σ is an automorphism of Rn (q). We complete the proof by announcing eight claims. Claim 1. Each Σ(s, t) is stabilized by σ for 1 ≤ s ≤ t ≤ n. For [A] ∈ Σ(s, t), suppose that σ([A]) ∈ Σ(k, l). Since σ is an automorphism, we have do (σ([A])) = do ([A]) and di (σ([A])) = di ([A]). Then, by Lemma 3.2, we have s = k and t = l. Hence, σ([A]) ∈ Σ(s, t). In other words, σ stabilizes Σ(s, t). Claim 2. There is a unit upper triangular matrix U such that σU · σ fixes each [Est ]. Firstly, we consider the action of σ on [E1t ] for 1 ≤ t ≤ n. By Claim 1, σ([E1t ]) ∈ n (t) (t) Σ(1, t). Clearly, σ([E1n ]) = [E1n ]. Assume that σ([E1t ]) = [ j=t a1j E1j ] with a1t = 1 for t = 1, 2, . . . , n − 1. Let Ut = I −
n
l=t+1
(t) a1l Etl
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for t = 1, 2, . . . , n − 1, and let U = U1 · U2 · · · Un−1 . Then σU · σ fixes each [E1t ] for t = 1, 2, . . . , n. Now we denote σU · σ by σ1 . Next, we prove σ1 fixes each [Esn ] for s = 1, 2, . . . , n. Clearly, σ1 fixes [E1n ]. For 2 ≤ s ≤ n, since σ1 ([Esn ]) ∈ Σ(s, n), we may assume that s (s)
σ1 [Esn ] = bkn Ekn k=1 (s)
with bsn = 1. If j < s, by [E1j ] → [Esn ], we have σ1 ([E1j ]) = [E1j ] → σ1 ([Esn ]), s (s) (s) which implies that E1j · ( k=1 bkn Ekn ) = 0. Thus, bjn = 0 for 1 ≤ j < s, which shows that σ1 ([Esn ]) = [Esn ] for s = 1, 2, . . . , n. Finally, we prove that σ1 fixes each [Est ] for 1 ≤ s ≤ t ≤ n. Suppose that σ1 ([Est ]) = [X] ∈ V (Rn (q)). By Claim 1, [X] ∈ Σ(s, t). For j = s, applying σ1 to [E1j ] → [Est ], we have [E1j ] → [X], or equivalently, E1j X = 0. It implies that the jth row of X is zero for j = s. Similarly, the kth column of X is zero for k = t. Hence, σ1 ([Est ]) = [Est ] for 1 ≤ s ≤ t ≤ n. Claim 3. Suppose σ1 ([A]) = [B] for A ∈ R(n, q). Then the kth row of [B] is zero if and only if so is [A], the lth column of [B] is zero if and only if so is [A]. In fact, the kth row of [A] is zero if and only if [E1k ] → [A], if and only if σ1 ([E1k ]) = [E1k ] → [B], if and only if E1k B = 0, if and only if the kth row of B is zero. The proof of the second assertion is similar. For 1 ≤ i < j ≤ n and a ∈ Fq , Claim 3 shows that σ1 sends [Eii + aEij ] to a vertex (i) of the form [Eii + bEij ] with b ∈ Fq . Thus we can define a permutation πij on Fq such (i)
that πij (0) = 0 and
(i) σ1 [Eii + aEij ] = Eii + πij (a)Eij . (j)
(j)
Similarly, we define a permutation πij on Fq such thatπij (0) = 0 and
(j) σ1 [Ejj + aEij ] = Ejj + πij (a)Eij . (j)
(i)
Claim 4. For 1 ≤ i < j ≤ n and a ∈ Fq , πij (−a) = −πij (a). (i)
By applying σ1 to [Eii + aEij ] → [Ejj − aEij ], we have [Eii + πij (a)Eij ] → [Ejj + (j)
(i)
(j)
πij (−a)Eij ], which implies that (Eii + πij (a)Eij ) · (Ejj + πij (−a)Eij ) = 0. Thus, (j)
(i)
πij (−a) = −πij (a). Claim 5 and Claim 6 are devoted to represent the action of σ1 on a vertex [A] by using these permutations. For a vertex [A] supported by (s, s), assume that A =
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s
249
n
akl Ekl with ass = 1, and assume that σ1 ([A]) = [A ]. By Claim 1, [A ] is also s n supported by (s, s). Thus, we may assume that A = k=1 l=s akl Ekl with ass = 1. Since A and A are both rank one matrices, k=1
l=s
akl = aks asl ,
akl = aks asl
for 1 ≤ k ≤ s − 1, s + 1 ≤ l ≤ n.
(1)
Claim 5. asl = πsl (asl ) for l = s + 1, . . . , n and aks = πks (aks ) for k = 1, 2, . . . , s − 1. s n (s) (s) (s) So σ1 ([A]) = [ k=1 l=s πks (aks )πsl (asl )Ekl ], where πss (1) = 1. (s)
(s)
By A · (Ell − asl Esl ) = 0 we have [A] → [Ell − asl Esl ]. By applying σ1 , we have (l) (l) [A ] → [Ell + πsl (−asl )Esl ], which implies that A · (Ell + πsl (−asl )Esl ) = 0, leading to asl = −πsl (−asl ). (l)
(2)
By using Claim 4, we have that asl = πsl (asl ). Similar discussion leads to the second result. The third assertion is obvious. s n For a vertex [B] supported by (s, t) with s < t, assume that B = k=1 l=t bkl Ekl with bst = 1 and assume that σ1 ([B]) = [B ], where B is written as B = s n k=1 l=t bkl Ekl with bst = 1. As B and B are rank one matrices, we have (s)
bkl = bkt bsl ,
bkl = bkt bsl
for 1 ≤ k ≤ s − 1, t + 1 ≤ l ≤ n.
(3)
Claim 6. bsl = (πst (1))−1 πsl (bsl ) and bkt = (πst (1))−1 πkt (bkt ), for l = t + 1, . . . , n and s n (t) (t) (s) (s) k = 1, 2, . . . , s − 1. So σ1 ([B]) = [ k=1 l=t ((πst (1))−1 πkt (bkt ) · (πst (1))−1 πsl (bsl )) · Ekl ]. (s)
(s)
(t)
(t)
The fact B · (Ell − bsl Etl ) = 0 shows that [B] → [Ell − bsl Etl ]. After applying σ1 to it, (l) (l) we have [B ] → [Ell + πtl (−bsl )Etl ], which implies that B · (Ell + πtl (−bsl )Etl ) = 0. It follows that bsl = −πtl (−bsl ). (l)
(4)
On the other hand, it follows from (Ess + Est + bsl Esl ) · (Ell − bsl Etl ) = 0 that [Ess + Est + bsl Esl ] → [Ell − bsl Etl ]. Applying σ1 , we have (s) (s) (l) Ess + πst (1)Est + πsl (bsl )Esl → Ell + πtl (−bsl )Etl , which implies that
(s) (s) (l) Ess + πst (1)Est + πsl (bsl )Esl · Ell + πtl (−bsl )Etl = 0. Consequently, (s)
(l)
(s)
πst (1)πtl (−bsl ) + πsl (bsl ) = 0.
(5)
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(4) and (5) together prove that (s) −1 (s) bsl = πst (1) πsl (bsl ).
(6)
(t) −1 (t) bkt = πst (1) πkt (bkt ).
(7)
Similar discussions show that
Thus the last assertion holds true. This completes the proof of Claim 6. The following claim describes the properties of these permutations and the relations between them. Claim 7. (1)
(1)
(k)
(i) π1j (ab) = π1k (a) · πkj (b) for a, b ∈ Fq and 2 ≤ k < j ≤ n. (ii) π1j (ab) = π1j (a) · π1j (b) · (π1j (1))−1 for a, b ∈ Fq and 2 ≤ j ≤ n. (1)
(1)
(1)
(1)
(iii) π1j (a−1 ) = (π1j (a))−1 · (π1j (1))2 for a ∈ Fq∗ and 2 ≤ j ≤ n. (1)
(1)
(1)
(iv) π1j (a) · (π1j (1))−1 = π12 (a) · (π12 (1))−1 for a ∈ Fq and 2 ≤ j ≤ n. (1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(v) π1j (a + b) = π1j (a) + π1j (b) for a, b ∈ Fq and 2 ≤ j ≤ n. (vi) π1j (−a) = −π1j (a) for a ∈ Fq and 2 ≤ j ≤ n. (k)
(l)
(vii) πkl = πkl for 1 ≤ k < l ≤ n. Suppose 2 ≤ k < j ≤ n. From (E11 + aE1k + abE1j ) · (Ejj − bEkj ) = 0, it follows that [E11 + aE1k + abE1j ] → [Ejj − bEkj ]. By applying σ1 , we have
(1) (1) (j) E11 + π1k (a)E1k + π1j (ab)E1j → Ejj + πkj (−b)Ekj ,
which implies that
(1) (1) (j) E11 + π1k (a)E1k + π1j (ab)E1j · Ejj + πkj (−b)Ekj = 0.
From which we have that (1)
(1)
(j)
(1)
(k)
π1j (ab) = −π1k (a) · πkj (−b) = π1k (a) · πkj (b).
(8)
This completes the proof of (i). From (8), we have (1)
(1)
(2)
π1j (ab) = π12 (a) · π2j (b)
for 3 ≤ j ≤ n,
(9)
and (1)
(1)
(2)
(1)
(2)
π1j (a) = π12 (a) · π2j (1) = π12 (1) · π2j (a)
for 3 ≤ j ≤ n.
(10)
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It follows that (2) −1 (1) (1) π12 (a) = π1j (a) · π2j (1)
(1) −1 (2) (1) and π2j (b) = π1j (b) · π12 (1)
(11)
for a, b ∈ Fq and 3 ≤ j ≤ n. Substituting (11) into (9), we have (1) −1 (1) (1) (1) π1j (ab) = π1j (a) · π1j (b) · π1j (1)
for 3 ≤ j ≤ n.
(12)
From π12 (a) = π1j (a) · (π2j (1))−1 , we have that (1)
(1)
(2)
(2) −1 (1) (1) π12 (ab) = π1j (ab) · π2j (1) (1) −1 (2) −1 (1) (1) = π1j (a) · π1j (b) · π1j (1) · π2j (1)
−1 (1) (1) (1) = π12 (a) · π1j (b) · π1j (1) (2) (1) −1 (1) (1) = π12 (a) · π12 (b) · π2j (1) · π1j (1) (1) −1 (1) (1) = π12 (a) · π12 (b) · π12 (1) . This equality together with (12) prove (1) −1 (1) (1) (1) π1j (ab) = π1j (a) · π1j (b) · π1j (1)
for 2 ≤ j ≤ n.
(13)
This completes the proof of (ii). (iii) follows immediately from (ii). For a ∈ Fq and 3 ≤ j ≤ n, by using (10), we have (1) −1 (1) −1 (1) −1 (1) (1) (2) (1) π1j (a) · π1j (1) = π12 (a) · π2j (1) · π1j (1) = π12 (a) · π12 (1) . Thus the result of (iv) follows. For j ≥ 3 and a, b ∈ Fq , from
E11 − bE12 + (a + b)E1j · (Ejj + E2j − aE1j ) = 0,
we have
E11 − bE12 + (a + b)E1j → [Ejj + E2j − aE1j ].
By applying σ1 , we arrive at (1) (1) (j) (j) E11 + π12 (−b)E12 + π1j (a + b)E1j → Ejj + π2j (1)E2j + π1j (−a)E1j , which leads to
(1) (1) (j) (j) E11 + π12 (−b)E12 + π1j (a + b)E1j · Ejj + π2j (1)E2j + π1j (−a)E1j = 0.
(14)
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Thus, (1)
(1)
(j)
(j)
π1j (a + b) + π12 (−b)π2j (1) + π1j (−a) = 0. (1)
(j)
(1)
(2)
(1)
(15) (j)
(1)
Since π12 (−b) · π2j (1) = −π12 (−b) · π2j (−1) = −π1j (b) and π1j (−a) = −π1j (a), (15) implies that (1)
(1)
(1)
π1j (a + b) = π1j (a) + π1j (b)
for a, b ∈ Fq , 3 ≤ j ≤ n.
(16)
Using (14) and (16), we also have (1)
(1)
(1)
for a, b ∈ Fq .
(17)
for a, b ∈ Fq , 2 ≤ j ≤ n.
(18)
π12 (a + b) = π12 (a) + π12 (b) Hence (1)
(1)
(1)
π1j (a + b) = π1j (a) + π1j (b)
This proves (v). (vi) follows immediately from (v). For any a ∈ Fq and 2 ≤ k < l ≤ n, using the results of (i), (vi) and applying Claim 4, we have (1)
(1)
(1)
(k)
(1)
(l)
−π1l (a) = π1l (−a) = π1k (1) · πkl (−a) = −π1k (1) · πkl (a). It follows that (1) −1 (1) (l) (k) πkl (a) = π1k (1) · π1l (a) = πkl (a). (k)
(l)
Consequently, πkl = πkl for 2 ≤ k < l ≤ n. For the special case that k = 1, from (1) (1) (l) (1) (l) π1l (a) = −π1l (−a) = π1l (a), we have π1l = π1l . This completes the proof of (vii). The proof of Claim 7 is completed. (k) (l) By (vii) of Claim 7, distinguishing πkl with πkl seems unnecessary, so we will denote both them by πkl . Now, let π = (π12 (1))−1 π12 . Claim 8. (i) π is an automorphism of the field Fq . (ii) Let π11 (1) = 1. Then πkl = (π1k (1))−1 π1l (1)π = πkl (1)π for 1 ≤ k < l ≤ n. For a, b ∈ Fq , since
−1
−1
−1 π(a + b) = π12 (1) π12 (a + b) = π12 (1) π12 (a) + π12 (1) π12 (b) = π(a) + π(b),
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and
−1
−1
−1
π(ab) = π12 (1) π12 (ab) = π12 (1) π12 (a)π12 (b) π12 (1) = π(a)π(b), π is an automorphism of the field Fq . Thus (i) of Claim 8 holds. (iv) of Claim 7 shows that π1l (a) = π1l (1)π(a) for any a and 2 ≤ l. For 2 ≤ k < l ≤ n, using (i) and (iv) of Claim 7, we have πkl (a) = (π1k (1))−1 π1l (a) = (π1k (1))−1 π1l (1)π(a), from which it follows that πkl = (π1k (1))−1 π1l (1)π. Since (π1k (1))−1 π1l (1) = πkl (1), we have πkl = πkl (1)π. Unifying the results for k = 1 and k ≥ 2, we obtain πkl = (π1k (1))−1 π1l (1)π = πkl (1)π for 1 ≤ k < l ≤ n, where π11 (1) = 1. The proof of Claim 8 is completed. Let D = diag(d1 , d2 , . . . , dn ), where dk = (π1k (1))−1 for 1 ≤ k ≤ n, and denote σD · σ1 by σ2 . Claim 9. σ2 identifies with σπ . We consider the action of σ2 on each vertex. s n For A = k=1 l=s akl Ekl supported by (s, s), where ass = 1 and akl = aks asl for 1 ≤ k ≤ s − 1 and s + 1 ≤ l ≤ n, Claim 5 shows that s n
σ1 [A] = πks (aks )πsl (asl )Ekl ,
k=1 l=s
where πss (1) = 1. Using Claim 8 we have s n
−1 σ1 [A] = π1k (1) π1l (1)π(aks )π(asl )Ekl .
k=1 l=s
Hence,
σ2 [A] = σD =
s n
k=1 l=s n s
−1 π1k (1) π1l (1)π(aks )π(asl )Ekl
π(aks )π(asl )Ekl .
k=1 l=s
This equality shows that the action of σ2 on each vertex [A] ∈ Σ(s, s) identifies with the action of σπ on [A].
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s n For B = k=1 l=t bkl Ekl supported by (s, t), where s < t, bst = 1, and bkl = bkt bsl for 1 ≤ k ≤ s − 1 and t + 1 ≤ l ≤ n, Claim 6 shows that s n
−1
−1
σ1 [B] = πst (1) πkt (bkt ) · πst (1) πsl (bsl ) · Ekl .
k=1 l=t
Using Claim 8 we have s n
−1
−1 σ1 [B] = πst (1) π1k (1) π1l (1)π(bkt )π(bsl ) · Ekl .
k=1 l=t
Hence, s n s n
−1 σ2 [B] = πst (1) π(bkt )π(bsl ) · Ekl = π(bkt )π(bsl ) · Ekl .
k=1 l=t
k=1 l=t
The last equality comes from the fact that [aA] = [A] for a ∈ Fq∗ and A ∈ R(n, q). Hence, the action of σ2 on each vertex [B] ∈ Σ(s, t), for s < t, identifies with the action of σπ on this vertex. Thus, σD · σU · σ = σπ . Finally, we have σ = σP · σπ , where P = U −1 D−1 . Now we turn to the case when n = 2. As in the case when n ≥ 3, we can find a 2 × 2 unit upper triangular matrix U such that σU · σ respectively fixes [E11 ], [E12 ] and [E22 ]. Denote σU · σ by σ1 . Suppose σ1 ([A]) = [B] for A ∈ V (R2 (q)). Then, as in the case when n ≥ 3, the kth row of [B] is zero if and only if so is [A], the kth column of [B] is zero if and only if so is [A], where k = 1 or 2. Now, we can define a permutation ω on Fq , with ω(0) = 0, such that
σ1 [E11 + aE12 ] = E11 + ω(a)E12
for a ∈ Fq .
Now for b ∈ Fq , suppose that σ1 sends [bE12 + E22 ] to [f (b)E12 + E22 ]. By (E11 − bE12 )(bE12 + E22 ) = 0, we have [E11 − bE12 ] → [bE12 + E22 ], from which it follows that E11 + ω(−b)E12 → f (b)E12 + E22 ,
E11 + ω(−b)E12 · f (b)E12 + E22 = 0. The last equality implies that f (b) + ω(−b) = 0. Thus f (b) = −ω(−b), and
σ1 [bE12 + E22 ] = −ω(−b)E12 + E22 .
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Observing the action of σ1 on each vertex, we find that σ1 identifies σω . Finally, we have σ = σU1 · σω with U1 = U −1 , as desired. 2 Finally, we consider the orbit partition of the vertex set and the order of the group of automorphisms. Corollary 3.4. The orbit partition of V (Rn (q)) under the automorphisms is V (Rn (q)) = n(n+1) . 1≤s≤t≤n Σ(s, t). The number of orbits is 2 Proof. Claim 1 of Theorem 3.3 (also holds for n = 2) shows that each Σ(s, t) is stabilized under Aut(Rn (q)). To complete the proof we only need to prove that, for any [A] ∈ Σ(s, t), there is an automorphism σ such that σ([Est ]) = [A]. Assume that s n A = a = 1 and akl = akt asl for 1 ≤ k ≤ s − 1 and k=1 l=t akl Ekl , where n st s−1 t + 1 ≤ l ≤ n. Let P = (I + l=t+1 asl Etl )(I − k=1 akt Eks ). Then A = P −1 Est P . So, σP ([Est ]) = [A]. The second result is obvious. 2 Corollary 3.5. Let q = pm . Then
n(n−1) Aut Rn (q) = m(q − 1)n−1 q 2 , when n ≥ 3; q
Aut Rn (q) = k , when n = 2. k=1
Proof. When n ≥ 3, by Theorem 3.3, each automorphism σ of Rn (q) can be written as σ = σP σπ for some P ∈ Tn (q), π ∈ Aut(Fq ). If σP1 σπ1 = σP2 σπ2 , then, σP0 = σπ0 , where P0 = P2−1 P1 and π0 = π2 π1−1 . Since each [Ess ] is fixed by σπ0 , we have σP0 ([Ess ]) = [Ess ] for s = 1, 2, . . . , n. That is [P0−1 Ess P0 ] = [Ess ] for s = 1, 2, . . . , n. Consequently, P0 must be a diagonal matrix. By σP0 ([Est ]) = [P0−1 Est P0 ] = σπ0 ([Est ]) = [Est ] for s < t, we have P0 must be a nonzero scalar matrix. Thus P1 = aP2 for certain a ∈ Rq∗ . Since σπ0 = σP0 , σπ0 must be the identity automorphism. It follows from [E11 + aE12 ] = σπ0 ([E11 + aE12 ]) = [E11 + π0 (a)E12 ] that π0 (a) = a for any a ∈ Fq . So π0 is the identity automorphism of Fq . Hence, π1 = π2 . Above discussions show that | Aut(Rn (q))| = |Tn (q)| q−1 · | Aut(Fq )|. n(n−1)
It is well known that |Tn (q)| = (q − 1)n q 2 and | Aut(Fq )| = m. Thus the first result follows. For the second result, since the number of permutations ω on Fq fixing q−1 0 is Πk=1 k, and the number of 2 × 2 unit upper triangular matrices is q, we have q | Aut(R2 (q))| = ( k=1 k). 2 By the proof of Corollary 3.5, the following corollary is obvious. T (q) Corollary 3.6. When n ≥ 3, Aut(Rn (q)) ∼ = nK Aut(Fq ); when n = 2, Aut(R2 (q)) ∼ = U2 (q) Sq−1 , where K = {aI|a ∈ Fq∗ }, U2 (q) denotes the set of 2 ×2 unit upper triangular matrices over Fq , and Sq−1 means the symmetric group on q − 1 elements.
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4. Automorphisms of Γ (R(n, q)) Applying Theorem 3.3, we can describe the automorphisms of Γ (R(n, q)) immediately. Firstly, we construct a standard automorphism for Γ (R(n, q)). Note that R(n, q) is the disjoint union of all [A] with A being rank one matrices. • Let τ be a permutation on R(n, q) which stabilizes each [A], or more definitely, τ is a bijection on R(n, q) such that τ (X) = aX X for any X ∈ R(n, q), where aX ∈ Fq∗ depends on X. It is easy to see that τ is an automorphism of Γ (R(n, q)), which is called a local scalar multiplication on Γ (R(n, q)). Now we show how to reduce an automorphism of Γ (R(n, q)) to that of Rn (q). Lemma 4.1. Let A, B ∈ R(n, q). Then Ni (A) = Ni (B) and No (A) = No (B) (in Γ (R(n, q))) if and only if B is a nonzero multiple of A. Proof. The sufficiency is obvious. For the necessity, suppose A is supported by (s, t). The proof of Lemma 3.2 shows that there exist P ∈ Tn (q) such that P −1 AP = aEst , with a ∈ Fq∗ . From Est · i=t Eii = 0, we have AP ( i=t Eii ) = 0. Further, by No (A) = No (B) we have BP ( i=t Eii ) = 0. It implies that all columns of BP are zero except for the tth column. The same assertion also holds for P −1 BP , since multiplying P −1 to BP from the left side is equivalent to applying elementary row transformation to BP . Similarly, all row vectors of P −1 BP are zero except for the sth row. Consequently, P −1 BP = bEst with b ∈ Fq∗ . Finally, we have B = bP Est P −1 = ba−1 A, as desired. 2 Let χ be an automorphism of Γ (R(n, q)). We define χ on Rn (q) by χ([A]) = [χ(A)]. Lemma 4.2. Let χ be an automorphism of Γ (R(n, q)). Then χ is an automorphism of Rn (q). Proof. If [B] = [A] ∈ V (Rn (q)), then B is a nonzero multiple of A. By Lemma 4.1, χ(B) is also a nonzero multiple of χ(A). Thus χ([A]) = χ([B]). It shows that χ is well defined. Clearly, χ is a bijection and χ([X]) → χ([Y ]) if and only if [X] → [Y ]. Hence, χ is an automorphism of Rn (q). 2 Next, we can use the characterization of automorphisms of Γ (R(n, q)) to determine the automorphisms of Rn (q). Theorem 4.3. (i) When n ≥ 3, σ is an automorphism of Γ (R(n, q)) if and only if σ = σP · σπ · τ , where σP : X → P −1 XP, ∀X ∈ R(n, q); σπ : [xij ] → [π(xij )], ∀[xij ] ∈ R(n, q); τ is a local scalar multiplication. (ii) When n = 2, σ is an automorphism of Rn (q) if and only if σ = σU · σω · τ , where σU : X → U −1 XU, ∀X ∈ R(2, q) with U a 2 × 2 unit upper triangular matrix (all
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diagonal elements are 1); ω is a permutation on Fq fixing 0 and σω : a0 cb → a0 dc with d = ω(a−1 b) if c = 0, a = 0, or d = −ω(−c−1 b) if a = 0, c = 0, or d = b if a = c = 0, b = 0; and τ is a local scalar multiplication. Proof. Let n ≥ 3 and let σ be an automorphism of Γ (R(n, q)). Then by Lemma 4.2, σ : [A] → [σ(A)] is an automorphism of Rn (q). Applying Theorem 3.3, we know there −1 is an invertible matrix P and an automorphism π of Fq such that σ −1 π · σ P · σ acts as the identity automorphism on Rn (q), or equivalently, σπ−1 · σP−1 · σ sends each rank one upper triangular matrix A to a scalar multiple of A. Thus σπ−1 · σP−1 · σ exactly is a local scalar multiplication of Rn (q), which completes the proof of the first part. The proof for n = 2 is similar, thus omitted. 2 Above result for n ≥ 3 can be stated in a more explicit way as follows. Theorem 4.4. A bijective map σ on R(n, q) is an automorphism of Γ (R(n, q)) if and only if σ(X) = aX P −1 π(xij ) P,
∀X = [xij ] ∈ R(n, q),
where P is an invertible upper triangular matrix; aX ∈ Fq∗ depends on X; π is an automorphism of the field Fq , [π(xij )] denotes the matrix whose (i, j)-entry is π(xij ). In this paper, the automorphisms of an induced subgraph of the zero-divisor graph of upper triangular matrices are determined, the general case that the zero-divisor graph is over all upper triangular matrices is left unsolved. Acknowledgement We thank the referee for his suggestion of developing relation between the graph of our paper and zero-divisor graphs of matrix ring, which have been received a lot of attention. References [1] S. Akbari, A. Mohammadian, On zero-divisor graphs of finite rings, J. Algebra 314 (2007) 168–184. [2] D.D. Anderson, M. Naseer, Beck’s coloring of a commutative ring, J. Algebra 159 (1993) 500–514. [3] D.F. Anderson, P.S. Livingston, The zero-divisor graph of a commutative ring, J. Algebra 217 (1999) 434–447. [4] D.F. Anderson, S.B. Mulay, On the diameter and girth of a zero-divisor graph, J. Pure Appl. Algebra 210 (2007) 543–550. [5] D.F. Anderson, R. Levy, J. Shapiro, Zero-divisor graphs, von Neumann regular rings and Boolean algebras, J. Pure Appl. Algebra 180 (2003) 221–241. [6] M. Axtell, J. Coykendall, J. Stickles, Zero-divisor graphs of polynomial and power series over commutative rings, Comm. Algebra 33 (2005) 2043–2050. [7] I. Beck, Coloring of commutative rings, J. Algebra 116 (1988) 208–226. [8] I. Bo˘ zić, Z. Petrović, Zero-Divisor Graphs of Matrices Over Commutative Rings, Comm. Algebra 37 (2009) 1186–1192.
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[9] Z. Gu, Subconstituents of symplectic graphs modulo pn , Linear Algebra Appl. 439 (2013) 1321–1329. [10] Z. Gu, Z. Wan, Orthogonal graphs of odd characteristic and their automorphisms, Finite Fields Appl. 14 (2008) 291–313. [11] J. Han, The zero-divisor graph under group actions in a noncommutative ring, J. Korean Math. Soc. 45 (2008) 1647–1659. [12] R. Levy, T. Shapiro, The zero-divisor graph of von Neumann regular rings, Comm. Algebra 30 (2002) 745–750. [13] B. Li, Zero-divisor graph of triangular matrix rings over commutative rings, Int. J. Algebra 5 (2011) 255–260. [14] A. Li, P.R. Tucci, Zero-divisor graphs of upper triangular matrix rings, Comm. Algebra 41 (2013) 4622–4636. [15] F. Li, Y. Wang, Subconstituents of symplectic graphs, European J. Combin. 29 (2008) 1092–1103. [16] F. Li, K. Wang, J. Guo, More on symplectic graphs modulo pn , Linear Algebra Appl. 438 (2013) 2651–2660. [17] T.G. Lucas, The diameter of a zero divisor graph, J. Algebra 301 (2006) 174–193. [18] Y. Meemark, T. Prinyasart, On symplectic graphs modulo pn , Discrete Math. 311 (2011) 1874–1878. [19] S. Park, J. Han, The group of graph automorphisms over a matrix rings, J. Korean Math. Soc. 48 (2011) 301–309. [20] S. Redmond, The zero-divisor graph of a noncommutative ring, Int. J. Commut. Rings 1 (2002) 203–211. [21] Z. Tang, Z. Wan, Symplectic graphs and their automorphisms, European J. Combin. 27 (2006) 38–50. [22] Z. Wan, K. Zhou, Unitary graphs and their automorphisms, Ann. Comb. 14 (2010) 367–395. [23] Z. Wan, K. Zhou, Orthogonal graphs of characteristic 2 and their automorphisms, Sci. China Math. 52 (2009) 361–380. [24] T. Wu, On directed zero-divisor graphs of finite rings, Discrete Math. 296 (2005) 73–86.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
The group of automorphisms of a zero-divisor graph based on rank one upper triangular matrices Dein Wong ∗,1 , Xiaobin Ma, Jinming Zhou 2 Department of Mathematics, China University of Mining and Technology, Xuzhou 221116, China
a r t i c l e
i n f o
Article history: Received 16 March 2014 Accepted 27 July 2014 Available online 14 August 2014 Submitted by R. Brualdi MSC: 05C25 05C60 05E15 11T99 Keywords: Zero-divisor graph Graph automorphisms Transitivity
a b s t r a c t Let Fq be a finite field with q elements, n(≥ 2) a positive integer, Matn (q) the set of all n × n matrices over Fq , R(n, q) the set of all rank one upper triangular matrices in Mat(n, q). The zero-divisor graph of Matn (q), written as Γ (Matn (q)), is a directed graph with vertex set all nonzero zero-divisors of Matn (q), and there is a directed edge from a vertex A to a vertex B, written as A → B, if and only if AB = 0. In this paper, we determine the automorphisms of an induced subgraph, written as Γ (R(n, q)), of Γ (Matn (q)) with vertex set R(n, q). The main theorem of this article proves that a bijective map σ on R(n, q) with n ≥ 3 is an automorphism of Γ (R(n, q)) if and only if σ(X) = aX P −1 π(xij ) P,
∀X = [xij ] ∈ R(n, q),
where aX ∈ Fq∗ depends on X; P is an invertible upper triangular matrix; π is an automorphism of the field Fq , [π(xij )] denotes the matrix whose (i, j)-entry is π(xij ). © 2014 Elsevier Inc. All rights reserved.
* Corresponding author. 1 2
E-mail address: [email protected] (D. Wong). Supported by the National Natural Science Foundation of China (No. 11171343). Supported by NSF grant BK20140177 of Jiangsu Province.
http://dx.doi.org/10.1016/j.laa.2014.07.041 0024-3795/© 2014 Elsevier Inc. All rights reserved.
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1. Introduction Let R be a commutative ring with identity 1. The zero-divisor graph of R, denoted by Γ (R), is an undirected graph with vertices Z ∗ (R), the set of nonzero zero-divisors of R, and for distinct elements x, y ∈ Z ∗ (R), there is an edge joining x and y if and only if xy = 0. The concept of zero-divisor graph was first defined and studied for commutative rings by Beck in [7], and further studied by many authors (see, e.g., Akbari et al. [1]; Anderson et al. [2–5]; Axtell et al. [6]; Levy et al. [12]; Lucas [17]; Wu [24]). A lot of results about the diameter, the girth of Γ (R) and so on have been obtained. The main idea of these researches is to study the interplay between the ring-theoretic properties of a commutative ring R and the graph-theoretic properties of Γ (R). Redmond [20] extended the concept of zero-divisor graphs of commutative rings to zero-divisor graphs of noncommutative rings. Recently, zero-divisor graphs of matrix rings attracted a lot of attention. Akbari and Mohammadian [1] studied the problem of determining when the zero-divisor graphs of rings are isomorphic, given that the zero-divisor graphs of their matrix rings are isomorphic. I. Bo˘zić and Z. Petrović [8] further investigated the properties of (directed) zero-divisor graphs of matrix rings, studied the relation between the diameter of the zero-divisor graph of a commutative ring R and that of the matrix ring Matn (R). Li [13] proved that when studying Γ (R), one can replace R with its total quotient ring, and they determined the girth of Γ (T ) of T , triangular matrices over a commutative ring. The diameter, the girth and the bounds for the number of edges of Γ (T ) were given in [14]. Generally, determining the full automorphisms of a graph is an important however a difficult problem both in graph theory and in algebraic theory. Searching the literature, we find that little is known for the automorphisms of zero-divisor graphs of rings. In [3], Anderson and Livingston have shown that Aut(Γ (Zn )) is a direct product of symmetric groups for n ≥ 4 a nonprime integer. For the noncommutative case, it was shown in [11] that Aut(Γ (R)) is isomorphic to the symmetric group of degree p+1, when R = Mat2 (Zp ) (p is a prime). S. Park and J. Han [19] proved that Aut(Γ (R)) ∼ = Sq+1 for R = Mat2 (Fq ) with Fq an arbitrary finite field. It is slightly regrettable that the authors of [11] and [19] only concerned with the special case when n = 2. Till now nothing is known about the automorphisms of the zero-divisor graphs of n × n matrices with n ≥ 3. This observation motivates us to do some work on this topic. It seems much difficult to determine the full automorphisms of Γ (Matn (q)), so we focus our attention on the subgraph of Γ (Matn (q)) induced by all rank one upper triangular matrices. Let T (n, q) be the set of all n × n upper triangular matrices over Fq , R(n, q) the set of all rank one matrices in T (n, q). The subgraph of Γ (Matn (q)) induced by all rank one upper triangular matrices is denoted by Γ (R(n, q)). The main theorem of this article proves that any graph automorphism of Γ (R(n, q)) with n ≥ 3 can be decomposed into the product of an inner automorphism, a field automorphism and a local scalar multiplication. For proving the main theorem of this paper we apply a technique from some recent papers which focused on determining the automorphisms of some graphs based on row
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vectors over finite fields. Here we list some papers closely related to our topic. Let Fq2v be the 2v-dimensional row vector space over a finite field Fq , and let K be a non-singular 2v × 2v skew symmetric matrix over Fq . The symplectic graph Sp(2v, q) with respect to K is defined to be the graph with vertices of all 1-dimensional subspace [α] of Fq2v , two vertices [α] and [β] are adjacent if and only if αKβ T = 0. Tang and Wan [21] determined the automorphisms of Sp(2v, q) and proved that this graph is strongly regular. Li and Wang [15] further showed that the subconstituents of the symplectic graph are strictly Deza graphs except the trivial case when υ = 2. Meemark and Prinyasart [18] studied the symplectic graph Sp(2υ, Zpn ) over a commutative ring Zpn and showed that it is strongly regular when υ = 1 and it is arc transitive when p is an odd prime. Li et al. [16] continued the research on Sp(2υ, Zpn ), showing that Sp(2υ, Zpn ) is arc transitive for any prime p, and it is a strictly Deza graph if υ ≥ 2 and n ≥ 2. Gu [9] studied the subconstituents of the symplectic graphs Sp(2υ, Zpn ). Wan and Zhou [22,23] studied the (isotropic) unitary graphs and the (isotropic) orthogonal graphs of characteristic 2. Gu [10] defined the (isotropic) orthogonal graph O(n, q), with respect to a nonsingular symmetric matrix H over Fq , with the vertex set to be all isotropic lines of Fqn , two isotropic lines [α] and [β] are adjacent if and only if αHβ T = 0. Gu determined the automorphisms of O(n, q) and proved that O(n, q) is vertex transitive and edge transitive. 2. Notations and preliminaries Let Fq be a finite field with q elements and Fq∗ = Fq \ {0}, where q is a power of a prime p. For a matrix M , we denote by M T the transpose of M . For n ≥ 2, the symbol T (n, q) denotes the set of all n × n upper triangular matrices over Fq , R(n, q) the set of all rank one matrices in T (n, q). All invertible ones in T (n, q) form a group under multiplication of matrices, which we denote by Tn (q). By I we denote the n × n identity matrix. Denote by Eij (1 ≤ i, j ≤ n) the matrix unit, whose (i, j) position is 1 and all others are 0. Since all Eij , for 1 ≤ i ≤ j ≤ n, form a basis of T (n, q), each A ∈ T (n, q) can be uniquely written as A = 1≤i≤j≤n aij Eij with aij ∈ Fq . Before studying graph automorphisms we give an elementary property about rank one upper triangular matrices. Definition 2.1. Let A = 1≤i≤j≤n aij Eij ∈ R(n, q) be a rank one matrix, if ast = 0, asl = 0 for l = 1, 2, . . . , t − 1 and ak,t = 0 for k = s + 1, . . . , n, then we say that A is supported by (s, t). Lemma 2.2. Let A =
1≤i≤j≤n
aij Eij be a rank one upper triangular matrix over Fq .
(i) There exists a unique pair (s, t) by which A is supported. (ii) akl = 0 if 1 ≤ l ≤ t − 1 or s + 1 ≤ k ≤ n.
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Proof. Since A is of rank ⎛ a1one, ⎞ it can be written as the product of a row vector and a a2
column vector as A = ⎝ .. ⎠ (b1 b2 . . . bn ). Thus aij = ai bj for 1 ≤ i ≤ j ≤ n. Suppose . an
that as = 0, bt = 0, and as+1 = . . . = an = b1 = . . . = bt−1 = 0. Since A is upper triangular, s ≤ t. Then, it is obvious that A is supported by (s, t). Further, the second assertion holds. 2
For example, if n = 6 and if A = 00 M is a 6 × 6 matrix over Fq , where 0 ad bd cd M = 0 0 0 with abcd = 0, then A is supported by (3, 4). a b c
For A ∈ R(n, q), denote by [A] the subspace of T (n, q) spanned by A. As [A] = [aA] for a ∈ Fq∗ , the representation of [A] is not unique. However, if we require the representative A, supported by (s, t), to take 1 at the (s, t) position, then the representation is unique. For the aim of describing all automorphisms of Γ (R(n, q)), we need firstly to study the automorphisms of a related graph. Let Rn (q) be the graph with vertex set {[A] | A ∈ R(n, q)}; there is a directed edge from [A] ∈ V (Rn (q)) to [B] ∈ V (Rn (q)), written as [A] → [B], if and only if AB = 0. The definition is well since AB = 0 ⇔ (aA)(bB) = 0 for a, b ∈ Fq∗ . We also say a vertex [A] ∈ V (Rn (q)) is supported by (s, t), if so is A. The automorphism group of Rn (q) is written as Aut(Rn (q)). In the last section we will show that the problem of determining automorphisms of Γ (R(n, q)) can be reduced to those of Rn (q). 3. Automorphisms of Rn (q) In this section, we determine the automorphisms of Rn (q). We begin with studying the vertex number of Rn (q), the in-degree and the out-degree of vertices. For 1 ≤ s ≤ t ≤ n, denote by Σ(s, t) the set of vertices of Rn (q) which are supported by (s, t). Lemma 3.1. (i) V (Rn (q)) is the disjoint union of all Σ(s, t) for 1 ≤ s ≤ t ≤ n. (ii) The number of vertices in Σ(s, t) is q s+n−t−1 . (iii) The vertex number of Rn (q) is 1≤s≤t≤n q s+n−t−1 . Proof. (i) follows immediately from Lemma 2.2. Let [A1 ], [A2 ] ∈ Σ(s, t). Suppose the (s, t) positions of Ai are taken to be 1. Then Ai (i = 1 or 2) can be uniquely written as ⎞ (i) a1 ⎜ (i) ⎟ ⎜ a2 ⎟ (i) (i)
(i) ⎟ Ai = ⎜ ⎜ .. ⎟ b1 b2 . . . bn , ⎝ . ⎠ (i) an ⎛
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(i)
(i)
(i)
(i)
(i)
(i)
where as − 1 = as+1 = . . . = an = 0 and b1 = . . . = bt−1 = bt − 1 = 0. It is (1) (2) (1) (2) clear that [A1 ] = [A2 ] if and only if ak = ak and bk = bk for all k. On the (1) (1) other hand, ak (1 ≤ k ≤ s − 1) and bl (t + 1 ≤ l ≤ n) can be chosen from Fq arbitrarily. Thus, |Σ(s, t)| = q s−1 · q n−t = q s+n−t−1 , which confirms (ii). By applying (i), |V (Rn (q))| = 1≤s≤t≤n |Σ(s, t)| = 1≤s≤t≤n q s+n−t−1 , which proves (iii). 2 Now, we consider the degree of each vertex. For [A] ∈ V (Rn (q)), we denote by No ([A]) (resp., Ni ([A])) the set of vertices [X] ∈ V (Rn (q)) for which [A] → [X] (resp., [X] → [A]). The out-degree (resp., in-degree) of [A], written as do ([A]) (resp., di ([A])), is the number of vertices contained in No ([A]) (resp., Ni ([A])). Lemma 3.2. (i) If [A] and [B] are contained in the same Σ(s, t), then do ([A]) = do ([B]) and di ([A]) = di ([B]). (ii) For [A] ∈ Σ(s, t) and [B] ∈ Σ(k, l), if t < l then do ([A]) < do ([B]), if s < k then di ([B]) < di ([A]). (iii) For [A] ∈ Σ(s, t) and [B] ∈ Σ(k, l), do ([A]) = do ([B]) and di ([A]) = di ([B]) if and only if s = k and t = l. Proof. To complete the proof of (i), it suffices to prove that do ([A]) = do ([Est ]) and s n di ([A]) = di ([Est ]) for all [A] supported by (s, t). Suppose that A = k=1 l=t akl Ekl with ast = 1. Since A is a rank one matrix we know that akl = akt asl for 1 ≤ k ≤ s −1 and n s−1 t + 1 ≤ l ≤ n. Let P = (I − l=t+1 asl Etl )(I + k=1 akt Eks ). Then P −1 AP = Est . Let ρ be the mapping from No ([A]) to No ([Est ]) defined by [X] → [P −1 X]. Obviously, the mapping is well defined and is bijective. Hence, do ([A]) = do ([Est ]). Similarly, di ([A]) = di ([Est ]). By (i) of this lemma, we only need to prove that do ([Est ]) < do ([Ekl ]) when t < l. Note that a vertex [X] lies in No ([Ekl ]) if and only if the lth row of X is zero. Now we define a mapping θ from No ([Est ]) to No ([Ekl ]) by sending [X] ∈ No ([Est ]) to [Ptl X], where Ptl denote the elementary matrix by permuting the tth and the lth rows of the n × n identity matrix. Since [Ptl X] ∈ No ([Ekl ]) for [X] ∈ No ([Est ]), the mapping is well defined. It is easy to see that θ is injective, since [Ptl X1 ] = [Ptl X2 ] if and only if [X1 ] = [X2 ]. We claim that [Ett ] ∈ No ([Ekl ]) is not the image of any element of No ([Est ]). Suppose on the contrary that θ([X]) = [Ett ] for certain [X] ∈ No ([Est ]). Then [X] = [Ptl Ett ] = [Elt ]. This contradicts the fact that X is upper triangular. Consequently, θ is not surjective, which implies do ([Est ]) < do ([Ekl ]). Similarly, di ([B]) < di ([A]) if s < k. (iii) follows from (i) and (ii) immediately. 2 Next, three standard automorphisms of Rn (q) are constructed. Based on them we can describe any automorphism of Rn (q).
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• Let P ∈ Tn (q) be invertible. We define σP from V (Rn (q)) to itself by [X] → [P −1 XP ]. Then it is easy to check that σP is an automorphism of Rn (q), which is called an inner automorphism of Rn (q). • Let n ≥ 3, and let π be an automorphism of the field Fq . Define σπ from V (Rn (q)) to itself by
ast Est →
1≤s≤t≤n
π(ast )Est .
1≤s≤t≤n
It is not difficult to check that σπ is an automorphism of Rn (q), which is called a field automorphism of Rn (q). • Let ω be a permutation on Fq such that ω(0) = 0. When n = 2, define σω on any vertex [A] ∈ V (R2 (q)) with A = aE11 + bE12 + cE22 as follows: if a = 1, c = 0, [A] is sent to [E11 +ω(b)E12 ]; if c = 1, a = 0, [A] is sent to [−ω(−b)E12 + E22 ]; if a = c = 0, b = 1, [A] is fixed by σω . It is trivial to verify that σω is an automorphism of R2 (q). The main result of this section is as follows. Theorem 3.3. When n ≥ 3, σ is an automorphism of Rn (q) if and only if σ = σP · σπ , where σP is an inner automorphism of Rn (q) and σπ is a field automorphism of Rn (q). When n = 2, σ is an automorphism of Rn (q) if and only if σ = σU · σω , where ω is a permutation on Fq fixing 0, U is a 2 × 2 unit upper triangular matrix (all diagonal elements are 1) over Fq . Proof. The sufficiency is obvious. Now we consider the necessity for the case when n ≥ 3. Suppose σ is an automorphism of Rn (q). We complete the proof by announcing eight claims. Claim 1. Each Σ(s, t) is stabilized by σ for 1 ≤ s ≤ t ≤ n. For [A] ∈ Σ(s, t), suppose that σ([A]) ∈ Σ(k, l). Since σ is an automorphism, we have do (σ([A])) = do ([A]) and di (σ([A])) = di ([A]). Then, by Lemma 3.2, we have s = k and t = l. Hence, σ([A]) ∈ Σ(s, t). In other words, σ stabilizes Σ(s, t). Claim 2. There is a unit upper triangular matrix U such that σU · σ fixes each [Est ]. Firstly, we consider the action of σ on [E1t ] for 1 ≤ t ≤ n. By Claim 1, σ([E1t ]) ∈ n (t) (t) Σ(1, t). Clearly, σ([E1n ]) = [E1n ]. Assume that σ([E1t ]) = [ j=t a1j E1j ] with a1t = 1 for t = 1, 2, . . . , n − 1. Let Ut = I −
n
l=t+1
(t) a1l Etl
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for t = 1, 2, . . . , n − 1, and let U = U1 · U2 · · · Un−1 . Then σU · σ fixes each [E1t ] for t = 1, 2, . . . , n. Now we denote σU · σ by σ1 . Next, we prove σ1 fixes each [Esn ] for s = 1, 2, . . . , n. Clearly, σ1 fixes [E1n ]. For 2 ≤ s ≤ n, since σ1 ([Esn ]) ∈ Σ(s, n), we may assume that s (s)
σ1 [Esn ] = bkn Ekn k=1 (s)
with bsn = 1. If j < s, by [E1j ] → [Esn ], we have σ1 ([E1j ]) = [E1j ] → σ1 ([Esn ]), s (s) (s) which implies that E1j · ( k=1 bkn Ekn ) = 0. Thus, bjn = 0 for 1 ≤ j < s, which shows that σ1 ([Esn ]) = [Esn ] for s = 1, 2, . . . , n. Finally, we prove that σ1 fixes each [Est ] for 1 ≤ s ≤ t ≤ n. Suppose that σ1 ([Est ]) = [X] ∈ V (Rn (q)). By Claim 1, [X] ∈ Σ(s, t). For j = s, applying σ1 to [E1j ] → [Est ], we have [E1j ] → [X], or equivalently, E1j X = 0. It implies that the jth row of X is zero for j = s. Similarly, the kth column of X is zero for k = t. Hence, σ1 ([Est ]) = [Est ] for 1 ≤ s ≤ t ≤ n. Claim 3. Suppose σ1 ([A]) = [B] for A ∈ R(n, q). Then the kth row of [B] is zero if and only if so is [A], the lth column of [B] is zero if and only if so is [A]. In fact, the kth row of [A] is zero if and only if [E1k ] → [A], if and only if σ1 ([E1k ]) = [E1k ] → [B], if and only if E1k B = 0, if and only if the kth row of B is zero. The proof of the second assertion is similar. For 1 ≤ i < j ≤ n and a ∈ Fq , Claim 3 shows that σ1 sends [Eii + aEij ] to a vertex (i) of the form [Eii + bEij ] with b ∈ Fq . Thus we can define a permutation πij on Fq such (i)
that πij (0) = 0 and
(i) σ1 [Eii + aEij ] = Eii + πij (a)Eij . (j)
(j)
Similarly, we define a permutation πij on Fq such thatπij (0) = 0 and
(j) σ1 [Ejj + aEij ] = Ejj + πij (a)Eij . (j)
(i)
Claim 4. For 1 ≤ i < j ≤ n and a ∈ Fq , πij (−a) = −πij (a). (i)
By applying σ1 to [Eii + aEij ] → [Ejj − aEij ], we have [Eii + πij (a)Eij ] → [Ejj + (j)
(i)
(j)
πij (−a)Eij ], which implies that (Eii + πij (a)Eij ) · (Ejj + πij (−a)Eij ) = 0. Thus, (j)
(i)
πij (−a) = −πij (a). Claim 5 and Claim 6 are devoted to represent the action of σ1 on a vertex [A] by using these permutations. For a vertex [A] supported by (s, s), assume that A =
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s
249
n
akl Ekl with ass = 1, and assume that σ1 ([A]) = [A ]. By Claim 1, [A ] is also s n supported by (s, s). Thus, we may assume that A = k=1 l=s akl Ekl with ass = 1. Since A and A are both rank one matrices, k=1
l=s
akl = aks asl ,
akl = aks asl
for 1 ≤ k ≤ s − 1, s + 1 ≤ l ≤ n.
(1)
Claim 5. asl = πsl (asl ) for l = s + 1, . . . , n and aks = πks (aks ) for k = 1, 2, . . . , s − 1. s n (s) (s) (s) So σ1 ([A]) = [ k=1 l=s πks (aks )πsl (asl )Ekl ], where πss (1) = 1. (s)
(s)
By A · (Ell − asl Esl ) = 0 we have [A] → [Ell − asl Esl ]. By applying σ1 , we have (l) (l) [A ] → [Ell + πsl (−asl )Esl ], which implies that A · (Ell + πsl (−asl )Esl ) = 0, leading to asl = −πsl (−asl ). (l)
(2)
By using Claim 4, we have that asl = πsl (asl ). Similar discussion leads to the second result. The third assertion is obvious. s n For a vertex [B] supported by (s, t) with s < t, assume that B = k=1 l=t bkl Ekl with bst = 1 and assume that σ1 ([B]) = [B ], where B is written as B = s n k=1 l=t bkl Ekl with bst = 1. As B and B are rank one matrices, we have (s)
bkl = bkt bsl ,
bkl = bkt bsl
for 1 ≤ k ≤ s − 1, t + 1 ≤ l ≤ n.
(3)
Claim 6. bsl = (πst (1))−1 πsl (bsl ) and bkt = (πst (1))−1 πkt (bkt ), for l = t + 1, . . . , n and s n (t) (t) (s) (s) k = 1, 2, . . . , s − 1. So σ1 ([B]) = [ k=1 l=t ((πst (1))−1 πkt (bkt ) · (πst (1))−1 πsl (bsl )) · Ekl ]. (s)
(s)
(t)
(t)
The fact B · (Ell − bsl Etl ) = 0 shows that [B] → [Ell − bsl Etl ]. After applying σ1 to it, (l) (l) we have [B ] → [Ell + πtl (−bsl )Etl ], which implies that B · (Ell + πtl (−bsl )Etl ) = 0. It follows that bsl = −πtl (−bsl ). (l)
(4)
On the other hand, it follows from (Ess + Est + bsl Esl ) · (Ell − bsl Etl ) = 0 that [Ess + Est + bsl Esl ] → [Ell − bsl Etl ]. Applying σ1 , we have (s) (s) (l) Ess + πst (1)Est + πsl (bsl )Esl → Ell + πtl (−bsl )Etl , which implies that
(s) (s) (l) Ess + πst (1)Est + πsl (bsl )Esl · Ell + πtl (−bsl )Etl = 0. Consequently, (s)
(l)
(s)
πst (1)πtl (−bsl ) + πsl (bsl ) = 0.
(5)
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(4) and (5) together prove that (s) −1 (s) bsl = πst (1) πsl (bsl ).
(6)
(t) −1 (t) bkt = πst (1) πkt (bkt ).
(7)
Similar discussions show that
Thus the last assertion holds true. This completes the proof of Claim 6. The following claim describes the properties of these permutations and the relations between them. Claim 7. (1)
(1)
(k)
(i) π1j (ab) = π1k (a) · πkj (b) for a, b ∈ Fq and 2 ≤ k < j ≤ n. (ii) π1j (ab) = π1j (a) · π1j (b) · (π1j (1))−1 for a, b ∈ Fq and 2 ≤ j ≤ n. (1)
(1)
(1)
(1)
(iii) π1j (a−1 ) = (π1j (a))−1 · (π1j (1))2 for a ∈ Fq∗ and 2 ≤ j ≤ n. (1)
(1)
(1)
(iv) π1j (a) · (π1j (1))−1 = π12 (a) · (π12 (1))−1 for a ∈ Fq and 2 ≤ j ≤ n. (1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(v) π1j (a + b) = π1j (a) + π1j (b) for a, b ∈ Fq and 2 ≤ j ≤ n. (vi) π1j (−a) = −π1j (a) for a ∈ Fq and 2 ≤ j ≤ n. (k)
(l)
(vii) πkl = πkl for 1 ≤ k < l ≤ n. Suppose 2 ≤ k < j ≤ n. From (E11 + aE1k + abE1j ) · (Ejj − bEkj ) = 0, it follows that [E11 + aE1k + abE1j ] → [Ejj − bEkj ]. By applying σ1 , we have
(1) (1) (j) E11 + π1k (a)E1k + π1j (ab)E1j → Ejj + πkj (−b)Ekj ,
which implies that
(1) (1) (j) E11 + π1k (a)E1k + π1j (ab)E1j · Ejj + πkj (−b)Ekj = 0.
From which we have that (1)
(1)
(j)
(1)
(k)
π1j (ab) = −π1k (a) · πkj (−b) = π1k (a) · πkj (b).
(8)
This completes the proof of (i). From (8), we have (1)
(1)
(2)
π1j (ab) = π12 (a) · π2j (b)
for 3 ≤ j ≤ n,
(9)
and (1)
(1)
(2)
(1)
(2)
π1j (a) = π12 (a) · π2j (1) = π12 (1) · π2j (a)
for 3 ≤ j ≤ n.
(10)
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It follows that (2) −1 (1) (1) π12 (a) = π1j (a) · π2j (1)
(1) −1 (2) (1) and π2j (b) = π1j (b) · π12 (1)
(11)
for a, b ∈ Fq and 3 ≤ j ≤ n. Substituting (11) into (9), we have (1) −1 (1) (1) (1) π1j (ab) = π1j (a) · π1j (b) · π1j (1)
for 3 ≤ j ≤ n.
(12)
From π12 (a) = π1j (a) · (π2j (1))−1 , we have that (1)
(1)
(2)
(2) −1 (1) (1) π12 (ab) = π1j (ab) · π2j (1) (1) −1 (2) −1 (1) (1) = π1j (a) · π1j (b) · π1j (1) · π2j (1)
−1 (1) (1) (1) = π12 (a) · π1j (b) · π1j (1) (2) (1) −1 (1) (1) = π12 (a) · π12 (b) · π2j (1) · π1j (1) (1) −1 (1) (1) = π12 (a) · π12 (b) · π12 (1) . This equality together with (12) prove (1) −1 (1) (1) (1) π1j (ab) = π1j (a) · π1j (b) · π1j (1)
for 2 ≤ j ≤ n.
(13)
This completes the proof of (ii). (iii) follows immediately from (ii). For a ∈ Fq and 3 ≤ j ≤ n, by using (10), we have (1) −1 (1) −1 (1) −1 (1) (1) (2) (1) π1j (a) · π1j (1) = π12 (a) · π2j (1) · π1j (1) = π12 (a) · π12 (1) . Thus the result of (iv) follows. For j ≥ 3 and a, b ∈ Fq , from
E11 − bE12 + (a + b)E1j · (Ejj + E2j − aE1j ) = 0,
we have
E11 − bE12 + (a + b)E1j → [Ejj + E2j − aE1j ].
By applying σ1 , we arrive at (1) (1) (j) (j) E11 + π12 (−b)E12 + π1j (a + b)E1j → Ejj + π2j (1)E2j + π1j (−a)E1j , which leads to
(1) (1) (j) (j) E11 + π12 (−b)E12 + π1j (a + b)E1j · Ejj + π2j (1)E2j + π1j (−a)E1j = 0.
(14)
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Thus, (1)
(1)
(j)
(j)
π1j (a + b) + π12 (−b)π2j (1) + π1j (−a) = 0. (1)
(j)
(1)
(2)
(1)
(15) (j)
(1)
Since π12 (−b) · π2j (1) = −π12 (−b) · π2j (−1) = −π1j (b) and π1j (−a) = −π1j (a), (15) implies that (1)
(1)
(1)
π1j (a + b) = π1j (a) + π1j (b)
for a, b ∈ Fq , 3 ≤ j ≤ n.
(16)
Using (14) and (16), we also have (1)
(1)
(1)
for a, b ∈ Fq .
(17)
for a, b ∈ Fq , 2 ≤ j ≤ n.
(18)
π12 (a + b) = π12 (a) + π12 (b) Hence (1)
(1)
(1)
π1j (a + b) = π1j (a) + π1j (b)
This proves (v). (vi) follows immediately from (v). For any a ∈ Fq and 2 ≤ k < l ≤ n, using the results of (i), (vi) and applying Claim 4, we have (1)
(1)
(1)
(k)
(1)
(l)
−π1l (a) = π1l (−a) = π1k (1) · πkl (−a) = −π1k (1) · πkl (a). It follows that (1) −1 (1) (l) (k) πkl (a) = π1k (1) · π1l (a) = πkl (a). (k)
(l)
Consequently, πkl = πkl for 2 ≤ k < l ≤ n. For the special case that k = 1, from (1) (1) (l) (1) (l) π1l (a) = −π1l (−a) = π1l (a), we have π1l = π1l . This completes the proof of (vii). The proof of Claim 7 is completed. (k) (l) By (vii) of Claim 7, distinguishing πkl with πkl seems unnecessary, so we will denote both them by πkl . Now, let π = (π12 (1))−1 π12 . Claim 8. (i) π is an automorphism of the field Fq . (ii) Let π11 (1) = 1. Then πkl = (π1k (1))−1 π1l (1)π = πkl (1)π for 1 ≤ k < l ≤ n. For a, b ∈ Fq , since
−1
−1
−1 π(a + b) = π12 (1) π12 (a + b) = π12 (1) π12 (a) + π12 (1) π12 (b) = π(a) + π(b),
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and
−1
−1
−1
π(ab) = π12 (1) π12 (ab) = π12 (1) π12 (a)π12 (b) π12 (1) = π(a)π(b), π is an automorphism of the field Fq . Thus (i) of Claim 8 holds. (iv) of Claim 7 shows that π1l (a) = π1l (1)π(a) for any a and 2 ≤ l. For 2 ≤ k < l ≤ n, using (i) and (iv) of Claim 7, we have πkl (a) = (π1k (1))−1 π1l (a) = (π1k (1))−1 π1l (1)π(a), from which it follows that πkl = (π1k (1))−1 π1l (1)π. Since (π1k (1))−1 π1l (1) = πkl (1), we have πkl = πkl (1)π. Unifying the results for k = 1 and k ≥ 2, we obtain πkl = (π1k (1))−1 π1l (1)π = πkl (1)π for 1 ≤ k < l ≤ n, where π11 (1) = 1. The proof of Claim 8 is completed. Let D = diag(d1 , d2 , . . . , dn ), where dk = (π1k (1))−1 for 1 ≤ k ≤ n, and denote σD · σ1 by σ2 . Claim 9. σ2 identifies with σπ . We consider the action of σ2 on each vertex. s n For A = k=1 l=s akl Ekl supported by (s, s), where ass = 1 and akl = aks asl for 1 ≤ k ≤ s − 1 and s + 1 ≤ l ≤ n, Claim 5 shows that s n
σ1 [A] = πks (aks )πsl (asl )Ekl ,
k=1 l=s
where πss (1) = 1. Using Claim 8 we have s n
−1 σ1 [A] = π1k (1) π1l (1)π(aks )π(asl )Ekl .
k=1 l=s
Hence,
σ2 [A] = σD =
s n
k=1 l=s n s
−1 π1k (1) π1l (1)π(aks )π(asl )Ekl
π(aks )π(asl )Ekl .
k=1 l=s
This equality shows that the action of σ2 on each vertex [A] ∈ Σ(s, s) identifies with the action of σπ on [A].
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s n For B = k=1 l=t bkl Ekl supported by (s, t), where s < t, bst = 1, and bkl = bkt bsl for 1 ≤ k ≤ s − 1 and t + 1 ≤ l ≤ n, Claim 6 shows that s n
−1
−1
σ1 [B] = πst (1) πkt (bkt ) · πst (1) πsl (bsl ) · Ekl .
k=1 l=t
Using Claim 8 we have s n
−1
−1 σ1 [B] = πst (1) π1k (1) π1l (1)π(bkt )π(bsl ) · Ekl .
k=1 l=t
Hence, s n s n
−1 σ2 [B] = πst (1) π(bkt )π(bsl ) · Ekl = π(bkt )π(bsl ) · Ekl .
k=1 l=t
k=1 l=t
The last equality comes from the fact that [aA] = [A] for a ∈ Fq∗ and A ∈ R(n, q). Hence, the action of σ2 on each vertex [B] ∈ Σ(s, t), for s < t, identifies with the action of σπ on this vertex. Thus, σD · σU · σ = σπ . Finally, we have σ = σP · σπ , where P = U −1 D−1 . Now we turn to the case when n = 2. As in the case when n ≥ 3, we can find a 2 × 2 unit upper triangular matrix U such that σU · σ respectively fixes [E11 ], [E12 ] and [E22 ]. Denote σU · σ by σ1 . Suppose σ1 ([A]) = [B] for A ∈ V (R2 (q)). Then, as in the case when n ≥ 3, the kth row of [B] is zero if and only if so is [A], the kth column of [B] is zero if and only if so is [A], where k = 1 or 2. Now, we can define a permutation ω on Fq , with ω(0) = 0, such that
σ1 [E11 + aE12 ] = E11 + ω(a)E12
for a ∈ Fq .
Now for b ∈ Fq , suppose that σ1 sends [bE12 + E22 ] to [f (b)E12 + E22 ]. By (E11 − bE12 )(bE12 + E22 ) = 0, we have [E11 − bE12 ] → [bE12 + E22 ], from which it follows that E11 + ω(−b)E12 → f (b)E12 + E22 ,
E11 + ω(−b)E12 · f (b)E12 + E22 = 0. The last equality implies that f (b) + ω(−b) = 0. Thus f (b) = −ω(−b), and
σ1 [bE12 + E22 ] = −ω(−b)E12 + E22 .
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Observing the action of σ1 on each vertex, we find that σ1 identifies σω . Finally, we have σ = σU1 · σω with U1 = U −1 , as desired. 2 Finally, we consider the orbit partition of the vertex set and the order of the group of automorphisms. Corollary 3.4. The orbit partition of V (Rn (q)) under the automorphisms is V (Rn (q)) = n(n+1) . 1≤s≤t≤n Σ(s, t). The number of orbits is 2 Proof. Claim 1 of Theorem 3.3 (also holds for n = 2) shows that each Σ(s, t) is stabilized under Aut(Rn (q)). To complete the proof we only need to prove that, for any [A] ∈ Σ(s, t), there is an automorphism σ such that σ([Est ]) = [A]. Assume that s n A = a = 1 and akl = akt asl for 1 ≤ k ≤ s − 1 and k=1 l=t akl Ekl , where n st s−1 t + 1 ≤ l ≤ n. Let P = (I + l=t+1 asl Etl )(I − k=1 akt Eks ). Then A = P −1 Est P . So, σP ([Est ]) = [A]. The second result is obvious. 2 Corollary 3.5. Let q = pm . Then
n(n−1) Aut Rn (q) = m(q − 1)n−1 q 2 , when n ≥ 3; q
Aut Rn (q) = k , when n = 2. k=1
Proof. When n ≥ 3, by Theorem 3.3, each automorphism σ of Rn (q) can be written as σ = σP σπ for some P ∈ Tn (q), π ∈ Aut(Fq ). If σP1 σπ1 = σP2 σπ2 , then, σP0 = σπ0 , where P0 = P2−1 P1 and π0 = π2 π1−1 . Since each [Ess ] is fixed by σπ0 , we have σP0 ([Ess ]) = [Ess ] for s = 1, 2, . . . , n. That is [P0−1 Ess P0 ] = [Ess ] for s = 1, 2, . . . , n. Consequently, P0 must be a diagonal matrix. By σP0 ([Est ]) = [P0−1 Est P0 ] = σπ0 ([Est ]) = [Est ] for s < t, we have P0 must be a nonzero scalar matrix. Thus P1 = aP2 for certain a ∈ Rq∗ . Since σπ0 = σP0 , σπ0 must be the identity automorphism. It follows from [E11 + aE12 ] = σπ0 ([E11 + aE12 ]) = [E11 + π0 (a)E12 ] that π0 (a) = a for any a ∈ Fq . So π0 is the identity automorphism of Fq . Hence, π1 = π2 . Above discussions show that | Aut(Rn (q))| = |Tn (q)| q−1 · | Aut(Fq )|. n(n−1)
It is well known that |Tn (q)| = (q − 1)n q 2 and | Aut(Fq )| = m. Thus the first result follows. For the second result, since the number of permutations ω on Fq fixing q−1 0 is Πk=1 k, and the number of 2 × 2 unit upper triangular matrices is q, we have q | Aut(R2 (q))| = ( k=1 k). 2 By the proof of Corollary 3.5, the following corollary is obvious. T (q) Corollary 3.6. When n ≥ 3, Aut(Rn (q)) ∼ = nK Aut(Fq ); when n = 2, Aut(R2 (q)) ∼ = U2 (q) Sq−1 , where K = {aI|a ∈ Fq∗ }, U2 (q) denotes the set of 2 ×2 unit upper triangular matrices over Fq , and Sq−1 means the symmetric group on q − 1 elements.
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4. Automorphisms of Γ (R(n, q)) Applying Theorem 3.3, we can describe the automorphisms of Γ (R(n, q)) immediately. Firstly, we construct a standard automorphism for Γ (R(n, q)). Note that R(n, q) is the disjoint union of all [A] with A being rank one matrices. • Let τ be a permutation on R(n, q) which stabilizes each [A], or more definitely, τ is a bijection on R(n, q) such that τ (X) = aX X for any X ∈ R(n, q), where aX ∈ Fq∗ depends on X. It is easy to see that τ is an automorphism of Γ (R(n, q)), which is called a local scalar multiplication on Γ (R(n, q)). Now we show how to reduce an automorphism of Γ (R(n, q)) to that of Rn (q). Lemma 4.1. Let A, B ∈ R(n, q). Then Ni (A) = Ni (B) and No (A) = No (B) (in Γ (R(n, q))) if and only if B is a nonzero multiple of A. Proof. The sufficiency is obvious. For the necessity, suppose A is supported by (s, t). The proof of Lemma 3.2 shows that there exist P ∈ Tn (q) such that P −1 AP = aEst , with a ∈ Fq∗ . From Est · i=t Eii = 0, we have AP ( i=t Eii ) = 0. Further, by No (A) = No (B) we have BP ( i=t Eii ) = 0. It implies that all columns of BP are zero except for the tth column. The same assertion also holds for P −1 BP , since multiplying P −1 to BP from the left side is equivalent to applying elementary row transformation to BP . Similarly, all row vectors of P −1 BP are zero except for the sth row. Consequently, P −1 BP = bEst with b ∈ Fq∗ . Finally, we have B = bP Est P −1 = ba−1 A, as desired. 2 Let χ be an automorphism of Γ (R(n, q)). We define χ on Rn (q) by χ([A]) = [χ(A)]. Lemma 4.2. Let χ be an automorphism of Γ (R(n, q)). Then χ is an automorphism of Rn (q). Proof. If [B] = [A] ∈ V (Rn (q)), then B is a nonzero multiple of A. By Lemma 4.1, χ(B) is also a nonzero multiple of χ(A). Thus χ([A]) = χ([B]). It shows that χ is well defined. Clearly, χ is a bijection and χ([X]) → χ([Y ]) if and only if [X] → [Y ]. Hence, χ is an automorphism of Rn (q). 2 Next, we can use the characterization of automorphisms of Γ (R(n, q)) to determine the automorphisms of Rn (q). Theorem 4.3. (i) When n ≥ 3, σ is an automorphism of Γ (R(n, q)) if and only if σ = σP · σπ · τ , where σP : X → P −1 XP, ∀X ∈ R(n, q); σπ : [xij ] → [π(xij )], ∀[xij ] ∈ R(n, q); τ is a local scalar multiplication. (ii) When n = 2, σ is an automorphism of Rn (q) if and only if σ = σU · σω · τ , where σU : X → U −1 XU, ∀X ∈ R(2, q) with U a 2 × 2 unit upper triangular matrix (all
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diagonal elements are 1); ω is a permutation on Fq fixing 0 and σω : a0 cb → a0 dc with d = ω(a−1 b) if c = 0, a = 0, or d = −ω(−c−1 b) if a = 0, c = 0, or d = b if a = c = 0, b = 0; and τ is a local scalar multiplication. Proof. Let n ≥ 3 and let σ be an automorphism of Γ (R(n, q)). Then by Lemma 4.2, σ : [A] → [σ(A)] is an automorphism of Rn (q). Applying Theorem 3.3, we know there −1 is an invertible matrix P and an automorphism π of Fq such that σ −1 π · σ P · σ acts as the identity automorphism on Rn (q), or equivalently, σπ−1 · σP−1 · σ sends each rank one upper triangular matrix A to a scalar multiple of A. Thus σπ−1 · σP−1 · σ exactly is a local scalar multiplication of Rn (q), which completes the proof of the first part. The proof for n = 2 is similar, thus omitted. 2 Above result for n ≥ 3 can be stated in a more explicit way as follows. Theorem 4.4. A bijective map σ on R(n, q) is an automorphism of Γ (R(n, q)) if and only if σ(X) = aX P −1 π(xij ) P,
∀X = [xij ] ∈ R(n, q),
where P is an invertible upper triangular matrix; aX ∈ Fq∗ depends on X; π is an automorphism of the field Fq , [π(xij )] denotes the matrix whose (i, j)-entry is π(xij ). In this paper, the automorphisms of an induced subgraph of the zero-divisor graph of upper triangular matrices are determined, the general case that the zero-divisor graph is over all upper triangular matrices is left unsolved. Acknowledgement We thank the referee for his suggestion of developing relation between the graph of our paper and zero-divisor graphs of matrix ring, which have been received a lot of attention. References [1] S. Akbari, A. Mohammadian, On zero-divisor graphs of finite rings, J. Algebra 314 (2007) 168–184. [2] D.D. Anderson, M. Naseer, Beck’s coloring of a commutative ring, J. Algebra 159 (1993) 500–514. [3] D.F. Anderson, P.S. Livingston, The zero-divisor graph of a commutative ring, J. Algebra 217 (1999) 434–447. [4] D.F. Anderson, S.B. Mulay, On the diameter and girth of a zero-divisor graph, J. Pure Appl. Algebra 210 (2007) 543–550. [5] D.F. Anderson, R. Levy, J. Shapiro, Zero-divisor graphs, von Neumann regular rings and Boolean algebras, J. Pure Appl. Algebra 180 (2003) 221–241. [6] M. Axtell, J. Coykendall, J. Stickles, Zero-divisor graphs of polynomial and power series over commutative rings, Comm. Algebra 33 (2005) 2043–2050. [7] I. Beck, Coloring of commutative rings, J. Algebra 116 (1988) 208–226. [8] I. Bo˘ zić, Z. Petrović, Zero-Divisor Graphs of Matrices Over Commutative Rings, Comm. Algebra 37 (2009) 1186–1192.
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