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A note on automorphisms of the zero-divisor graph of upper triangular matrices
Linear Algebra and its Applications 465 (2015) 214–220
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
A note on automorphisms of the zero-divisor graph of upper triangular matrices ✩ Long Wang School of Mathematical Sciences, Anhui University, Hefei 230039, PR China
a r t i c l e
i n f o
Article history: Received 1 September 2014 Accepted 24 September 2014 Available online xxxx Submitted by R. Brualdi MSC: 16S50 13A99 05C99 Keywords: Zero-divisor graphs Directed graphs Graph automorphisms
a b s t r a c t Let Fq be a finite field with q elements, n(≥ 3) a positive integer, T (n, q) the set of all n × n upper triangular matrices over Fq . In [13], the zero-divisor graph of T (n, q), written as T , is defined to be a graph with all nonzero zero-divisors in T (n, q) as vertices, and there is a directed edge from a vertex X to a vertex Y if and only if XY = 0. The subgraph of T induced by all rank one matrices in T (n, q) is denoted by R. Wong et al. (2014) in [13] determined the automorphisms of R and left the automorphisms of T unsolved. In this note, we solve this problem. © 2014 Elsevier Inc. All rights reserved.
1. Introduction Let R be a ring with identity 1. The zero-divisor graph of R, denoted by Γ (R), is a directed graph with vertices Z ∗ (R), the set of nonzero zero-divisors of R, and for distinct elements x, y ∈ Z ∗ (R), there is a directed edge from x to y, written as x → y, if and only if xy = 0. The concept of zero-divisor graph was first defined and studied for ✩
Supported by National Natural Science Foundation of China (11371028; 11471016; 11401003). E-mail address: [email protected].
http://dx.doi.org/10.1016/j.laa.2014.09.035 0024-3795/© 2014 Elsevier Inc. All rights reserved.
L. Wang / Linear Algebra and its Applications 465 (2015) 214–220
215
commutative rings by Beck in [7], and further studied by many authors (see, e.g., Akbari and Mohammadian [1], Anderson et al. [2–5], Axtell et al. [6], Bo˘zić and Petrović [8], Levy and Shapiro [10], Lucas [11], Wu [14]), even for noncommutative rings. A lot of results about the diameter, the girth of Γ (R) and so on have been obtained. For a directed graph G, a bijection σ on V (G) is called an automorphism of G if σ(x) → σ(y) if and only if x → y. All automorphisms of G, denoted by Aut(G), forms a group under composition of transformations. 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 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 [9] that Aut(Γ (R)) is isomorphic to the symmetric group of degree p + 1, when R = Mat2 (Zp ) (p is a prime). Park and Han [12] proved that Aut(Γ (R)) ∼ = Sq+1 for R = Mat2 (Fq ) with Fq an arbitrary finite field. Let T (n, q) be the set of all n × n upper triangular matrices over Fq . More recently, Wong et al. in [13] determined the automorphisms of a subgraph of T induced by all rank one matrices in T (n, q), and left the automorphisms of T open, where T was defined to be a directed graph with vertex set all nonzero zero-divisors in T (n, q), and there is a directed edge from a vertex X to a vertex Y if and only if XY = 0. In this paper, we are devoted to solve this open problem. Although the authors of [13] also called T the zero-divisor graph of T (n, q), one should note that T differs slightly from Γ (T (n, q)), since X → X is possible in T but it is not allowed in Γ (T (n, q)). Indeed, Γ (T (n, q)) is the spanned subgraph of T obtained from T by deleting all loops. When n = 2, A ∈ T (n, q) is a nonzero zero-divisor of T (n, q) if and only if A is of rank one, thus T identifies with the subgraph induced by rank one matrices, the automorphisms of which have been determined in [13]. So in this paper, we only consider the case when n ≥ 3. Hereafter, n ≥ 3 is always assumed. Now, we construct three types of standard automorphisms for T . Based on these known automorphisms any automorphism of T can be characterized. • For a directed graph G and x, y ∈ V (G), we write x → y to denote there is a − → ← − directed edge from x to y, and we denote by N G (x) (resp., N G (x)) the set of vertices y ∈ V (G) for which x → y (resp., y → x). We write NG (x) = NG (y) to denote that − → − → ← − ← − N G (x) = N G (y) and N G (x) = N G (y) for x, y ∈ V (G). If NG (x) = NG (y) then x, y are called twin points. If a bijection ρ on V (G) permutes twin points, then it is an automorphism of G, which is called a regular automorphism of G. The set of all regular automorphisms of G is denoted by Reg(G). • Let P ∈ T (n, q) be invertible. We define φP from V (T ) to itself by X → P −1 XP . Then it is easy to check that φP is an automorphism of T , which is called an inner automorphism of T .
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• Let π be an automorphism of the field Fq . For a matrix A = (aij )n×n ∈ T (n, q), we denote (π(aij ))n×n by π(A). Define φπ from V (T ) to itself by A → π(A). It is not difficult to check that φπ is an automorphism of T , which is called a field automorphism of T . The main result of this article is as follows. Theorem 1.1. A mapping φ on V (T ) is an automorphism of T if and only if it can be uniquely decomposed into the product of an inner automorphism, a field automorphism and a regular automorphism. 2. Proof of Theorem 1.1 Let R(n, q) be the set of all rank one matrices in T (n, q). For A ∈ R(n, q), denote by [A] the subspace of T (n, q) spanned by A. In [13], R was defined to be a subgraph of T induced by R(n, q), and Rn (q) was defined to be a graph with vertex set {[A]: A ∈ R(n, q)} and there is a directed edge from [A] to [B] if and only if AB = 0. For proof of the main result of this paper, we need to apply two results obtained in [13]. Lemma 2.1. (See [13, Lemma 4.1].) Let A, B ∈ R(n, q). Then NR (A) = NR (B) if and only if B is a nonzero multiple of A. Lemma 2.2. (See [13, Theorem 3.3].) σ is an automorphism of Rn (q) if and only if σ can be decomposed into σ = σP ◦ σπ , where σP : [A] → [P −1 AP ], ∀A ∈ R(n, q) and σπ : A → [π(A)], ∀A ∈ R(n, q). Lemma 2.3. Let G be an arbitrary graph. Then Reg(G), consisting of regular automorphisms of G, is a normal subgroup of Aut(G), and it is isomorphic to the direct product of some symmetric groups. Proof. By the definition of Reg(G), it is easy to see that Reg(G) forms a subgroup of Aut(G), and it is isomorphic to the direct product of some symmetric groups. Let σ ∈ Aut(G) and ρ ∈ Reg(G), we need to show σ −1 ρσ ∈ Reg(G). It suffices to show that NG σ −1 ρσ (x) = NG (x),
∀x ∈ V (G).
Suppose that x → y, then σ(x) → σ(y), further (ρσ)(x) → σ(y) and (σ −1 ρσ)(x) → y, − → − → from which it follows that N G (x) ⊆ N G ((σ −1 ρσ)(x)). As σ −1 ρσ is an automorphism − → − → − → − → of G, |N G ((σ −1 ρσ)(x))| = |N G (x)|, from which N G ((σ −1 ρσ)(x)) = N G (x) follows. ← − ← − Similarly, N G ((σ −1 ρσ)(x)) = N G (x). Hence, NG ((σ −1 ρσ)(x)) = NG (x) for any x ∈ V (G). 2
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Lemma 2.4. Let A ∈ V (T ) be a nonzero zero-divisor of T (n, q). Then r(A) ≥ 2 if and − → − → only if there is certain B ∈ V (T ) such that N T (A) N T (B), where r(A) refers to the rank of A. Proof. “Necessity”: As r(A) ≥ 2, A has a nonzero row, say α. Let B be the n × n upper triangular matrix which has α as its first row and all other rows of B are zero. Then − → − → − → from AC = 0 it follows that BC = 0 for C ∈ N T (A). Consequently, N T (A) ⊆ N T (B). Denote the solution spaces of the linear equations AX = 0 and BX = 0 respectively by SA and SB . Obviously, SA ⊆ SB . Since r(B) < r(A), the dimension of SB is larger than that of SA , from which it follows that SA SB . Let β ∈ SB \ SA , and let D be the n × n upper triangular matrix which has β as its last column and all other columns of D are − → − → zero. Then AD = 0 and BD = 0, from which it follows that N T (A) N T (B). “Sufficiency”: SA and SB are as above. To achieve the goal, we need first to show that SA SB . Let 0 = β ∈ SA , and let D be the n × n upper matrix which has β as its last column and all other columns of D are zero. From AD = 0 it follows that − → − → D ∈ N T (A). Further, D ∈ N T (B), i.e., BD = 0. Consequently, Bβ = 0, i.e., β ∈ SB . − → − → Hence, SA ⊆ SB . Suppose C ∈ N T (B) \ N T (A). Thus, AC = 0 and BC = 0, which implies that C has a column, say ξ, such that Aξ = 0 and Bξ = 0. Hence, SA SB , as required. Consequently, r(A) > r(B) ≥ 1. 2 Lemma 2.5. Let φ be an automorphism of T . Then φ stabilizes the set of rank one matrices in T (n, q), i.e., φ(R(n, q)) = R(n, q). Proof. Suppose for a contradiction that φ(A) ∈ / R(n, q) for A ∈ R(n, q). Then − → − → r(φ(A)) ≥ 2. By Lemma 2.4, there is certain B ∈ V (T ) such that N T (φ(A)) N T (B). − → − → Let C be the pre-image of B under φ, i.e., φ(C) = B. Now, from N T (φ(A)) N T (φ(C)) − → − → it follows that N T (A) N T (C), which further implies that r(A) ≥ 2, a contradiction. 2 Lemma 2.6. Let A, B ∈ V (T ) be rank one matrices, and let φ be an automorphism of T . Then φ(B) is a nonzero multiple of φ(A) if and only if B is a nonzero multiple of A. Proof. By Lemma 2.5, φ(B) and φ(A) both are rank one matrices. Suppose B = aA for certain a ∈ Fq∗ . Then B and A have the same neighbors in T , i.e., NT (B) = NT (A). Further, NT (φ(B)) = NT (φ(A)). As vertices of R, φ(B) and φ(A) also have the same neighbors in R. By Lemma 2.1, φ(B) is a nonzero multiple of φ(A). By considering the action of φ−1 on φ(B) and φ(A), we can prove that if φ(B) is a nonzero multiple of φ(A), then B is a nonzero multiple of A. 2 The main idea of this paper is to reduce the study of automorphisms of T to that of Rn (q). So it is necessary to study the connection between Aut(T ) and Aut(Rn (q)). Theorem 2.7.
Aut(T ) Reg(T )
is isomorphic to Aut(Rn (q)).
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Proof. Lemma 2.3 has shown that Reg(T ) is a normal subgroup of Aut(T ). For an automorphism φ of T , we define φ from V (Rn (q)) to itself by φ ([A]) = [φ(A)] for any A ∈ R(n, q). Lemma 2.5 tells us φ(A) ∈ R(n, q) for A ∈ R(n, q); Lemma 2.6 proves that if [B] = [A] for B, A ∈ R(n, q) then [φ(B)] = [φ(A)], namely, φ ([B]) = φ ([A]), from which it follows that φ is well defined. If φ ([B]) = φ ([A]), i.e., [φ(B)] = [φ(A)], then φ(B) is a nonzero multiple of φ(A). By Lemma 2.6, B is a nonzero multiple of A, i.e., [B] = [A], which proves that φ is injective. For any [B] ∈ V (Rn (q)) suppose that φ(A) = B, then φ ([A]) = [B], which proves that φ is surjective. If [A] → [B] in Rn (q), i.e., AB = 0, then φ(A)φ(B) = 0, from which it follows that φ ([A]) → φ ([B]). Conversely, φ ([A]) → φ ([B]) also implies that [A] → [B]. Consequently, φ is an automorphism of Rn (q). We define a mapping f from Aut(T ) to Aut(Rn (q)) by f (φ) = φ for any φ ∈ Aut(T ). Let φ1 , φ2 ∈ Aut(T ). From (φ1 φ2 ) ([A]) = [(φ1 φ2 )(A)] = φ1 (φ2 ([A])) it follows that (φ1 φ2 ) = φ1 φ2 , which implies that f is a homomorphism from Aut(T ) to Aut(Rn (q)). To complete the proof we need to prove that the kernal of f is Reg(Γ (T )) and the image of f is Aut(Rn (q)). Let φ lie in the kernal of f , i.e., φ acts as the identity on V (Rn (q)), or equivalently, φ sends any A ∈ R(n, q) to a nonzero multiple of A. We need first to show that φ ∈ Reg(T ), or equivalently, NT (φ(A)) = NT (A) for any nonzero zero-divisor A in T (n, q). − → Let B ∈ N T (A), then AB = 0. Suppose the i-th column vector of B is βi , and let Bi be the triangular matrix which has βi as its i-th column and all other columns are zero. Then Bi is a rank one matrix or a zero matrix and ABi = 0. If Bi is of rank one, then we have φ(A)φ(Bi ) = 0. Since φ(Bi ) is a nonzero multiple of Bi , we further have φ(A)Bi = 0 and φ(A)βi = 0. If Bi = 0, φ(A)βi = 0 also holds. From φ(A)βi = 0 − → for i = 1, 2, . . . , n, it follows that φ(A)B = 0. Consequently, B ∈ N T (φ(A)). Hence, − → − → N T (A) ⊆ N T (φ(A)). Similarly, by considering the action of φ−1 on φ(A), we have − → − → − → − → − → N T (φ(A)) ⊆ N T (φ−1 (φ(A))) = N T (A). Therefore, N T (φ(A)) = N T (A). An analogous ← − ← − discussion leads to N T (φ(A)) = N T (A). Consequently, NT (φ(A)) = NT (A) for A ∈ V (T ). Now, it is proved that Ker(f ) ⊆ Reg(T ). Conversely, we need to prove Reg(T ) ⊆ Ker(f ). Suppose that φ ∈ Reg(T ). Then NT (φ(A)) = NT (A) for any A ∈ V (T ). Further, for any rank one matrix A0 , by NR (φ(A0 )) = NR (A0 ), we have φ(A0 ) is a nonzero multiple of A0 (using Lemma 2.1). Thus, φ acts as the identity on Rn (q), i.e., φ ∈ Ker(f ). Hence, Reg(T ) ⊆ Ker(f ). The proof for Ker(f ) = Reg(T ) is completed. Now we are in a position to show the image of f is Aut(Rn (q)). Let σ be an automorphism of Rn (q). By Lemma 2.2, there exists an invertible P in T (n, q) and an automorphism π of the field Fq such that σ([A]) = [P −1 π(A)P ] for any A ∈ R(n, q). Let φP : A → P −1 AP , ∀A ∈ V (T ), the inner automorphism of T induced by P ,
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and let φπ : A → π(A), ∀A ∈ V (T ), a field automorphism of T . Then we find that f (φP φπ ) = φP φπ = σ. Consequently, f is surjective. Aut(T ) ∼ Finally, we obtain Reg(T ) = Aut(Rn (q)) by applying the homomorphism theorem. 2 Proof of Theorem 1.1. The sufficient condition is obvious. Let φ be an automorphism of T . Then φ : [A] → φ(A) ,
∀A ∈ R(n, q)
is an automorphism of Rn (q). By Lemma 2.2, there is an invertible matrix P ∈ T (n, q) and an automorphism π of the field Fq such that φ [A] = P −1 π(A)P ,
∀A ∈ R(n, q).
Clearly, (φπ−1 ◦ φP −1 ◦ φ) acts as the identity on Rn (q). So φπ−1 ◦ φP −1 ◦ φ lies in the kernal of f , from which it follows that φπ−1 ◦ φP −1 ◦ φ acts as a regular automorphism, say ρ, on V (T ). Hence, φ = φP ◦ φπ ◦ ρ. Now we prove the uniqueness of the decomposition. Suppose φ = φP1 ◦ φπ1 ◦ ρ1 = φP2 ◦ φπ2 ◦ ρ2 be two decompositions of φ, where φPi , φπi , ρi are respectively inner automorphisms, field automorphisms and regular automorphisms of T . Recalling the definition of σP and σπ in Lemma 2.2, we have (φP ) = σP and (φπ ) = σπ . Applying f on two sides of −1 −1 φ−1 π2 ◦ φP2 ◦ φP1 ◦ φπ1 = ρ2 ◦ ρ1 ,
we have σπ−1 ◦ σP−1 ◦ σP1 ◦ σπ1 = I, 2 2 where I denotes the identity automorphism on Rn (q), from which it follows that σP−1 ◦ σP1 = σπ2 ◦ σπ−1 . 1 2 From the proof of Corollary 3.5 in [13], we have P2 is just a nonzero multiple of P1 and π2 = π1 . Consequently, φP2 = φP1 and φπ2 = φπ1 . Furthermore, ρ2 = ρ1 . 2 By Theorem 2.7 and Lemma 2.3, the following corollary is obvious. Corollary 2.8. Aut(T ) ∼ = Reg(T ) Aut(Rn (q)).
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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. zić, Z. Petrović, Zero-divisor graphs of matrices over commutative rings, Comm. Algebra 37 [8] I. Bo˘ (2009) 1186–1192. [9] J. Han, The zero-divisor graph under group actions in a noncommutative ring, J. Korean Math. Soc. 45 (2008) 1647–1659. [10] R. Levy, T. Shapiro, The zero-divisor graph of von Neumann regular rings, Comm. Algebra 30 (2002) 745–750. [11] T.G. Lucas, The diameter of a zero divisor graph, J. Algebra 301 (2006) 174–193. [12] S. Park, J. Han, The group of graph automorphisms over a matrix rings, J. Korean Math. Soc. 48 (2011) 301–309. [13] D. Wong, X. Ma, J. Zhou, The group of automorphisms of a zero-divisor graph based on rank one upper triangular matrices, Linear Algebra Appl. 460 (2014) 242–258. [14] 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
A note on automorphisms of the zero-divisor graph of upper triangular matrices ✩ Long Wang School of Mathematical Sciences, Anhui University, Hefei 230039, PR China
a r t i c l e
i n f o
Article history: Received 1 September 2014 Accepted 24 September 2014 Available online xxxx Submitted by R. Brualdi MSC: 16S50 13A99 05C99 Keywords: Zero-divisor graphs Directed graphs Graph automorphisms
a b s t r a c t Let Fq be a finite field with q elements, n(≥ 3) a positive integer, T (n, q) the set of all n × n upper triangular matrices over Fq . In [13], the zero-divisor graph of T (n, q), written as T , is defined to be a graph with all nonzero zero-divisors in T (n, q) as vertices, and there is a directed edge from a vertex X to a vertex Y if and only if XY = 0. The subgraph of T induced by all rank one matrices in T (n, q) is denoted by R. Wong et al. (2014) in [13] determined the automorphisms of R and left the automorphisms of T unsolved. In this note, we solve this problem. © 2014 Elsevier Inc. All rights reserved.
1. Introduction Let R be a ring with identity 1. The zero-divisor graph of R, denoted by Γ (R), is a directed graph with vertices Z ∗ (R), the set of nonzero zero-divisors of R, and for distinct elements x, y ∈ Z ∗ (R), there is a directed edge from x to y, written as x → y, if and only if xy = 0. The concept of zero-divisor graph was first defined and studied for ✩
Supported by National Natural Science Foundation of China (11371028; 11471016; 11401003). E-mail address: [email protected].
http://dx.doi.org/10.1016/j.laa.2014.09.035 0024-3795/© 2014 Elsevier Inc. All rights reserved.
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215
commutative rings by Beck in [7], and further studied by many authors (see, e.g., Akbari and Mohammadian [1], Anderson et al. [2–5], Axtell et al. [6], Bo˘zić and Petrović [8], Levy and Shapiro [10], Lucas [11], Wu [14]), even for noncommutative rings. A lot of results about the diameter, the girth of Γ (R) and so on have been obtained. For a directed graph G, a bijection σ on V (G) is called an automorphism of G if σ(x) → σ(y) if and only if x → y. All automorphisms of G, denoted by Aut(G), forms a group under composition of transformations. 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 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 [9] that Aut(Γ (R)) is isomorphic to the symmetric group of degree p + 1, when R = Mat2 (Zp ) (p is a prime). Park and Han [12] proved that Aut(Γ (R)) ∼ = Sq+1 for R = Mat2 (Fq ) with Fq an arbitrary finite field. Let T (n, q) be the set of all n × n upper triangular matrices over Fq . More recently, Wong et al. in [13] determined the automorphisms of a subgraph of T induced by all rank one matrices in T (n, q), and left the automorphisms of T open, where T was defined to be a directed graph with vertex set all nonzero zero-divisors in T (n, q), and there is a directed edge from a vertex X to a vertex Y if and only if XY = 0. In this paper, we are devoted to solve this open problem. Although the authors of [13] also called T the zero-divisor graph of T (n, q), one should note that T differs slightly from Γ (T (n, q)), since X → X is possible in T but it is not allowed in Γ (T (n, q)). Indeed, Γ (T (n, q)) is the spanned subgraph of T obtained from T by deleting all loops. When n = 2, A ∈ T (n, q) is a nonzero zero-divisor of T (n, q) if and only if A is of rank one, thus T identifies with the subgraph induced by rank one matrices, the automorphisms of which have been determined in [13]. So in this paper, we only consider the case when n ≥ 3. Hereafter, n ≥ 3 is always assumed. Now, we construct three types of standard automorphisms for T . Based on these known automorphisms any automorphism of T can be characterized. • For a directed graph G and x, y ∈ V (G), we write x → y to denote there is a − → ← − directed edge from x to y, and we denote by N G (x) (resp., N G (x)) the set of vertices y ∈ V (G) for which x → y (resp., y → x). We write NG (x) = NG (y) to denote that − → − → ← − ← − N G (x) = N G (y) and N G (x) = N G (y) for x, y ∈ V (G). If NG (x) = NG (y) then x, y are called twin points. If a bijection ρ on V (G) permutes twin points, then it is an automorphism of G, which is called a regular automorphism of G. The set of all regular automorphisms of G is denoted by Reg(G). • Let P ∈ T (n, q) be invertible. We define φP from V (T ) to itself by X → P −1 XP . Then it is easy to check that φP is an automorphism of T , which is called an inner automorphism of T .
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• Let π be an automorphism of the field Fq . For a matrix A = (aij )n×n ∈ T (n, q), we denote (π(aij ))n×n by π(A). Define φπ from V (T ) to itself by A → π(A). It is not difficult to check that φπ is an automorphism of T , which is called a field automorphism of T . The main result of this article is as follows. Theorem 1.1. A mapping φ on V (T ) is an automorphism of T if and only if it can be uniquely decomposed into the product of an inner automorphism, a field automorphism and a regular automorphism. 2. Proof of Theorem 1.1 Let R(n, q) be the set of all rank one matrices in T (n, q). For A ∈ R(n, q), denote by [A] the subspace of T (n, q) spanned by A. In [13], R was defined to be a subgraph of T induced by R(n, q), and Rn (q) was defined to be a graph with vertex set {[A]: A ∈ R(n, q)} and there is a directed edge from [A] to [B] if and only if AB = 0. For proof of the main result of this paper, we need to apply two results obtained in [13]. Lemma 2.1. (See [13, Lemma 4.1].) Let A, B ∈ R(n, q). Then NR (A) = NR (B) if and only if B is a nonzero multiple of A. Lemma 2.2. (See [13, Theorem 3.3].) σ is an automorphism of Rn (q) if and only if σ can be decomposed into σ = σP ◦ σπ , where σP : [A] → [P −1 AP ], ∀A ∈ R(n, q) and σπ : A → [π(A)], ∀A ∈ R(n, q). Lemma 2.3. Let G be an arbitrary graph. Then Reg(G), consisting of regular automorphisms of G, is a normal subgroup of Aut(G), and it is isomorphic to the direct product of some symmetric groups. Proof. By the definition of Reg(G), it is easy to see that Reg(G) forms a subgroup of Aut(G), and it is isomorphic to the direct product of some symmetric groups. Let σ ∈ Aut(G) and ρ ∈ Reg(G), we need to show σ −1 ρσ ∈ Reg(G). It suffices to show that NG σ −1 ρσ (x) = NG (x),
∀x ∈ V (G).
Suppose that x → y, then σ(x) → σ(y), further (ρσ)(x) → σ(y) and (σ −1 ρσ)(x) → y, − → − → from which it follows that N G (x) ⊆ N G ((σ −1 ρσ)(x)). As σ −1 ρσ is an automorphism − → − → − → − → of G, |N G ((σ −1 ρσ)(x))| = |N G (x)|, from which N G ((σ −1 ρσ)(x)) = N G (x) follows. ← − ← − Similarly, N G ((σ −1 ρσ)(x)) = N G (x). Hence, NG ((σ −1 ρσ)(x)) = NG (x) for any x ∈ V (G). 2
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Lemma 2.4. Let A ∈ V (T ) be a nonzero zero-divisor of T (n, q). Then r(A) ≥ 2 if and − → − → only if there is certain B ∈ V (T ) such that N T (A) N T (B), where r(A) refers to the rank of A. Proof. “Necessity”: As r(A) ≥ 2, A has a nonzero row, say α. Let B be the n × n upper triangular matrix which has α as its first row and all other rows of B are zero. Then − → − → − → from AC = 0 it follows that BC = 0 for C ∈ N T (A). Consequently, N T (A) ⊆ N T (B). Denote the solution spaces of the linear equations AX = 0 and BX = 0 respectively by SA and SB . Obviously, SA ⊆ SB . Since r(B) < r(A), the dimension of SB is larger than that of SA , from which it follows that SA SB . Let β ∈ SB \ SA , and let D be the n × n upper triangular matrix which has β as its last column and all other columns of D are − → − → zero. Then AD = 0 and BD = 0, from which it follows that N T (A) N T (B). “Sufficiency”: SA and SB are as above. To achieve the goal, we need first to show that SA SB . Let 0 = β ∈ SA , and let D be the n × n upper matrix which has β as its last column and all other columns of D are zero. From AD = 0 it follows that − → − → D ∈ N T (A). Further, D ∈ N T (B), i.e., BD = 0. Consequently, Bβ = 0, i.e., β ∈ SB . − → − → Hence, SA ⊆ SB . Suppose C ∈ N T (B) \ N T (A). Thus, AC = 0 and BC = 0, which implies that C has a column, say ξ, such that Aξ = 0 and Bξ = 0. Hence, SA SB , as required. Consequently, r(A) > r(B) ≥ 1. 2 Lemma 2.5. Let φ be an automorphism of T . Then φ stabilizes the set of rank one matrices in T (n, q), i.e., φ(R(n, q)) = R(n, q). Proof. Suppose for a contradiction that φ(A) ∈ / R(n, q) for A ∈ R(n, q). Then − → − → r(φ(A)) ≥ 2. By Lemma 2.4, there is certain B ∈ V (T ) such that N T (φ(A)) N T (B). − → − → Let C be the pre-image of B under φ, i.e., φ(C) = B. Now, from N T (φ(A)) N T (φ(C)) − → − → it follows that N T (A) N T (C), which further implies that r(A) ≥ 2, a contradiction. 2 Lemma 2.6. Let A, B ∈ V (T ) be rank one matrices, and let φ be an automorphism of T . Then φ(B) is a nonzero multiple of φ(A) if and only if B is a nonzero multiple of A. Proof. By Lemma 2.5, φ(B) and φ(A) both are rank one matrices. Suppose B = aA for certain a ∈ Fq∗ . Then B and A have the same neighbors in T , i.e., NT (B) = NT (A). Further, NT (φ(B)) = NT (φ(A)). As vertices of R, φ(B) and φ(A) also have the same neighbors in R. By Lemma 2.1, φ(B) is a nonzero multiple of φ(A). By considering the action of φ−1 on φ(B) and φ(A), we can prove that if φ(B) is a nonzero multiple of φ(A), then B is a nonzero multiple of A. 2 The main idea of this paper is to reduce the study of automorphisms of T to that of Rn (q). So it is necessary to study the connection between Aut(T ) and Aut(Rn (q)). Theorem 2.7.
Aut(T ) Reg(T )
is isomorphic to Aut(Rn (q)).
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Proof. Lemma 2.3 has shown that Reg(T ) is a normal subgroup of Aut(T ). For an automorphism φ of T , we define φ from V (Rn (q)) to itself by φ ([A]) = [φ(A)] for any A ∈ R(n, q). Lemma 2.5 tells us φ(A) ∈ R(n, q) for A ∈ R(n, q); Lemma 2.6 proves that if [B] = [A] for B, A ∈ R(n, q) then [φ(B)] = [φ(A)], namely, φ ([B]) = φ ([A]), from which it follows that φ is well defined. If φ ([B]) = φ ([A]), i.e., [φ(B)] = [φ(A)], then φ(B) is a nonzero multiple of φ(A). By Lemma 2.6, B is a nonzero multiple of A, i.e., [B] = [A], which proves that φ is injective. For any [B] ∈ V (Rn (q)) suppose that φ(A) = B, then φ ([A]) = [B], which proves that φ is surjective. If [A] → [B] in Rn (q), i.e., AB = 0, then φ(A)φ(B) = 0, from which it follows that φ ([A]) → φ ([B]). Conversely, φ ([A]) → φ ([B]) also implies that [A] → [B]. Consequently, φ is an automorphism of Rn (q). We define a mapping f from Aut(T ) to Aut(Rn (q)) by f (φ) = φ for any φ ∈ Aut(T ). Let φ1 , φ2 ∈ Aut(T ). From (φ1 φ2 ) ([A]) = [(φ1 φ2 )(A)] = φ1 (φ2 ([A])) it follows that (φ1 φ2 ) = φ1 φ2 , which implies that f is a homomorphism from Aut(T ) to Aut(Rn (q)). To complete the proof we need to prove that the kernal of f is Reg(Γ (T )) and the image of f is Aut(Rn (q)). Let φ lie in the kernal of f , i.e., φ acts as the identity on V (Rn (q)), or equivalently, φ sends any A ∈ R(n, q) to a nonzero multiple of A. We need first to show that φ ∈ Reg(T ), or equivalently, NT (φ(A)) = NT (A) for any nonzero zero-divisor A in T (n, q). − → Let B ∈ N T (A), then AB = 0. Suppose the i-th column vector of B is βi , and let Bi be the triangular matrix which has βi as its i-th column and all other columns are zero. Then Bi is a rank one matrix or a zero matrix and ABi = 0. If Bi is of rank one, then we have φ(A)φ(Bi ) = 0. Since φ(Bi ) is a nonzero multiple of Bi , we further have φ(A)Bi = 0 and φ(A)βi = 0. If Bi = 0, φ(A)βi = 0 also holds. From φ(A)βi = 0 − → for i = 1, 2, . . . , n, it follows that φ(A)B = 0. Consequently, B ∈ N T (φ(A)). Hence, − → − → N T (A) ⊆ N T (φ(A)). Similarly, by considering the action of φ−1 on φ(A), we have − → − → − → − → − → N T (φ(A)) ⊆ N T (φ−1 (φ(A))) = N T (A). Therefore, N T (φ(A)) = N T (A). An analogous ← − ← − discussion leads to N T (φ(A)) = N T (A). Consequently, NT (φ(A)) = NT (A) for A ∈ V (T ). Now, it is proved that Ker(f ) ⊆ Reg(T ). Conversely, we need to prove Reg(T ) ⊆ Ker(f ). Suppose that φ ∈ Reg(T ). Then NT (φ(A)) = NT (A) for any A ∈ V (T ). Further, for any rank one matrix A0 , by NR (φ(A0 )) = NR (A0 ), we have φ(A0 ) is a nonzero multiple of A0 (using Lemma 2.1). Thus, φ acts as the identity on Rn (q), i.e., φ ∈ Ker(f ). Hence, Reg(T ) ⊆ Ker(f ). The proof for Ker(f ) = Reg(T ) is completed. Now we are in a position to show the image of f is Aut(Rn (q)). Let σ be an automorphism of Rn (q). By Lemma 2.2, there exists an invertible P in T (n, q) and an automorphism π of the field Fq such that σ([A]) = [P −1 π(A)P ] for any A ∈ R(n, q). Let φP : A → P −1 AP , ∀A ∈ V (T ), the inner automorphism of T induced by P ,
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and let φπ : A → π(A), ∀A ∈ V (T ), a field automorphism of T . Then we find that f (φP φπ ) = φP φπ = σ. Consequently, f is surjective. Aut(T ) ∼ Finally, we obtain Reg(T ) = Aut(Rn (q)) by applying the homomorphism theorem. 2 Proof of Theorem 1.1. The sufficient condition is obvious. Let φ be an automorphism of T . Then φ : [A] → φ(A) ,
∀A ∈ R(n, q)
is an automorphism of Rn (q). By Lemma 2.2, there is an invertible matrix P ∈ T (n, q) and an automorphism π of the field Fq such that φ [A] = P −1 π(A)P ,
∀A ∈ R(n, q).
Clearly, (φπ−1 ◦ φP −1 ◦ φ) acts as the identity on Rn (q). So φπ−1 ◦ φP −1 ◦ φ lies in the kernal of f , from which it follows that φπ−1 ◦ φP −1 ◦ φ acts as a regular automorphism, say ρ, on V (T ). Hence, φ = φP ◦ φπ ◦ ρ. Now we prove the uniqueness of the decomposition. Suppose φ = φP1 ◦ φπ1 ◦ ρ1 = φP2 ◦ φπ2 ◦ ρ2 be two decompositions of φ, where φPi , φπi , ρi are respectively inner automorphisms, field automorphisms and regular automorphisms of T . Recalling the definition of σP and σπ in Lemma 2.2, we have (φP ) = σP and (φπ ) = σπ . Applying f on two sides of −1 −1 φ−1 π2 ◦ φP2 ◦ φP1 ◦ φπ1 = ρ2 ◦ ρ1 ,
we have σπ−1 ◦ σP−1 ◦ σP1 ◦ σπ1 = I, 2 2 where I denotes the identity automorphism on Rn (q), from which it follows that σP−1 ◦ σP1 = σπ2 ◦ σπ−1 . 1 2 From the proof of Corollary 3.5 in [13], we have P2 is just a nonzero multiple of P1 and π2 = π1 . Consequently, φP2 = φP1 and φπ2 = φπ1 . Furthermore, ρ2 = ρ1 . 2 By Theorem 2.7 and Lemma 2.3, the following corollary is obvious. Corollary 2.8. Aut(T ) ∼ = Reg(T ) Aut(Rn (q)).
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