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Automorphisms of a nilpotent Liealgebra over a commutative ringShikun Ou a , Dengyin Wang a & Chunguang Xia aa Department of Mathematics , China University of Mining andTechnology , Xuzhou, 221008, People's Republic of ChinaPublished online: 23 Oct 2008.

To cite this article: Shikun Ou , Dengyin Wang & Chunguang Xia (2009) Automorphisms of anilpotent Lie algebra over a commutative ring, Linear and Multilinear Algebra, 57:1, 75-85, DOI:10.1080/03081080701571224

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Linear and Multilinear Algebra, Vol. 57, No. 1, January 2009, 75–85

Automorphisms of a nilpotent Lie algebra

over a commutative ring

SHIKUN OU*, DENGYIN WANG and CHUNGUANG XIA

Department of Mathematics, China University of Mining and Technology,Xuzhou, 221008, People’s Republic of China

Communicated by C.-K. Li

(Received 31 May 2007; in final form 26 June 2007)

Suppose that m� 5 and that R is a commutative ring with identity in which 2 is invertible. Thispaper determines all automorphisms of a nilpotent subalgebra of the orthogonal Lie algebrao(2m,R).

Keywords: Orthogonal Lie algebra; Automorphisms; Commutative rings

AMS 2000 Mathematics Subject Classifications: 17B; 15A

1. Introduction

Let R be a commutative ring with identity, R* the group consisting of all invertibleelements in R, E(n) the n� n identity matrix (E(m) is abbreviated to E), Rm�n the set of allm� n matrices over R, gl(m,R) the general linear Lie algebra consisting of all m�mmatrices over R with bracket: [X,Y]¼XY�YX. For A2Rm�n, A0 denotes the transposeof A. Let T(m,R) (respectively, S(m,R)) be the subalgebra of gl(m,R) consisting of allupper triangular (respectively, strictly upper triangular) matrices. Set I ¼ 0 E

E 0

� �. The

orthogonal Lie algebra o(2m,R) over R is defined to be the subalgebra of gl(2m,R),consisting of all X2 gl(2m,R) satisfying X0I¼�IX. The condition for A B

C D

� �(A,B,C,D2Rm�m) to be orthogonal is that B0 ¼�B, C0 ¼�C and D0 ¼�A. Let

LðRÞ ¼A B

0 �A0

� ���A 2 Sðm,RÞ,B 2 Rm�m satisfies B0 ¼ �B

� �:

It is a nilpotent subalgebra of o(2m,R).The automorphisms of a linear Lie algebra (or sometimes R-algebra) over

commutative rings were recently studied in [1–8]. Dokovic [5] in 1994 and Cao [3]

*Corresponding author. Email: ok6096@126.com

Linear and Multilinear AlgebraISSN 0308-1087 print/ISSN 1563-5139 online � 2009 Taylor & Francis

www.informaworld.comDOI: 10.1080/03081080701571224

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in 1998 characterized the automorphisms group of the solvable Lie algebra T(m,R).In 2003, Cao [2] characterized the automorphisms group of the nilpotent Lie algebraS(m,R). In 2007, Wang [1] described the automorphisms group of the standard Borelsubalgebra of o(2m,R). Motivated by Cao [2] and Wang [1], we intend to describe allthe automorphisms of the maximal nilpotent subalgebra L(R) of o(2m,R).

In this article, by using the main theorem of [2], we shall describe the automorphismsof L(R) when m� 5 and R is a commutative ring with identity in which 2 is invertible.Different from [1], in this article, we construct two new types of automorphisms ofL(R) – central and subcentral automorphisms, and we use the localization ideal toconstruct the graph automorphism in the proof of the main theorem.

2. Preliminaries

In the following, we always suppose that m� 5 and 22R*. Set

pðRÞ ¼A 0

0 �A0

� ���A 2 Sðm,RÞ

� �;

wðRÞ ¼0 B

0 0

� ���B 2 Rm�m, B0 ¼ �B

� �;

sðRÞ ¼ fdiagðA, 0, � A0, 0ÞjA 2 Sðm� 1,RÞg:

We use the following notations. For 1� i, j�m, let Eij denote the 2m� 2m matrix,whose (i, j)-entry is 1, all other entries are 0; Ei,�j the 2m� 2m matrix, whose (i,mþ j)-entry is 1, all other entries are 0; E�i, j the 2m� 2m matrix, whose (mþ i, j)-entry is 1,all other entries are 0; E�i,�j the 2m� 2m matrix, whose (iþm, jþm)-entry is 1, allother entries are 0. For a2R, 1� i5 j�m, we denote

TijðaÞ ¼ aðEij � E�j,�iÞ, Ti,�jðaÞ ¼ aðEi,�j � Ej,�iÞ,

Tij ¼ fTijðaÞja 2 Rg, Ti,�j ¼ fTi,�jðaÞja 2 Rg:

It is well known that the Lie algebra L(R) is generated by the set {Ti,iþ 1(1),Tm� 1,�m(1) j i¼ 1, 2, . . . ,m� 1} and for any Y in L(R), we may write Y ¼

P1�i5j�m

TijðaijÞ þP

1�k5l�m Tk,�lðak,�lÞ. Assume that

LtðRÞ ¼

( X1�i5j�mj�i�tþ1

TijðaijÞ þX

1�k5l�mkþl�2m�t�1

Tk,�lðak,�lÞ���aij, ak,�l 2 R

), 1 � t � 2m� 4,

and denote

xðRÞ ¼X

1�i5j�m�1

Ti,�j;

yðRÞ ¼ xðRÞ þ T1m þ T1,�m;

zðRÞ ¼ wðRÞ þX

1�i�m�1

Ti,m;

qðRÞ ¼ zðRÞ þ T1,m�1:

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It is clear that Lt(R) (1� t� 2m� 4), x(R), y(R), z(R), and q(R) leave stable under anyautomorphism of L(R) [1], and the center of the Lie algebra L(R) is L2m�4(R)¼T1,�2.

For R-modules L andM, we denote by HomR(L,M) the set of all homomorphisms ofR-module L to M.

3. Standard automorphisms of p(R)

It is obvious that p(R) is isomorphic to S(m,R). Cao [2] has described theautomorphisms of S(m,R), we now transfer them to p(R) for later use. p(R) hasthe following standard automorphisms.

(i) Inner AutomorphismsFor invertible A2T(m,R), set X ¼ A 0

0 A0�1

� �and define IntpX : p(R)! p(R), sending

P2 p(R) to XPX�1. Then IntpX is an automorphism of p(R), called the innerautomorphism of p(R) induced by X.

(ii) Graph automorphismsLet "¼ "2 be an idempotent in R, J ¼ E

ðmÞ1m þ E

ðmÞ2,m�1 þ � � � þ E

ðmÞm1 . Define �p," : p(R)!

p(R) by sending any�A 00 �A0

�to�"A�ð1�"ÞJA0J 00 �"A0þð1�"ÞJAJ

�. Then �p," is an automorphism

of p(R), called the graph automorphism of p(R) induced by ".(iii) Extremal automorphismsFor b¼ (b1, b2)2R

2 and P ¼P

1�i<j�m aijTij 2 pðRÞ, define linear maps �p,b on p(R) by

�p, bðPÞ ¼ Pþ b1a12T2m þ b2am�1,mT1,m�1:

Then �p,b is an automorphism of p(R), called the extremal automorphism of p(R)induced by b.

(iv) Central automorphismsLet F¼ { f2HomR(p(R),R)j f(X)¼ 0 for X2 s(R)}. For any f2F, the map �p, f :

p(R)! p(R), P�Pþ f(P)T1m(1) is an automorphism of p(R), which is called a centralautomorphism of p(R) induced by f.

Theorem 3.1 [2] If R is a commutative ring with identity, 22R* and m� 4. Then everyautomorphism �p of p(R) can be written in the form

�p ¼ �p, " � �p, b � �p, f � IntpX,

where �p,", �p,b, �p, f and IntpX are graph, extremal, central and inner automorphismsof p(R) defined above.

4. Standard automorphisms of L(R)

We now define some standard automorphisms of L(R).

(i) Inner automorphismsFor invertible A2T(m,R) and B0 ¼�B2Rm�m, set X ¼ A AB

0 A0�1

� �and define IntX :

L(R)!L(R), sending Y2L(R) to XYX�1. Then IntX is an automorphism of L(R),called the inner automorphism of L(R) induced by X.

(ii) Graph automorphismsLet !¼E(2m)

�Emm�E�m,�mþEm,�mþE�m,m, p¼ p2 be an idempotent in R.We define �p : L(R)!L(R), sending any Y2L(R) to pYþ (1� p)!Y!. Then �p is an

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automorphism of L(R)(note that �2� is the identity), called the graph automorphism

of L(R) induced by p.(iii) Extremal automorphismsFor g2R*, we define �g : L(R)!L(R), sending A B

0 �A0

� �2 LðRÞ to A gB

0 �A0

� �. Then �g

is an automorphism of L(R), called the extremal automorphism of L(R) induced by g.(iv) Central automorphismsFor any c¼ (c1, c2, . . . , cm)2R

m, any Y ¼P

1�i5j�m TijðaijÞ þP

1�k5l�m Tk,�lðak,�lÞ

2 LðRÞ, we define the map �c : L(R)!L(R) by

�cðYÞ ¼ YþX

1�i�m�1

ciai, iþ1 þ cmam�1,�m

!T1,�2ð1Þ:

Then it is easy to check that �c is an automorphism of L(R), called the central

automorphism of L(R) induced by c.(v) Subcentral automorphismsFor h2R, Y ¼

P1�i5j�m TijðaijÞ þ

P1�k5l�m Tk,�lðak,�lÞ 2 LðRÞ, we define lh:

L(R)!L(R) by

lhðYÞ ¼ Yþ hT1,�3ða23Þ:

Then lh is an automorphism of L(R), called the subcentral automorphism of L(R)

induced by h.

Lemma 4.1 Let X ¼ A AB0 A0�1

� �with A ¼

P1�i�m aieii þ

P1�i5j�m aijeij ðai 2 R�Þ,

B0 ¼�B. Let h2R, c¼ (c1, c2, . . . , cm)2Rm. Then

(i) Int�1X ��c � IntX¼�c0 for c0 ¼ a�11 a�12 ðc1a1a�12 , c2a2a

�13 , . . . , cm�1am�1a

�1m ,

cmam�1amÞ 2 Rm;(ii) IntX � lh0 � Int

�1X¼ lh0 ��d for h0 ¼ ha1a

�12 a23, d ¼ ð0, ha1a

�12 a3a23, 0, . . . , 0Þ 2 Rm.

Proof

(i) For any Y ¼P

1�i�m�1 Ti, iþ1 (bi,iþ1)þTm�1,�m(bm�1,�m)þZ2L(R), where

Z2L1(R), we have the following equality.

Int�1X � �c � IntXðYÞ

¼ Int�1X XYX�1 þX

1�i�m�1

ciaia�1iþ1bi, iþ1 þ cmam�1ambm�1,�m

!T1,�2ð1Þ

!

¼ Yþ a�11 a�12

X1�i�m�1

ciaia�1iþ1bi, iþ1 þ cmam�1ambm�1,�m

!T1,�2ð1Þ

¼ �c0 ðYÞ:

(ii) For any Y ¼P

1�i5j�m TijðbijÞ þW 2 LðRÞ, where W2w(R), we see that

IntX � lhðYÞ ¼ IntXðYþ hb23T1,�3ð1ÞÞ ¼ XYX�1 þ hb23ða1a23T1,�2ð1Þ þ a1a3T1,�3ð1ÞÞ

¼ lh0 � �d � IntXðYÞ:

This shows that IntX � lh � Int�1X¼ lh0 ��d. g

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5. Automorphisms of L(R)

Let R be a commutative ring with identity, SpecR the prime spectrum of R. If

M2 SpecR, we denote RM ¼ fav j a 2 R, v 2 RnMg, which is the localization of R at M.

For any r2R and any Y ¼P

1�i5j�m TijðaijÞ þP

1�k5l�m Tk,�lðak,�lÞ 2 LðRÞ, we denote

by ðr=1Þ the image of r under the canonical homomorphism R!RM, and YM the

matrixP

1�i5j�m Tijðaij=1Þ þP

1�k5l�m Tk,�lðak,�l=1Þ in L(RM). It is not difficult to

verify that in L(RM) each matrix may be expressed as ð1=vÞXM with v2R/M and

X2L(R), i.e., LðRMÞ ¼ f1v XMj v 2 RnM,X 2 LðRÞg. For convenience sake, we denote

(�(A))2 by �(A)2 where A2L(R), � an automorphism of L(R).First we prove an auxiliary lemma.

Lemma 5.1 Let 22R*, � be an automorphism of L(R), Tij (1) (1� i5 j�m) any

canonical basis element of p(R). Then for any canonical basis element Tkl (1) or Tk,�l (1)

(1� k5 l�m) of L(R), we have

(i) �ðTklð1ÞÞ2Tijð1Þ, �ðTk,�lð1ÞÞ

2Tijð1Þ 2P

1�t�i REt,�i;

(ii) Tijð1Þ�ðTklð1ÞÞ2, Tijð1Þ�ðTk,�lð1ÞÞ

22P

1�t�i REi,�t;

(iii) �ðTklð1ÞÞTijð1Þ�ðTk, lð1ÞÞ,�ðTk,�lð1ÞÞTijð1Þ�ðTk,�lð1ÞÞ 2P

1�t�i�1ðREt,�iþ REi,�tÞ.

Proof By applying � on

½Tklð1Þ, ½Tklð1Þ,��1ðTijð1ÞÞ�� ¼ 0, ½Tk,�lð1Þ, ½Tk,�lð1Þ,�

�1ðTijð1ÞÞ�� ¼ 0,

we have that

�ðTklð1ÞÞ2Tijð1Þ � 2�ðTklð1ÞÞTijð1Þ�ðTklð1ÞÞ þ Tijð1Þ�ðTklð1ÞÞ

2¼ 0,

�ðTk,�lð1ÞÞ2Tijð1Þ � 2�ðTk,�lð1ÞÞTijð1Þ�ðTk,�lð1ÞÞ þ Tijð1Þ�ðTk,�lð1ÞÞ

2¼ 0:

On the other hand, for any Y2L(R), we see that

�ðYÞ2Tijð1Þ 2X1�s5i

REsj þX1�s5j

REs,�i þX

j5t�m

ðRE�t,�i þ REt,�iÞ;

Tijð1Þ�ðYÞ22X

j5t�m

ðREit þ REi,�tÞ þX1�s5i

RE�j,�s þX1�s5j

REi,�s;

�ðYÞTijð1Þ�ðYÞ 2X

1�s5ij5t�m

ðREst þ REs,�t þ REt,�s þ RE�t,�sÞ þX1�s5i1�t5j

ðREs,�t þ REt,�sÞ:

These show that Lemma 5.1 holds. g

Theorem 5.2 Let R be a commutative ring with identity, 22R* and m� 5. Then every

automorphism � of L(R) can be written in the form

� ¼ �� � �g � lh � �c � IntX,

where �p, xig, lh, �c, and IntX are the automorphisms of L(R), defined as in section 4.

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Proof Let � be an automorphism of L(R). We shall give the proof by steps.

Step 1 There exists X1¼ diag(A, 1,A0�1, 1) with A2T(m� 1,R) invertible such that

� � Int�1X1ðTi, iþ1ð1ÞÞ ¼ Ti, iþ1ð1Þ ðmod qðRÞÞ, 1 � i � m� 2:

Because z(R) is stable under �, � induces an automorphism � of L(R)/z(R) by

�ðYÞ ¼ �ðYÞ, Y2L(R). Since L(R)/z(R) is isomorphic to s(R), we now directly view

L(R)/z(R) as s(R). By section 3, we know that

� ¼ �s, " � �s, b � �s, f � IntsX1,

where �s,", �s,b, �s, f, and IntsX1 are the graph, extremal, central and inner

automorphisms of s(R), respectively. It is easy to see that IntsX1 ¼ IntX1.

So � � Int�1X1 ¼ �s, " � �s, b � �s, f. Denote �1¼� � Int�1X1.

In the following of this step, we first assert that "¼ 1. Since T1,�2¼L2m�4,

T1,�3L2m�5, T2,�3L2m�6, we may assume

�1ðT1,�2ð1ÞÞ ¼ T1,�2ðaÞ where a 2 R�,

�1ðT1,�3ð1ÞÞ ¼ T1,�2ða1Þ þ T1,�3ða2Þ where a1, a2 2 R,

�1ðT2,�3ð1ÞÞ ¼ T2,�3ðb1Þ þ T1,�4ðb2Þ þ Z where b1, b2 2 R, Z 2 L2m�5ðRÞ:

By applying �1 on [T23(1),T1,�3(1)]¼T1,�2(1), we get that a2¼ a2R*. By applying �1on [T12(1),T2,�3(1)]¼T1,�3(1), we have that "b1¼ a2, which shows that "2R*, leadingto "¼ 1. Hence �s," is the identity. Now we may assume that

�1ðT12ð1ÞÞ ¼ T12ð1Þ þ T2,m�1ðr1Þ þQ1,

�1ðTi, iþ1ð1ÞÞ ¼ Ti, iþ1ð1Þ þQi, 2 � i � m� 3,

�1ðTm�2,m�1ð1ÞÞ ¼ Tm�2,m�1ð1Þ þ T1,m�2ðr2Þ þQm�2,

where ri2R (i¼ 1, 2), Qj2 q(R) (1� j�m� 2).Next, we will prove that r1¼ r2¼ 0. Since �1(T12(1))

2Tm�1,m(1)¼ r1E1mþ

d1E1,�(m�1)þ d2E2,�(m�1) (where d1, d22R) and �1(Tm�2,m�1(1))2 Tm�1,m(1)¼

r2E1mþ d3E1,�(m�1)þ d4Em�2,�(m�1) (where d3, d42R), by Lemma 5.1 (i) we get that

r1¼ r2¼ 0, as desired.

Step 2 There exist an idempotent p2R and an invertible element g2R such that

��1g � �� � �1ðT1,�mð1ÞÞ T1,�mð1Þ ðmod xðRÞÞ:

Because x(R) and y(R) are stable under �1, we may assume that

�1ðT1,�mð1ÞÞ T1,�mða1Þ þ T1mða2Þ ðmod xðRÞÞ,

�1ðT1mð1ÞÞ T1,�mðb1Þ þ T1mðb2Þ ðmod xðRÞÞ,

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where ai, bi2R (i¼ 1, 2). Given M2 specR, we define e�1 : LðRMÞ ! LðRMÞ bye�1ð1v YMÞ ¼1v ð�1ðYÞÞM for v2R\M and Y2L(R). It is not difficult to check that e�1

is a Lie automorphism of L(RM) (which implies that x(RM) and y(RM) are stable

under e�1), ande�1 T1,�m

1

2

� �� � T1,�m

a11

þ T1m

a22

ðmod xðRMÞÞ,

e�1 T1m1

1

� �� � T1,�m

b11

� �þ T1m

b21

� �ðmod xðRMÞÞ:

These follow that e�1 may induce an automorphism e�1 of y(RM)/x(RM) (e�1 : yðRMÞ=xðRMÞ ! yðRMÞ=xðRMÞ,

1v YM 7! e�1ð1v YMÞ for any

1v YM 2 yðRMÞ) such that

e�1 T1,�m1

1

� �¼ T1,�m

a11

þ T1m

a21

,

e�1 T1m1

1

� �¼ T1,�m

b11

� �þ T1m

b21

� �:

Denote D ¼�a1=1 b1=1a2=1 b2=1

�, which is the coefficients matrix of e�1 on canonical basis

T1,�mð11Þ, T1mð

11Þ of y(RM)/x(RM). Since e�1 is the automorphism of y(RM)/x(RM), D is

invertible in gl(2,RM). Then by RM is a local ring, we see that each row of D has an

entry in R�M (otherwise, if each entry of certain row of D is in RM=R�M, which is the

maximal ideal of R, then detD 2 RM=R�M. This implies detD is not invertible in RM, a

contradiction).Next, we will prove that a1

1 �a21 ¼ 0. It is easy to see that T2,�mð

11Þ 2 zðRMÞ

TLm�3ðRMÞ,

following that e�1ðT2,�mð11ÞÞ 2 zðRMÞ

TLm�3ðRMÞ yðRMÞ þ T2mðRMÞ þ T2,�mðRMÞ.

Suppose

e�1 T2,�m1

1

� �� � T2,�m

r11

þ T2m

r21

ðmod yðRMÞÞ,

where r1, r22R. Since e�1ðT12ð11ÞÞ T12ð

11Þ (mod q(R)), by applying e�1 on T1,�mð

11Þ ¼

½T12ð11Þ,T2,�mð

11Þ�, we have a1

1 ¼r11 and a2

1 ¼r21 . On the other hand, by e�1ðT2,�mð

11ÞÞ

2�

T12ð11Þ ¼

21 �

r11 �

r21 ðE2,�1ÞM þ

d1 ðE1,�1ÞM (where d2R) and Lemma 5.1 (ii), we see

r11 �

r21 ¼ 0. Thus, a1

1 �a21 ¼ 0.

Thus, we see that in RM either

a11

2 RMð Þ

� anda21¼ 0

or

a11¼ 0 and

a21

2 ðRMÞ

�:

Then by [2, Lemma 3.2] there exists an idempotent p in R such that a12 pR,a22 (1�p)R, and a1þ a22R*. Let g¼ a1þ a2. We now construct the graph

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automorphism �p and the extremal automorphism �g of L(R), and denote �2 ¼ ��1g �

�� � �1, we see that �2ðT1,�mð1ÞÞ T1,�mð1Þ (modx(R)).

Step 3 There exist X2¼E(2m)�P1 with P12T2,m� 1 and X3¼E(2m)

þP2 with

P2 2P

1�i�m�1 T1i, such that w(R) is stable under Int�1X3 � Int�1X2 ��2; in particular,

Int�1X3 � Int�1X2 � �2ðTi,�mð1ÞÞ Ti,�mð1Þ ðmod xðRÞÞ, 1 � i � m� 1:

If we can prove the latter assertion, then we shall get the one before by applying

Int�1X3 � Int�1X2 ��3 on Ti,�j (1)¼�[Tim(1),Tj,�m(1)], 1� i5 j�m.

By step 1 and step 2, we can get that �2(Tk,kþ1(1))Tk,kþ1(1) (mod q(R)),

1� k�m� 2, �2(T2,m�1(1))T2,m�1(1) (mod q(R)). Suppose that �2(T12(1))

T12(1)þT1,m� 1(a) (mod z(R)). Set X2¼E(2m)�T2,m�1(a) and denote �3¼ Int�1X2 ��2,

we see that �3(T12(1))T12(1) (mod z(R)). Since Ti,�m2 z(R), we may assume that

�3ðTi,�mð1ÞÞ X

1�t�m�1

ðTtmðaðiÞtmÞ þ Tt,�mða

ðiÞt,�mÞÞ ðmodxðRÞÞ, 2 � i � m� 1: ð5:2:1Þ

For 2� k�m� 2 and k 6¼ i, by applying �3 on [Tk� 1,k(1),Ti,�m(1)]¼ 0, we have that

in equation (5.2.1) aðiÞkm ¼ a

ðiÞk,�m ¼ 0.

For 2� k�m� 2, by applying �3 on

½T2,m�1ð1Þ,Tk,�mð1Þ� ¼ 0 and ½Tk�1, kð1Þ,Tk,�mð1Þ� ¼ T1,�mð1Þ,

we get that aðkÞm�1,m ¼ a

ðkÞm�1,�m ¼ 0, and a

ðkÞkm ¼ 0, a

ðkÞk,�m ¼ 1 in equation (5.2.1).

By applying �3 on[T2,m�1(1),Tm�1,�m(1)]¼T2,�m(1), we get aðm�1Þm�1,m ¼ 0, a

ðm�1Þm�1,�m ¼ 1

in equation (5.2.1).Thus, equation (5.2.1) can be rewritten as

�3ðTi,�mð1ÞÞ T1mðaðiÞ1mÞ þ T1,�mða

ðiÞ1,�mÞ þ Ti,�mð1Þ ðmod xðRÞÞ, 2 � i � m� 1:

For 2� k�m� 2, then by applying �3 on [Tk,�m(1),Tm�1,�m(1)]¼ 0, we obtain aðkÞ1m ¼ 0,

aðm�1Þ1m ¼ 0, and a

ðm�1Þm�1,m ¼ 0. These show that

�3ðTi,�mð1ÞÞ T1,�mðaðiÞ1,�mÞ þ Ti,�mð1Þ ðmod xðRÞÞ, 2 � i � m� 1:

Set X3 ¼ Eð2mÞ þP

1�i�m�1 T1iðaðiÞ1,�mÞ and denote �4¼ Int�1X3 ��3, we see that

�4(Ti,�m(1))Ti,�m (1) (modx(R)), 2� i�m� 1.

Step 4 There exist X4¼ diag(A,A�1) with A2T(m,R) invertible and X5¼E(2m)þP3

with P32 p(R) such that Int�1X5 ��4 � Int�1X4 fixes each Pþw(R) for P2 p(R).

Since w(R) is stable under �4, thus �4 induces an automorphism �4 of

L(R)/w(R) by �4ðXÞ ¼ �4ðXÞ, X 2 LðRÞ. Since L(R)/w(R) is isomorphic to p(R),

we may directly view L(R)/w(R) as p(R). Thus by Theorem 3.1, �4 can be written in

the form

�4 ¼ �p, � � �p, b � �p, f � IntpX4,

where X4¼ diag(A,A0�1) with A2T(m,R) invertible, f2HomR(p(R),R) such that

f(S)¼ 0 for any S2 s(R), b¼ (b1, b2)2R2, � is an idempotent in R. The fact

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that �4(T12(1))¼T12(1) (mod q(R)) shows that �¼ 1. It is easy to know that

IntpX4 ¼ IntX4. Denote �5¼�4 � Int�1X4. Then we may assume that

�5ðT12ð1ÞÞ ¼ T12ð1Þ þ T1mðað1Þ1mÞ þ T2mðb1Þ þW1,

�5ðTi, iþ1ð1ÞÞ ¼ Ti, iþ1ð1Þ þ T1mðaðiÞ1mÞ þWi, 2 � i � m� 2,

�5ðTm�1,mð1ÞÞ ¼ Tm�1,mð1Þ þ T1,m�1ðb2Þ þ T1mðaðm�1Þ1m Þ þWm�1,

where aðiÞ1m 2 R and Wi2w(R), 1� i�m� 1.

By applying �5 on [T12(1),Tm�1,�m(1)]¼ 0, we get that b1¼ 0.For 2� k�m� 1, by applying �5 on [Tk,kþ1(1),Tm�1,�m(1)]¼ 0, we see that a

ðkÞ1m ¼ 0,

b2¼ 0.Now we set X5 ¼ Eð2mÞ � T2mða

ð1Þ1mÞ and denote �6¼ Int�1X5 ��5. Then we have

�6(Ti,iþ1(1))¼Ti,iþ1(1)þWi, 1� i�m� 1. Since p(R) is generated by {Ti,iþ1 j i¼

1, 2, . . . ,m� 1}, we see that �6 (P)P (modw(R)) for any P2 p(R).

Step 5 There exist h2R and X6¼E (2m)þW with W2w(R) such that

l�1h � Int�1X6 � �6ðTi, iþ1ð1ÞÞ ¼ Ti, iþ1ð1Þ ðmodL2m�4ðRÞÞ, 1 � i � m� 1:

Suppose that

�6ðTi, iþ1ð1ÞÞ ¼ Ti, iþ1ð1Þ þX

1�k5l�m

Tk,�lðaðiÞk,�lÞ, 1 � i � m� 1, ð5:2:2Þ

where aðiÞk,�l 2 R. For 1� t�m� 1 and t 6¼ i� 1, i, iþ 1, by applying �6 on [Ti,iþ1(1),

Tt,tþ1(1)]¼ 0, we haveXiþ1<l

Ti,�lðaðtÞiþ1,�lÞ þ

Xk<i

Tk,�iðaðtÞk,�ðiþ1ÞÞ �

Xtþ1<l

Tt,�lðaðiÞtþ1,�lÞ �

Xk<t

Tk,�tðaðiÞk,�ðtþ1ÞÞ ¼ 0:

which shows that aðiÞtþ1,�l ¼ 0 for tþ 15 l, l 6¼ i and a

ðiÞk,�ðtþ1Þ ¼ 0 for k5 t, k 6¼ i.

Thus, equation (5.2.2) can be rewritten as

�6ðT12ð1ÞÞ ¼ T12ð1Þ þ T1;�2ðað1Þ1;�2Þ þ

X3�t�m

T1;�tðað1Þ1;�tÞ þ T2;�3ða

ð1Þ2;�3Þ þ T3;�4ða

ð1Þ3;�4Þ;

�6ðT23ð1ÞÞ ¼ T23ð1Þ þ T1;�2ðað2Þ1;�2Þ þ T1;�3ða

ð2Þ1;�3Þ þ T1;�4ða

ð2Þ1;�4Þ þ

X3�t�m

T2;�tðað2Þ2;�tÞ,

þ T3;�4ðað2Þ3;�4Þ þ T4;�5ða

ð2Þ4;�5Þ

�6ðTi;iþ1ð1ÞÞ ¼ Ti;iþ1ð1Þ þ T1;�2ðaðiÞ1;�2Þ þ

X1�t�i�1

Tt;�iðaðiÞt;�iÞ þ T1;�ðiþ1Þða

ðiÞ1;�ðiþ1ÞÞ

þ T1;�ðiþ2ÞðaðiÞ1;�ðiþ2ÞÞ þ

Xiþ1�t�m

Ti;�tðaðiÞi;�tÞ þ Tiþ1;�ðiþ2Þða

ðiÞiþ1;�ðiþ2ÞÞ

þ Tiþ2;�ðiþ3ÞðaðiÞiþ2;�ðiþ3ÞÞ; 3 � i � m� 1

(where Tk,�l(ak,�l) is defined to be zero when l4m).By �6ðT12ð1ÞÞT23ð1Þ�6ðT12ð1ÞÞ ¼ a

ð1Þ1,�3E1,�1 þ a

ð1Þ3,�4T1,�4 � a

ð1Þ2,�3T1,�2 and Lemma 5.1

(iii), we see that að1Þ1,�3 ¼ a

ð1Þ3,�4 ¼ 0. By applying �6 on [T12(1),T13(1)]¼ 0, we have

að1Þ2,�3 ¼ 0.

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By applying �6 on [T23(1),T13(1)]¼ 0, [T23(1),T14(1)]¼ 0, and [T23(1),T24(1)]¼ 0,

we have að2Þ2,�3 ¼ a

ð2Þ3,�4 ¼ 0, a

ð2Þ2,�4 ¼ a

ð2Þ4,�5 ¼ 0, and a

ð2Þ1,�4 ¼ 0.

For 3� k�m� 1, by applying �6 on [T2,kþ1(1),Tk,kþ1(1)]¼ 0, we get

aðkÞ1,�ðkþ1Þ ¼ a

ðkÞk,�ðkþ1Þ ¼ a

ðkÞkþ1,�ðkþ2Þ ¼ 0.

For 3� k�m� 2, by applying �6 on [Tk,kþ1(1),Tk�1,kþ2(1)]¼ 0, we obtain

aðkÞ1,�ðkþ2Þ ¼ a

ðkÞk,�ðkþ2Þ ¼ a

ðkÞkþ2,�ðkþ3Þ ¼ 0.

These show that

�6ðT12ð1ÞÞ ¼ T12ð1Þ þ T1;�2ðað1Þ1;�2Þ þ

X4�t�m

T1;�tðað1Þ1;�tÞ;

�6ðT23ð1ÞÞ ¼ T23ð1Þ þ T1;�2ðað2Þ1;�2Þ þ T1;�3ða

ð2Þ1;�3Þ þ

X5�t�m

T2;�tðað2Þ2;�tÞ;

�6ðTi;iþ1ð1ÞÞ ¼ Ti;iþ1ð1Þ þ T1;�2ðaðiÞ1;�2Þ þ

X1�t�i�1

Tt;�iðaðiÞt;�iÞ

þX

iþ3�t�m

Ti;�tðaðiÞi;�tÞ; 3 � i � m� 1:

We choose

Z1 ¼ Eð2mÞ �X

4�t�m

T2;�tðað1Þ1;�tÞ;

Z2 ¼ Eð2mÞ �X

5�t�m

T3;�tðað2Þ2;�tÞ;

Zi ¼ ðEð2mÞ �

Xiþ3�t�m

Tiþ1;�tðaðiÞi;�tÞÞðE

ð2mÞ �X

1�t�i�1

Tt;�ðiþ1ÞðaðiÞt;�iÞÞ; 3 � i � m� 1

and set X6 ¼Qm�1

i¼1 Zi, h ¼ að2Þ1,�3. Then

l�1h � Int�1X6 � �6ðTi, iþ1ð1ÞÞ ¼ Ti, iþ1ð1Þ þ T1,�2ða

ðiÞ1,�2Þ, 1 � i � m� 1:

In the following, we denote �7 ¼ l�1h � Int�1X6 � �6.

Step 6 There exists X7¼E(2m)þP4 with P42 p(R) such that

Int�1X7 � �7ðTm�1,�mð1ÞÞ Tm�1,�mð1Þ ðmodL2m�4ðRÞÞ:

Suppose that

�7ðTm�1,�mð1ÞÞ ¼ Tm�1,�mð1Þ þX

1�k5l�m�1

Tk,�lðak,�lÞ, where ak,�l 2 R: ð5:2:3Þ

For 2� t�m� 2, by applying �7 on [Tt�1,t(1),Tm�1,�m(1)]¼ 0, we haveXt<l�m�1

Tt�1,�lðat,�lÞ þX

1�k<t�1

Tk,�ðt�1Þðak,�tÞ ¼ 0:

which shows that at,�l¼ 0 for t5 l�m� 1 and ak,�t¼ 0 for 1� k5 t� 1. Thus,

equation (5.2.3) can be rewritten as

�7ðTm�1,�mð1ÞÞ ¼ Tm�1,�mð1Þ þ T1,�2ða1,�2Þ þ T1,�ðm�1Þða1,�ðm�1ÞÞ þ T1,�mða1,�mÞ:

Now set X7¼E(2m)�T1m(a1,�(m�1))þT1,m�1(a1,�m) and denote �8¼ Int�1X7 ��7, it is

easy to check that �8 (Tm�1,�m(1))¼Tm�1,�m(1)þT1,�2(a1,�2).

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Step 7 �8 exactly is a central automorphism of L(R).

Suppose that

�8ðTi, iþ1ð1ÞÞ ¼ Ti, iþ1ð1Þ þ T1,�2ðciÞ, 1 � i � m� 1,

�8ðTm�1,�mð1ÞÞ ¼ Tm�1,�mð1Þ þ T1,�2ðcmÞ:

where ci2R (1� i�m). Set c0¼ (c1, c2, . . . , cm)2Rm. Then we see that �8��c0

sendsTi,iþ1(1) (1� i�m� 1) and Tm�1,�m(1) to 0. Since L(R) is generated by the set Ti,iþ1(1),Tm�1,�m(1) j i¼ 1, 2, . . . ,m� 1}, �8 ¼ �c0 sends any element in L(R) to 0, forcing�8 ¼ �c0 .

Above discussion shows that

� ¼ �� � �g � IntðX2X3X5X6Þ � lh � IntX7 � �c0 � IntðX4X1Þ:

By this one can easily obtain the desired expression for �. This completes theproof. g

References

[1] Wang, D., Yu, Q. and Zhao, Y., 2007, Automorphisms of a linear Lie algebra over a commutative ring.Linear Algebra and its Applications, 423, 324–331.

[2] Cao, Y. and Tang, Z., 2003, Automorphisms of the Lie algebras of strictly upper triangular matrices overa commutative ring. Linear Algebra and its Applications, 360, 105–122.

[3] Cao, Y., 1997, Automorphisms of the Lie algebras of strictly upper triangular matrices over acommutative ring. Journal of Algebra, 189, 506–513.

[4] Jøndrup, S., 1995, Automorphisms and deviations of upper triangular matrix rings. Linear Algebra and itsApplications, 221, 205–218.

[5] Dokovic, D.Z., 1994, Automorphisms of the Lie algebras of strictly upper triangular matrices over acommutative ring. Journal of Algebra, 170, 101–110.

[6] Jøndrup, S., 1991, The group of automorphisms of certain subalgebras of matrix algebras. Journal ofAlgebra, 141, 106–114.

[7] Jøndrup, S., 1987, Automorphisms of upper triangular matrix rings. Archiv der Mathematik, 49, 497–502.[8] Benkart, G.M. and Osborn, J.M., 1981, Derivations and automorphisms of non-associative matrix

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