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1、Acta Mathematica Academiae Paedagogicae Ny regyh aziensis 24 (2008), 221225 www.emis.de/journals ISSN 1786-0091 THE SPECIAL CIRCULANT MATRIX AND UNITS IN GROUP RINGS JOE GILDEA Abstract. We introduce a new n n circulant matrix over Fpk. Under certain conditions, we show that this matrix generates a

2、copy of Cpk1. We conclude with determining the element of U(FpkCn) that corresponds to this matrix and providing explicit generators for U(FpkC2) when p 6= 2. 1. Introduction Let Fpk be the Galois fi eld of pkelements where p is a prime. To begin we defi ne a special n n circulant matrix over Fpk. W

3、e show that this matrix generates a copy of Cpk1when p does not divide n. The set of all the invertible elements of a ring S form a group called the unit group of S, denoted by U(S). Let RG denote the group ring G of the group G over the ring R. In 3, an explicit isomorphism between RG and a certain

4、 ring of n n matrices is given. Using this isomorphism we determine the element of U(FpkCn) that correspond to the above mentioned matrix when p does not divide n. We provide explicit generators for U(FpkC2) when p 6= 2, using the special circulant matrix and another matrix. The description of our m

5、ethod allows its straightforward implementation using the LAGUNA package 1 for the GAP system 4. Defi nition 1. A circulant matrix over a ring R is a square nn matrix, which takes the form circ(a1,a2,.,an) = a1a2a3.an ana1a2.an1 an1ana1.an2 . . . . . . . . . . . a2a3a4.a1 2000 Mathematics Subject Cl

6、assifi cation. 15A33, 20C05, 16S34. Key words and phrases. special circulant matrix, group ring, group algebra, unit group. 221 222JOE GILDEA where ai R. For further details on circulant matrices see Davis 2. Let R be a ring and Cnbe the cyclic group of order n. There exists an isomorphism between R

7、Cnand a certain ring of n n matrices over R given by : n1 X i=0 aixi7 circ(a0,a1,.,an1). See 3 for further details. 2. The Special Circulant Matrix Defi nition 2. Let a Fpkwhere a generates U(Fpk ) and p is a prime. Defi ne the special circulant matrix of order n over Fpkby Gn= circ(1 + (n 1)a,1 a,.

8、,1 a |z (n1)times ). Example 3. G2= circ(1 + a,1 a) and G3= circ(1 + 2a,1 a,1 a). Proposition 4. Let A = circ(, ,., | z (n1)times ), where A is a n n matrix and , Fpkand p is a prime. Then |A| = ( + (n 1)( )n1. Proof. Let A = circ(,.,), where A is a n n matrix and , Fpk. Then |A| = ( + (n 1) fl fl f

9、l fl fl fl fl fl 11.1 . . . . . . . . . . fl fl fl fl fl fl fl fl = ( + (n 1) fl fl fl fl fl fl fl fl fl fl 10.0 .0 0.0 . . . . . . . . 0. fl fl fl fl fl fl fl fl fl fl = ( + (n 1)( )n1. Proposition 5. Suppose that p does not divide n. Then (i) Gnis invertible. (ii) GnN= nN1circ(1 + (n 1)aN,1 aN,.,1

10、 aN |z (n1)times ). THE SPECIAL CIRCULANT MATRIX AND UNITS IN GROUP RINGS223 Proof. (i) |Gn| = (1 + (n 1)a + (n 1)(1 a)(1 + (n 1)a (1 a)n1 = (1 + na a + n na 1 + a)(1 + na a 1 + a)n1 = (n)(na)n1 = nnan1. Therefore Gnis invertible if p does not divide n. (ii) We prove this by induction on N. Let N =

11、1, clearly Gn1= Gn. Now lets assume that it holds for N = k. i.e. Gnk= nk1circ(1 + (n 1)ak,1 ak,.,1 ak |z (n1)times ). We must show that Gnk+1= nkcirc(1 + (n 1)ak+1,1 ak+1,.,1 ak+1 |z (n1)times ). Gnk+1= Gnk Gn1= = nk1 1 + (n 1)ak1 ak1 ak.1 ak 1 ak1 + (n 1)ak1 ak.1 ak 1 ak1 ak1 + (n 1)ak.1 ak . . .

12、. . . . . . . . 1 ak1 ak1 ak.1 + (n 1)ak 1 + (n 1)a1 a1 a.1 a 1 a1 + (n 1)a1 a.1 a 1 a1 a1 + (n 1)a.1 a . . . . . . . . . . . 1 a1 a1 a.1 + (n 1)a . When we multiply these matrices every diagonal entry will be of the form ()(1 + (n 1)ak)(1 + (n 1)a) + (n 1)(1 ak)(1 a) and every off diagonal entry ha

13、s the form () (1 + (n 1)ak)(1 a) + (1 ak)(1 + (n 1)a) + (n 2)(1 ak)(1 a). (1 + (n 1)ak)(1 + (n 1)a) + (n 1)(1 ak)(1 a) = 1 + (n 1)a + (n 1)ak+ (n 1)2ak+1+ (n 1)(1 a ak+ ak+1) = n + (n 1)2ak+1+ (n 1)ak+1 = n + (n2 2n + 1 + n 1)ak+1 = n + (n2 n)ak+1 = n(1 + (n 1)ak+1). 224JOE GILDEA (1 + (n 1)ak)(1 a)

14、 + (1 ak)(1 + (n 1)a) + (n 2)(1 ak)(1 a) = 1 a + (n 1)ak (n 1)ak+1+ 1 + (n 1)a ak (n 1)ak+1 + (n 2) (n 2)a (n 2)ak+ (n 2)ak+1 = n + (2(n 1) + (n 2)ak+1 = n + (2n + 2 + n 2)ak+1 = n nak+1 = n(1 ak+1). Therefore GnkGn1= nkcirc(1+(n1)ak+1,1 ak+1,.,1 ak+1 |z (n1)times ) = Gnk+1. Theorem 6. For a prime p

15、 which does not divide n, hGni = Cpk1. Proof. Gnp k1 = np k2 circ(1 + (n 1)ap k1,1 apk1,.,1 apk1 |z (n1)times ) = np k2 circ(1 + (n 1).1,1 1,.,1 1) = np k1I n = Insince a generates U(Fpk) and n U(Fpk). Consider nN1(1 aN ), which is an off diagonal entry of GnN. Now nN1(1 aN) = 0 1aN= 0 aN= 1 N = pk1

16、 since a generates U(Fpk) and n U(Fpk). Therefore hGni = Cpk1. Let An= diagn(a) where a generates U(Fpk) and p is a prime.Clearly hAni = Cpk1. Also hAni hGni = In since 1 aN = 0 iff N = pk 1. Corollary 7. Let n= (1 + (n 1)a) + (1 a) n1 X i=1 xi ! FpkCnwhere a generates U(Fpk), p is a prime and Cn= h

17、x|xn= 1i. Suppose p - n, then (i) hni = Cpk1. (ii) 2and a generate U(FpkC2). Proof. (i) () = Gn. (ii) (a) = A2, (2) = G2and hA2,G2i = Cpk1 Cpk1. References 1 V. Bovdi, A. Konovalov, R. Rossmanith, and C. Schneider. Laguna lie algebras and units of group algebras. http:/www.cs.st-andrews.ac.uk/ alexk/laguna.htm, 2007. Ver- sion 3.4. 2 P. J. Davis. Circulant matrices. John Wiley & Sons, New York-Chichester-Brisbane, 1979. A Wiley-Interscience Publication, Pure and Applied Mathematics. THE SPECIAL CIRCULANT MATRIX AND UNITS IN GROUP RINGS225 3 T. Hurley. Group rings and rings o