International Journal of Algebra, Vol. 6, 2012, no. 14, 663 - 666 A Note on Nilpotent Elements in Quaternion Rings over Zp Michael Aristidou American University of Kuwait Department of Science and Engineering P.O.Box 3323, Safat 13034, Kuwait [email protected] Andy Demetre Seattle University Department of Mathematics 901 12th Ave., P.O. Box 222000 Seattle, WA 98122-1090, USA [email protected] Abstract In this paper, we discuss nilpotent elements in the finite ring H/Zp. We provide examples and we establish conditions for nilpotency in H/ Zp. We also comment on the number of nilpotent elements. Mathematics Subject Classification: 15A33, 15A30, 20H25, 15A03 Keywords: quaternion, ring, nilpotent. 1. Introduction The quaternions, denoted by H, were first invented by W. R. Hamilton in 1843 as an extension of the complex numbers into four dimensions [8]. Algebraically speaking, H forms a division algebra (skew field) over R of dimension 4 ([8], 1 p.195-196). In [1], we studied the finite ring H/Zp, where p is a prime, looking into its structure and some of its properties. A more detailed description of the structure H/Zp was given by Miguel and Serodio in [5]. Among others, they found the number of zero-divisors, the number of idempotent elements, and provided an interesting description of the zero-divisor graph. In particularly, 2 they showed that the number of idempotent elements in H/Zp is p + p +2, 664 M. Aristidou and A. Demetre for p odd prime. In [2], we studied idempotent elements and gave conditions for idempotency. In the sections that follow, we examine nilpotent elements in H/Zp and provide conditions for nilpotency in H/Zp. 2. Nilpotent Elements in H/Zp Recall that an element x in a ring R is called nilpotent if xk = 0, for some k ∈ Z. In the ring H/Zp, p prime, an element x is of the form: x = a0 + a1i + a2j + a3k 2 2 2 where ai ∈ Zp,pprime, and i = j = k = p −1=−1. It is not hard to show 2 that for any x ∈ H/Zp, one has: 2 x − 2a0x + N(x) = 0 (1) 2 2 2 2 where N(x) is the norm of x. Namely, N(x)=a0 + a1 + a2 + a3. Considering (1), we have now the following Lemma: Lemma 2.1: Let x ∈ H/Zp.Ifx is nilpotent, then N(x)=0. Proof. If x is nilpotent, then xk = 0 for some k. From (1) above, we have: 2 x − 2a0x + N(x)=0⇒ x(x − 2a)=−N(x) ⇒ xk(x − 2a)k = −N(x)k ⇒ 0=N(x)k ⇒ N(x)=0 because Zp is a field. Theorem 2.2: Let x ∈ H/Zp.Ifx is nilpotent, then x = a1i+a2j +a3k with N(x)=0.3 Furthermore, x2 =0. Proof. From (1) and Lemma 2.1, we have: (i) k = even: 2 2 k/2 k/2 k/2 x =2ax ⇒ (x ) =(2a0) x k k/2 k/2 ⇒ x =(2a0) x k/2 k/2 ⇒ 0=(2a0) x ⇒ a0 =0. Nilpotent elements in quaternion rings 665 (ii) k = odd: 2 2 (k+1)/2 (k+1)/2 (k+1)/2 x =2ax ⇒ (x ) =(2a0) x k+1 (k+1)/2 (k+1)/2 ⇒ x =(2a0) x (k+1)/2 (k+1)/2 ⇒ 0=(2a0) x ⇒ a0 =0. The fact that x2 = 0 follows immediately. Example 2.3: Let p = 3. Then, x =2i +4j +2k is nilpotent. Notice that N(x) = 0 and x2 =0. Remark 2.4: Regarding the number of nilpotent elements, Fine and Herstein in [3] show that the probability that an n × n matrix over a Galois field having pα elements be nilpotent is p−α.n. In our case, n = 2 and α =1. −2 So, the probability that a 2 × 2 matrix over Zp be nilpotent is p . In other words, the (# of 2×2 nilpotent matrices)/(# of 2×2 matrices) = p−2. Hence, (# of 2 × 2 nilpotent matrices)/p4 = p−2. Which gives that the (# of 2 × 2 nilpotent matrices) = p2. Conclusion We discussed nilpotent elements in H/Zp and gave conditions for their existence, as well as some examples. An interesting and possibly harder project is to look at the structure of O/Zp, where O is the octonion division algebra, and discuss idempotent and nilpotent elements in that finite ring. Notes 1. Recall that addition and multiplication on H/Zn are defined as follows: x + y =(a0 + a1i + a2j + a3k)+(b0 + b1i + b2j + b3k) =(a0 + b0)+(a1 + b1)i +(a2 + b2)j +(a3 + b3)k x · y =(a0 + a1i + a2j + a3k) · (b0 + b1i + b2j + b3k) = a0b0 +(n − 1)a1b1 +(n − 1)a2b2 +(n − 1)a3b3 + (a0b1 + a1b0 + a2b3 +(n − 1)a3b2)i + (a0b2 +(n − 1)a1b3 + a2b0 + a3b1)j + (a0b3 + a1b2 +(n − 1)a2b1 + a3b0)k 666 M. Aristidou and A. Demetre 2. Over H the same equation holds. See Exercise 37(g) (p.214) in [6]. x x 1 a a i a j a k N x 3. If is idempotent, then = 2 0 + 1 + 2 + 3 , with ( ) = 0. This is essentially our Theorem 2.5 in [2] (putting p = 0). Over H the same equation holds. See Exercise 37(g) (p.214) in [6]. References [1] M. Aristidou, A. Demetre, A Note on Quaternion Rings over Zp, Interna- tional Journal of Algebra, Vol.3 (2009), No.15, 725-728. [2] M. Aristidou, A. Demetre, Idempotent Elements in Quaternion Rings over Zp, International Journal of Algebra, Vol.6 (2012), No.5-8, 249-254. [3] N. J. Fine, I. N. Herstein, The Probability that a Matrix be Nilpotent, Illinois Journal of Mathematics, Vol.2 (1958), No.4A, 499-504. [4] I. N. Herstein, Topics in Algebra, 2nd ed., Wiley, New York, 1975. [5] C. J. Miguel, R. Serodio, On the Structure of Quaternion Rings over Zp, International Journal of Algebra, Vol.5 (2011), No.27, 1313-1325. [6] K. Nicholson, Introduction to Abstract Algebra, 2nd ed., Wiley, New York, 1999. [7] R. S. Pierce, Associative Algebras, Springer, New York, 1982. [8] R. Remmert et all, Numbers, Springer, New York, 1991. Received: January, 2012.