International Journal of Algebra, Vol. 6, 01, no. 5, 9-5 Idemotent Elements in Quaternion Rings over Z Michael Aristidou American University of Kuwait Deartment of Science and Engineering P.O. Box 333, Safat 1303, Kuwait maristidou@auk.edu.kw Andy Demetre Seattle University Deartment of Mathematics 901 1th Ave., P.O. Box 000 Seattle, WA 981-1090, USA demetra@seattleu.edu Abstract In this aer, we discuss idemotent elements in the finite ring H/ Z. The number of idemotents in H/Z was found recently in []. We rovide examles and we establish conditions for idemotency in H/Z. Mathematics Subject Classification: 15A33, 15A30, 0H5, 15A03 Keywords: quaternion, ring, idemotent 1. Introduction The quaternions, denoted by H, were first invented by W. R. Hamilton in 183 as an extension of the comlex numbers into four dimensions [6]. Algebraically seaking, H forms a division algebra (skew field) over R of dimension ([6],.195-196). In [], we studied the finite ring 1 H/Z, where is a rime, looking into its structure and some of its roerties. A more detailed descrition of the structure H/Z was given recently by Miguel and Serodio in []. Among others, they found the number of zero-divisors, the number of idemotent elements, and rovided an interesting descrition of the zero-divisor grah. In articularly, they showed that the number of idemotent elements in H/Z is ++, for odd rime. In the sections that follow, we examine idemotent elements in H/Z and rovide conditions for idemotency in H/Z.
50 M. Aristidou and A. Demetre. Idemotent Elements in H/Z Recall that an element x in a ring R is called idemotent if x = x. In the ring H/Z, rime, an element x is of the form: x = a 0 + a 1 i + a j + a 3 k where a i Z, rime, and i = j = k = 1= 1. In the secial case where x = a 0, a 0 0, (i.e., x is a non-zero scalar in H/Z ), one quickly observes that if x is idemotent, then x = 1, for x {1,,..., 1}, since (x, ) = 1. Therefore, the only scalar idemotent in H/Z is 1. (We ommit the case x = 0 as trivial). Another simle case is the case where x = ai, aj or ak, a 0 (i.e., a non-zero scalar multile of the imginary units). Then, x =(ai) = a i = a ai = x, which shows that there are no idemotents of the form ai, aj or ak. (Again, we ommitted the case x = 0 as trivial). The next non-simle cases are catured by the following Proositions and Theorem. Proosition.1: Let x H/Z,, be of the form x = a 0 + a 1 i.ifx is idemotent, then x is of the form x = +1 + 1 i. Proof. We have (a 0 + a 1 i) =(a 0 + a 1 i)(a 0 + a 1 i)=a 0 +a 0a 1 i a 1 =(a 0 a 1 )+a 0a 1 i. Since x is idemotent, the RHS of the last equation must equal a 0 + a 1 i. Hence, we have: a 0 a 1 = a 0 (1) a 0 a 1 = a 1. () From (), we get a 0 a 1 = a 1 a 0 =1 a 0 = +1. From (1), we get a 0 a 1 = a 0 (+1) a 1 = (+1) a 1 = (+1) (+1) a 1 = 1. To see if 1 is a square mod, we calculate the Legendre Symbol ( multilicative roerty of the Legendre Symbol gives us ( But, ( 1/ ) = 1, since 1/ =(1/). Hence: 1 ( 1 )=( 1 1 ). The )=( 1 )( 1/ ). { )=( 1) 1 =( 1) 1 1, if 1(mod) = 1, if 3(mod). Therefore, there are no idemotents of the form a 0 + a 1 i,if 3(mod). Elements of the form a 0 + a 1 i are idemotent if 1(mod) and, in that case, a 0 = +1 and a 1 = 1.
Idemotent elements in quaternion rings 51 Remark.: For =, notice that (1) and () in the roof above imly that an idemotent must satisfy a 1 = 0 and a 0 = 1. Hence, x = 1 is the only idemotent element in this case. For = 3, clearly there are no idemotents of the form a 0 + a 1 i. An idemotent not of that form though is + i + j. Examle.3: Let = 13. Then, a 0 = +1 = 7 and a 1 = 1 == 3(mod13). So, a 1 = or 9. Therefore, the idemotents are 7 + i and 7 + 9i. Proosition.: Let x H/Z be of the form x = a 1 i + a j + a 3 k. Then, x is not an idemotent. Proof. We have that: (a 1 i + a j + a 3 k) =(a 1 i + a j + a 3 k)(a 1 i + a j + a 3 k) = a 1 + a 1a k a 1 a 3 j a a 1 k a + a a 3 i + a 3 a 1 j a 3 a i a 3 = (a 1 + a + a 3)+(a a 3 a a 3 )i +(a 1 a 3 a 1 a 3 )j +(a 1 a a 1 a )k = (a 1 + a + a 3 )+0i +0j +0k = (a 1 + a + a 3 ). As the outcome of the above is a scalar (instead of a 1 i + a j + a 3 k), we clearly have that x is not an idemotent. The above Proosition hels obtaining the following Theorem which generalizes Pro..1. Theorem.5: Let x = a 0 + a 1 i + a j + a 3 k H/Z. idemotent if a 0 = +1 and a 1 + a + a 3 = 1. Proof. We have that: Then, x is (a 0 + a 1 i + a j + a 3 k) =(a 0 + a 1 i + a j + a 3 k)(a 0 + a 1 i + a j + a 3 k) = a 0 + a 0a 1 i + a 0 a j + a 0 a 3 k + a 0 a 1 i a 1 + a 1a k a 1 a 3 j + a 0 a j a 1 a k a + a a 3 i + a 0 a 3 k + a 1 a 3 j a a 3 i a 3 = a 0 a 1 a a 3 +a 0 a 1 i +a 0 a j +a 0 a 3 k. As the last ortion of the above must equal a 0 + a 1 i + a j + a 3 k (if one is to have idemotency), then we get that: a 0 = a 0 a 1 a a 3 (1) a 1 =a 0 a 1 () a =a 0 a (3) a 3 =a 0 a 3. ()
5 M. Aristidou and A. Demetre Equations (), (3) and () imly: a 1 = 0 or a 0 =1 a = 0 or a 0 =1 a 3 = 0 or a 0 =1 from which (last art) we derive that a 0 = +1. Using this fact in (1), we also get: +1 a 1 + a + a 3 = ( +1) = ( +1) a 1 + a + a 3 = 1. a 1 a a 3 +1 Remark.6: Just as in Rem.., for =, (1) and () in the roof above imly that an idemotent must satisfy a 1 = 0 and a 0 = 1. Hence, x = 1 is the only idemotent element in this case. Examle.7: Let = 5. Then, a 0 = +1 = 3 and a 1 + a + a 3 = 1 = 6=1(mod5). One solution is a 1 =5,a =11,a 3 = 5. Therefore, an idemotent is 3 + 5i +11j +5k. Another is 3 + 6i +10j +15k, and basically one can find them all by solving a 1 + a + a 3 =1(mod5). Remark.8: To find the number of idemotents in H/Z one could naturally find how many ways 1 can be written as a sum of three or fewer squares. From [], we know that the number is + +, for odd rime. The equation a 1 + a + a 3 = 1 brings to mind the classical Sum of Three Squares Theorem which was roved by Gauss in his Disquisitones Arithmeticae (S.91) in 1801. 3 As that theorem says, an integer n can be the sum of three squares if and only if n m (8k +7),m,k 0. So, clearly, when n =7 one does not have solutions to the equation a 1 + a + a 3 = n. But, in our case (in this secial mod version), one does get solutions for = 7 to the equation a 1 + a + a 3 = 1. In articularly, (, 1, 3, ) is a solution and hence x =+i +3j +k is an idemotent in H/Z 7. More interestingly, we get solutions even if 1 = m (8k +7),m,k 0. For examle, for = 31: 1 = 0 = (8.1 + 7), but 0 = 09 = 3mod31 = 8 +9 +8 (and the idemotent is 16 + 8i +9j +8k). Finally, notice that + i + j is an idemotent in H/Z 3, but 1 = 3 1 =isnot the sum of three squares in Z 3.
Idemotent elements in quaternion rings 53 Conclusion We discussed idemotent elements in H/Z and gave conditions for their existence, as well as some examles. A natural question is to examine nilotent elements in H/Z, but we leave that as a future roject. Another is to study the grou structure of the units of H/Z in more detail. Miguel and Serodio already ointed out in [] that the grou can be generated by two elements. Finally, an interesting and ossibly harder roject is to look at the structure of O/Z, where O is the octonion division algebra, and discuss idemotent and nilotent elements in that finite ring. Notes 1. Recall that addition and multilication on H/Z n are defined as follows: x + y =(a 0 + a 1 i + a j + a 3 k)+(b 0 + b 1 i + b j + b 3 k) =(a 0 + b 0 )+(a 1 + b 1 )i +(a + b )j +(a 3 + b 3 )k x y =(a 0 + a 1 i + a j + a 3 k) (b 0 + b 1 i + b j + b 3 k) = a 0 b 0 +(n 1)a 1 b 1 +(n 1)a b +(n 1)a 3 b 3 + (a 0 b 1 + a 1 b 0 + a b 3 +(n 1)a 3 b )i + (a 0 b +(n 1)a 1 b 3 + a b 0 + a 3 b 1 )j + (a 0 b 3 + a 1 b +(n 1)a b 1 + a 3 b 0 )k. The Legendre Symbol ( a ) is defined as follows: ( a 1 )=a = 1, if a is a qudratic residue mod 1, if a is not a qudratic residue mod 0, if a. 3. For a roof see [8]. Also [1] for a more elementary roof. References [1] N. C. Ankeny, Sum of Three Squares, Proceedings of the AMS, Vol.8, (1957), 316-319.
5 M. Aristidou and A. Demetre [] M. Aristidou and A. Demetre, A Note on Quaternion Rings over Z, International Journal of Algebra, Vol.3, 15 (009), 75-78. [3] I. N. Herstein, Toics in Algebra, nd ed., Wiley, 1975. [] C. J. Miguel and R. Serodio, On the Structure of Quaternion Rings over Z, International Journal of Algebra, Vol.5, 7 (011), 1313-135. [5] R. S. Pierce, Associative Algebras, Sringer, 198. [6] R. Remmert et all, Numbers, Sringer, 1991. [7] R. Schafer, An Introduction to Nonassociative Algebras, Academic Press, 1996. [8] J. P. Serre, A Course in Arithmetic, Sringer,.5, 1973. Received: October, 011