Journal of Mathematical Sciences, VoL 75, No. 3, 1995 ON QUATERNION ALGEBRAS A.A. Tuganbaev UDC 512.55 It is proved that the distributiveness of the right ideals lattice for a quaternion algebra over a commutative ring A is equivalent to the following property: the equation x 2 -t- y2 + z 2 = 0 is uniquely solvable in the field AIM for any maxima/ideals M of A, the lattice of the ideals of A being distributive. Bibliography: 5 titles. All rings are assumed to be associative and with a nonzero identity, and all modules are unital. A module is called distributive (uniserial) if the lattice of its submodules is distributive (is a chain). A quaternion algebra Q(A) over a commtative ring A is the free A-module with free generators 1,i, j, k in which multiplication is defined by the rules: 1 is the common identity of rings Q(A) and A, i 2 = j2 = k 2 = -1, ij = -ji = k, jk = -kj = i, ki = -ik = j. It is known [1, Section 1.6] that if F is a field, then Q(F) is a skew field if and only if the equation z 2 + y2 + z 2 = 0 has the unique solution x = y = z = 0 in F. The main result of the present work is theorem i. Theorem 1. Let Q be a quaternion algebra over a commutative ring A. Then the following conditions are equivalent: (a) Q is a right distributive ring; (b) for each maximal ideal M of A a quaternion algebra over a localization AM is a right uniserial ring; (c) for each maxima] ideal M of A a localization AM is a uniserial ring and the equation x 2 + y2 + z 2 = 0 has the unique solution z = y = z = 0 in a field AM/J(AM); (d) A group ring A[G] over a quaternion group G is a right distributive ring; (e) A is a distributive ring and for each its maximal ideal M the equation z 2 + y2 + z 2 = 0 has the unique solution z = y = z = O in a field A/M. The Jacobson radical, the center, the group of invertible elements, and the set of all maximal ideals of a ring R are denoted by C(R), U(R), and max(R) respectively. If A is a unitary subring of C(R), M E max(A), then for each module NR its module of quatients with respect to the multiplicative subset A\M of R, regarded as a RM-module, is denoted by NM. A quaternion group is a group with 8 elements, with the generators a, b, and with defining relations a 4 -~- 1 , a 2 -- b 2, b-Zab = a -1. A ring will be called special if the equation z 2 + y-~ + z 2 = 0 has the unique solution z=y=z=0init. Lemma 1. For a module MA the following conditions are equivalent: (a) M is a distributive module; (b) for all elements m, n E M there exist elements a, b E A such that 1 = a + b, ma E mAN nA, nb E mA N rib. Proof. (a)=C,(b). Let f = re+n, T = mANnA. Since fA = fANmA+fANnA, there are elements b, d E A such that fbEmA, fdEnA, f=fb+fd. Thennb=fb-mbET, md=fd-ndET. Let a= 1-b,z=a-d=l-b-d. Then l=a+b, fz=f-fb-fd=O. Thereforema=md+mz=md+fz-nz=md-nz, andnz=-mzET. Then ma E T, and a, b are desired elements. (b)::#(a). It is sufficient to prove that if F, G, H are submodutes of M, f E FN(G+H) then f E FNG+FNH. Let f = m + n, where m E G, n E H. By our condition there are elements a, b E A such that 1 = a + b, ma E nA, nb E mA. Then fa = ma + na E fANnA C_ FflH, fb = mb+ nb E lAMInA C_ FNG, f = fb+ fa E FNG + FNH. Lemma 2. Let R be a right distributive ring. Then: (a) if R/J(R) is a simple artinian ring, then R is a right uniserial ring and in particular R/J(R) is a skew field; (b) ifA is a unitary subring of R and a module AA is isomorphic to a direct summand of a module AR, then A is a ring distributive ring; (c) if A is a unitary subring of R, and a module AR is free, then A is a right distributive ring. Proof. Part (a) was proved in [2]. Translated from Trudy Seminara imeni I. G. Petrovskogo, No. 17, pp. 209-214, 1994. Original article submitted June 11, 1990. 1750 1072-3374/95/7503-1750512.50 9 Plenum Publishing Corporation