The power series ring over an Ore domain need not be Ore

Abstract

A ring R satisfies the right Ore condition iff for all nonzero divisors, r, s E R, rR f? sR f 0. If5 in addition to satisfying the right Ore condition, R has no zero divisors, then R is said to be a right Ore domain. Ore’s famous theorem states that R satisfies the right Ore condition iff R has a right quotient ring. It has long been an open question whether the property of being a right Ore domain iifts to the power series ring. This is known to hold if the right Ore domain is right Noetherian or if it satisfies a polynomial identity. In this paper we construct a right Ore domain R such that its power series ring Rijtj fails to satisfy the right Ore condition. It is well known that all polynomial rings and matrix rings over a right Ore domain satisfy the right Ore condition (see [3, p. 439j). To gain insight into the construction of our example, we next review the proof for the polynomial ring R[t] over a right Ore domain R. Let Q be the quotiem division ring of R. Using the division algorithm we see that Q[r] is a right Ore domain. Let q(t) be an arbitrary polynomial in Q[t]. Because R satisfies the right Ore condition, the coefficients of q(f) have a common denominator. Hence q(r) has the form r(t) SC’, where r(t) E R[t], s E R. Let ,f(,rj, g(i) br polynomials in R[t]. Then, because Q[t] is a right Ore domain? there exist q!(r) = rl(tj s;‘, qz(t) = Y~(~)s;~ in Q[l] such that f(t) r,(i) ST” = g(r) P?(L) s; ‘. Using that R satisfies the right Ore condition, we immediately have S(t) R[t] 9 g(t) R[t] # 0. Hence, R[t] is a right Ore domain. A similar approach yields a proof for the satisfaction of the right Ore condition by the matrix rings over a right Ore domain. Heavily used in the preceding argument is the finite number of coefficients in q(t). In a power series ring we lose this finiteness. Our example exploits that fact. Of crucial importance for the construction of our example is Cohn’s theorem which states that every fir is embeddable in a division ring [Z]. Recall that the free product of any family of division rings over a common subdivision ring is a fir [ 11. We construct the power series ring example by