We construct a counterexample to settle simultaneously the following questions all in the negative: (1) Is a regular subdirect product of simple artinian rings unit-regular? (2) If R is a regular ring such that every nonzero ideal of R contains a nonzero ideal of bounded index, is R unit-regular? (3) Is a regular ring with a Hausdorff family of pseudo-rank functions unit-regular? (4) If R is a regular ring which contains no infinite direct sum of nonzero pairwise isomorphic right ideals, is R unit-regular? (5) Is a regular Schur ring unit-regular?In [1] Goodearl proposed a list of open problems on regular rings. Some involve potential sufficient conditions for a regular ring to be unit-regular. The primary aim of this paper is to construct a counterexample for the questions 6, 7, 8, 9 (second part) and 11 in GoodearΓs book.Among others the sixth question asks: Is a regular subdirect product of simple artinian rings always unit-regularΊ In [4] Tyukavkin has shown that any regular algebra over an uncountable field, which is a subdirect product of countably many simple artinian rings, is unitregular. Recently, Goodearl and Menal [2] have generalized this result by showing that any regular algebra over an uncountable field, which has no uncountable direct sums of nonzero right or left ideals, must be unit-regular; in particular, any regular algebra over an uncountable field, which has a rank function, is unit-regular. In this paper we shall construct an example of a regular ring which is a subdirect product of countably many simple artinian rings but is not unit-regular.Let F be a countable field, F[t] the ring of polynomials over F in an indeterminate t, and F{t) the quotient field of F [t]. Define an exponential valuation d on F(t) by dr{t) = +oo if r{t) = 0 and dr(t) = n if r{t) = t n f(t)/g(t) where n is an integer and f(t), g(t) e
F[t] with t \ f(t)g(t).Let V be the valuation ring associated with d, namely, V = {r(ή e F(t)\dr(t) > 0}. Note that F[t], F(t) and V are all countable. Consequently, V is a countable-dimensional vector space over F. Let VQ, v\ 9 ..., υ n ,... be a basis of V over F. First, we may assume that dVi Φ dVj for i φ j. Suppose that n is the least integer such that dv n = dvi for some i < n. Choose α f G F so that f«/^/ -α z G ίF; then 9(v Λ -α, v, ) > <9t>/. If <9(^« -α/V/) = <9^ for some j < n, then d(v n -α/V/ -CLJVJ) > dVj for some α y G /\ Continuing this process we get a t^ such that dv' n φ dVi for all i < n and that {vo,vu...,v n _ u υ' n } spans the same subspace as {vQ 9 υ\ 9 ... 9 v n -\ 9 v n } does. Next, we assume, by reordering, that dv 0 < dvi < dv 2
18 CHEN-LIAN CHUANG AND PJEK-HWEE LEEwith a k Φ 0, we see that dv = dv k . Since Vo 9 v\ 9 V2,... span the whole space V, we must have dvo = 0, dv\ = 1, 9^2 = 2 and so on.We begin by constructing a ring which is similar to that in Bergman's example [1; Example 4.26]. Let S be the set of those x G E = End/r(F) such that (JC -ά)t n V = 0 for some a G .F(ί) and some nonnegative integer n. As in [1; p. 47] we observe that a depends...