Polynomial rings over Goldie rings are often Goldie

Here, we prove a result that has as a consequence the fact that if the ring R is an algebra over an uncountable field the a.c.c. on annihilators is preserved under polynomial extensions by any number of variables. Recently Jeanne Kerr [1] has given an example of a ring R with ascending chain condition on annihilators that has the property that the polynomial ring R[X] does not have the ascending chain condition on annihilators (a.c.c.). This answered a questin of long standing duration on the behavior of the classical Goldie conditions under polynomial extensions. Kerr's example is an algebra over Z2. LEMMA. Let R be a ring which contains an uncountable set V in the center of R having the property that if u, v E V and u $4 v, then u v is a nonzero divisor. Let S be a countable subring of R. Then, there is an infinite subset V' of V such that V' is algebraically independent over S. PROOF. Let V' be a maximal subset of V which is algebraically independent over S. Suppose V' is finite. Then, any v E V satisfies a nonzero polynomial in the ring (S[V'])[X]. Since there are uncountably many elements in V and only countably many polynomials, there is a polynomial f (x) with infinitely many roots in V. Let u and v be two of these. Since we may divide by monics in any ring, f (x) = (x u)h(x), but f (v) = 0 = (v u)h(v) and v u is regular so h(v) = 0, and continuing, we obtain a contradiction since the number of roots is larger than the degree of f (x). THEOREM. Let R be a ring containing an uncountable set V in the center of R, so that if u and v are distinct elements of V, u v is a nonzero divisor. Let P be a property so that (*) A ring T satisfies P if and only if every countable subring of T satisfies P (e.g., the a.c.c. on annihilators). Then R satisfies P if and only if R[X] satisfies P, where X is any set of variables. PROOF. One way is trivial. Conversely, we show R[X] satisfies P by showing that every countable subring Ro of R [X] does. Clearly, 1?o C S [XI, X2)... ] for some countable subring S of R and countable set of variables. Choose V' as in the lemma so that S[X1,X2,...] S[V'] c R. Then, since R0 C S[Xl,X2,.], R0 has P, so that by * R[X] does. Received by the editors November 11, 1985. 1980 Mathematics Subject Cksification (1985 Revision). Primary 16A34, 16A33; Secondary 16A99. (@)1986 American Mathematical Society 0002-9939/86 $1.00 + $.25 per page