Let S be a cancellative monoid with quotient group of torsion-free rank a.We show that the monoid ring R[S] is a Hilbert ring if and only if the polynomial ring R[{ X, },s/] is a Hilbert ring, where |/| = a.Assume that R is a commutative unitary ring and G is an abelian group.The first research problem listed in [K, Chapter 7] is that of determining equivalent conditions in order that the group ring R[G] should be a Hilbert ring.Matsuda has considered this question in [Ml, 6].His results show that for G finitely generated, R[G] and R are simultaneously Hilbert rings; if G is not finitely generated, then R[G] a Hilbert ring implies that R is a Hilbert ring, but the converse fails.In [M2, 5] Matsuda considers briefly the corresponding Hilbert-ring-characterization problem for a monoid ring R[S], where S is torsion-free and cancellative.Let a be the torsion-free rank of G and let {Xj)lsl be a set of indeterminates over R of cardinality a.In Corollary 1 we show that R[G] and P[{^,},e/] are simultaneously Hilbert rings.While the problem of determining equivalent conditions for P[{Ar,}/e/] to be a Hilbert ring has not been completely resolved, it has been worked on extensively [Kr, Go, L, Gl, H], and a significant body of positive results exists concerning this problem.The proof of Theorem 1 suggests the following conjecture: if S is a cancellative commutative monoid with quotient group G, then R[S] is a Hilbert ring if and only if R[G] is a Hilbert ring; this result is established in Theorem 2.All monoids considered are assumed to be commutative, and rings are assumed to be commutative and unitary.The statement of Theorem 1 uses the following terminology.For a cardinal number a, an extension ring T of R is said to be a-generated over R if T is generated over R by a set of cardinality at most a.Theorem 1. Assume that X = {XA is a set of indeterminates of cardinality a over the ring R. Denote by Za the direct sum of a copies of the additive group Z of integers.The following conditions are equivalent.(1) R[X] is not a Hilbert ring.(2) There exists a prime ideal P of R such that R/P admits an a-generated extension ring that is a G-domain, but not afield.(3) The group ring R[Za] is not a Hilbert ring._