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We prove a strong version of a theorem of Balcerzak-RoslanowskiShelah by showing, in ZFC, that there exists a simply definable Borel CT-ideal for which both the ccc and property (M) fail. The proof involves Polish group actions. Definition. A Borel ideal is an ideal I on a Polish space X with the following property: for all A E I, there exists a Borel subset B of X such that A C B and B E I. We consider three types of properties for a Borel ideal I on a Polish space X. (1) I has a H' definition if the following set is Hl: C.1 = {c c 2': c is a Borel code for a subset B, of X and B, c I}. (2) For K a cardinal, I satisfies the /c-cc if there does not exist a family of i I-almost disjoint Borel sets that are not in I. The w1-cc is usually called the ccc. (3) I satisfies property (M) if there exists a Borel-measurable function f : X -? 2W with f-1(y) ? I for all y E 2W. Property (M) was introduced in Balcerzak [1]. Obviously, an ideal satisfying property (M) violates the ccc. Both the above paper and the later paper of Balcerzak-Roslanowski-Shelah [2] are concerned with circumstances in which it is possible for both the ccc and property (M) to fail. We refer the reader to these two references for information on these properties-and several other properties of ideals. The latter paper contains the following result. Theorem 1 (Balcerzak-Roslanowski-Shelah [2, 5.6]). Assume that either CH fails or every Al set of reals has the Baire property. Then there exists a Borel a-ideal I*, containing all singletons, such that: (a) I* has a H1 definition; (b) the ccc fails for I*; (c) property (M) fails for I*; (d) I* satisfies the w2-cc. Andrzej Roslanowski presented this result in a seminar talk at Ohio State University in 1993, and asked whether or not it is provable in ZFC. I thank him for bringing this question to my attention, and I thank Ohio State for their support. Received by the editors October 23, 1998 and, in revised form, December 3, 1998. 2000 Mathematics Subject Classification. Primary 03E15.

Ideals without ccc and without property $ \boldsymbol( \mathbf{M} \boldsymbol)$