Abstract.Let E and F be Riesz spaces and Tx, Ti, .. ■ , Tn be linear lattice homomorphisms (henceforth called lattice homomorphisms) from E to F . If T = Yl/!=\ T¡, then it is easy to check that T is positive and that if xq , xx, ... xn € E and x, A jc, = 0 for all i / j , then /\/Lo Tx¡ = 0 . The purpose of this note is to show that if F is Dedekind complete, the above necessary condition for T to be be the sum of n lattice homomorphisms is also sufficient. The result extends to sums of disjointness preserving operators, thereby leading to a characterization of the ideal of order bounded operators generated by the lattice homomorphisms.Throughout this note we assume that E and F are real Riesz spaces. For unexplained terminology we refer the reader to references [1,6,8,9]. Definition 1. Let « be a positive integer and T be a linear operator from E to F. We say that T is «-disjoint if T is order bounded, and for all Xq , xx, ... , xn G E such that \x¡\ A \Xj\ = 0 for all i ^ j, we have /\"=01Tx¡\ = 0.Clearly a 1-disjoint positive linear operator from £ to F is a lattice homomorphism, and a general 1-disjoint linear operator is order bounded and disjointness preserving (i.e., if |w| A |u| = 0, then \Tu\ A \Tv\ = 0). We can readily check that every (positive) linear operator on K" is the sum of n (lattice homomorphisms) disjointness preserving operators. To see this take the matrix representation relative to the standard basis of W (or any positive pairwise disjoint basis if greater generally is wanted), and write the matrix as the sum of n matrices each with at most one nonzero column.Before stating our first result we recall some facts about disjointness preserving operators. These can be found in [2,4]. Note, however, that the term disjointness preserving in [2] has a different definition and that [2] considers self-maps of E rather than maps from E to F . The relevant proofs in [2] all take place in the range of the operator and apply without change. The results we need are given by the following: -