Abstract. In this note we show that if two Toeplitz operators on a Bergman space commute and the symbol of one of them is analytic and nonconstant, then the other one is also analytic.Let Ω be a bounded open domain in the complex plane and let dA denote area measure on Ω. The Bergman space L 2 a (Ω) is the subspace of L 2 (Ω, dA) consisting of the square-integrable functions that are analytic on Ω. For a bounded measurable function ϕ on Ω, the Toeplitz operator T ϕ with symbol ϕ is the operator on L 2 a (Ω) defined by More general results concerning which operators, not necessarily Toeplitz, commute with an analytic Hardy space Toeplitz operator are due to Thompson ([10] and [11]) and Cowen [6]. On the Bergman space, the situation is more complicated. The Brown-Halmos result mentioned above fails. For example, if Ω is the unit disk, then any two Toeplitz operators whose symbols are radial functions commute (proof: an easy calculation shows that every Toeplitz operator with radial symbol has a diagonal matrix with respect to the usual orthonormal basis; any two diagonal matrices commute).