CAN THE WEYL ALGEBRA BE A FIXED RING?

Abstract. If a finite soluble group acts as automorphisms of a domain, then the invariant subring is not isomorphic to the first Weyl algebra C[r, d/dt]. Let R = C[t,d/dt] be the first Weyl algebra. We prove the following result. Theorem. Let G be a finite solvable group. Let S d R be a C-algebra such that (a) SR and RS are finitely generated, (b) S is a domain, (c) G acts as automorphisms of S, and S = R. Then S = R. We will prove a rather more general result, from which the theorem follows. The original proof was improved by comments of T. J. Hodges. I would like to thank him for his interest, and for allowing his improvements to be included here. Let B bean R- /?-bimodule. We call B an invertible bimodule, if there exists another bimodule, C say, such that B®RC is isomorphic to R as a bimodule. The invertible bimodules form a group under the operation of tensor product ®R; this group is called the Picard group, denoted Pic(/Î). If o, x G A.ut(R) are C-linear algebra automorphisms of R, then we write aRT for the invertible bimodule which is R as an abelian group, and for which the right A-module action is given by b-x = bx(x) for x G R,b e aRT and the left A-module action is given by x-b = o(x)b for x G R,b e aRT. There is a map Aut(/?)- * Pic(R) given by a>- * Rx. This is a group homo-morphism. A key point in our analysis is the following result of J. T. Stafford