1. Introduction. Let S/R be an extension of commutative rings always assumed to be associative and possessing identity elements. Let G be a finite group of R-algebra automorphisms of S such that R = S G := {s ∈ S | g.s = s, g ∈ G}. Recall that an R-basis for S is called normal , with respect to G, if it is the G-orbit of some element s in S. In that case s is called a normal basis generator. Normal bases do not always exist. In fact, by a result of Noether [18] , if S/R is a finite extension of

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... Dedekind domains, G = Aut R (S) and R is a discrete valuation ring, then the extension has a normal basis precisely when it is tamely ramified. On the other hand, if R is semilocal and S/R is a Galois ring extension with finite group G, that is, if S/R is a separable ring extension and for all g, g ′ ∈ G, g = g ′ , and all nonzero idempotents e ∈ S, there is s ∈ S such that (g.s)e = (g ′ .s)e, then the extension always has a normal basis (see [5] ). In particular, a finite Galois field extension always has a normal basis. A Galois ring extension S/R with finite group G is called odd if the order of G is odd. The trace function tr S/R : S → R, defined by tr S/R (s) = g∈G g.s for all s ∈ S, induces a symmetric bilinear form q S : S × S → R by the relation q S (s, s ′ ) = tr S/R (ss ′ ) for all s, s ′ ∈ S. The bilinear form q S is also a G-form, that is, it is invariant under the action of G. If a normal basis {g.s | g ∈ G} is self-dual with respect to q S , that is, if q S (g.s, g.s) = 1 and q S (g.s, g ′ .s) = 0 if g = g ′ for all g, g ′ ∈ G, then it is called a self-dual normal basis and s is called a self-dual normal basis generator. Note that the existence of such a basis can alternatively be formulated by saying that (S, q S ) and (R[G], q 0 ) are isomorphic as G-forms, where q 0 is the unit Gform, that is, the R-bilinear map R[G] × R[G] → R defined by q 0 (g, g) = 1 and q 0 (g, g ′ ) = 0 if g = g ′ for all g, g ′ ∈ G. The problem of when a self-dual