Abstract
Noether’s Problem concerns fields F(xg 1 g E G)G = F(G), where F is a field, G is a finite group, G acts by permuting the indeterminants x, in the obvious way, and F( )G refers to the field of invariant elements. In showing the F(G)‘s were nonrational for certain G, the author was led to consider multiplicative invariant fields ([SS]) which are defined as follows. Let G be a finite group and Q a G lattice. That is, Q is a finitely generated Z[G] module which is free as an abelian group. Form the group algebra F[G] and its field of fractions F(Q). G acts naturally on F(Q) and the multiplicative invariant field is F(Q)“. In [SS], it was shown that for certain Q, F(Q)G was essentially equivalent to (actually stably isomorphic to) F(G’) for G’ a split extension of G with abelian kernel. This leads one to ask whether F(G’) can be expressed as a kind of invariant field for G when G’ is a nonsplit extension of G. The answer is yes and is the core of this paper. In fact, F(G’) will be shown to be a so called “a-twisted” multiplicative invariant field of G. The goal of this paper is to present this fact and then draw a series of conclusions using it. Such a-twisted invariant fields have appeared explicitly in previous works. In [SS], these fields arise as another way of describing the invariant fields of reductive algebraic groups. Note that in [S6] the finite group is the Weyl group and the lattice is derived from the root lattice. In [S7], the unramified Brauer groups of a-twisted invariant fields are calculated.
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