This conjecture is known to be true when the base field is of characteristic zero. In fact we can then find a linear F and this is an immediate consequence of the complete reducibility of finite dimensional rational G-modules (theorem of H. Weyl). In this paper we present an attempt at the solution of the above conjecture. We have partial success. To state our main result, one first observes (cf. Remark 1.1) that (A) is equivalent to the following stronger assertion.

Quotient Spaces Modulo Reductive Algebraic Groups

Let G be a semi-simple algebraic group over an algebraically closed field, k. Let G act rationally by automorphisms on the finitely generated k-algebra, R. The problem of proving that the ring of invariants, RG, is finitely generated originates with the invariant theorists of the nineteenth century. When k = C, the complex numbers, and G GL (n, C) the question is answered affirmatively by Hilbert's "fundamental theorem of invariant theory". The proof involved constructing a G equivariant projection from R to RG and then using it to prove the result algebraically. When k is of characteristic 0 and G is any semi-simple group, by a theorem of H. Weyl, every finite dimensional representation of G is completely reducible. In the 1950's D. Mumford and others (Cartier, Iwahori, Nagata) applied Weyl's theorem to construct a projection from R to RG for any semi-simple group. This made it possible to generalize Hilbert's proof to an arbitrary semi-simple group. Certain geometric applications, particularly to the theory of moduli, made a generalization to groups over fields of positive characteristic highly desirable. In positive characteristic, complete reducibility definitely fails. Hence attempts were made to replace complete reducibility with a weaker condition which would at once hold for all semi-simple groups and make a proof of finite generation of RG possible. The weakest way to state complete reducibility is the following. If V is a finite dimensional G-module containing a G-stable sub-space of co-dimension one, VT, then there is a G-stable line LC V such that V0 E L = V. Mumford conjectured a weaker version of this statement by seeking a complement only in a higher symmetric power of V, SI( V). This is the conjecture as it is stated in the preface to [16]:

https://www.jstor.org/stable/pdfplus/1970974.pdf

Reductive groups are geometrically reductive

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Note on semi-reductive groups

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Note on semi-reductive groups

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Quotient Spaces Modulo Reductive Algebraic Groups

Reductive groups are geometrically reductive

Induced modules and affine quotients

Decomposition of exterior and symmetric powers of indecomposable Z/pZ-modules in characteristic p and relations to invariants

Hilbert's Fourteenth Problem

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