The multiplicative formal group \(X+Y+XY\) is linearly reductive in \(characteristic\quad p>0\).
Let k be a field and let \(F=F(X,Y)=(F_ 1(X,Y),...,F_ n(X,Y))\), \(X=(X_ 1,...,X_ n)\), \(Y=(Y_ 1,...,Y_ n)\) be an n-dimensional formal group over k. Let \(\phi:\quad V\to Vx\) be an F-module and let \(V^ F=:\{v\in V| \phi (v)=v\}.\) (Here V is a vector space and \(\phi\) a linear map such that if \(\phi (v)=\sum_{u}\phi_ u(v)X^ u \), then \(\phi_ 0(v)=v\) and \(\sum_{u,w}\phi_ u\circ \phi_ w(v)X^ uY^ w =\sum_{s}\phi_ s(v)F(X,Y)^ s \) for all \(v\in V.)\) Definition. A formal group F is said to be linearly reductive if for each finite dimensional F-module V and \(0\neq v_ 0\in V^ F\), there exists \(f\in (V^*)^ F\) such that \(f(v_ 0)\neq 0\). (Here \(V^*\), the dual vector space of V, is viewed as an F-module.) Main result: Theorem. If \(char(k)=p>0\), then the multiplicative group \(F^{(m)}=X+Y+XY=(X_ 1+Y_ 1+X_ 1Y_ 1,...,X_ n+Y_ n+X_ nY_ n)\) is linearly reductive for all \(n\geq 1\). - This is proved by passing to the Hopf algebras and showing that the corresponding Hopf algebra to \(F^{(m)}\) is linearly coreductive. Corollary. If k is algebraically closed, then every finite dimensional \(F^{(m)}\)-module is a direct sum of one-dimensional submodules. - The proof rests on the criterion that F is linearly reductive if and only if every finite dimensional F-module is a direct sum of simple submodules [\textit{A. Tyc}, J. Algebra 101, 166-187 (1986; Zbl 0603.14031)], and on the observation that every finite dimensional simple \(F^{(m)}\)-module is one-dimensional.
scientific article; zbMATH DE number 4095647
scientific article