Geometrically reductive algebraic group

For every finite-dimensional rational representation $V$ and every nonzero $v\in V^G$, a geometrically reductive algebraic group admits a homogeneous invariant polynomial $F\in k[V{]}^G$ of positive degree with $F(v)\ne 0$. A linearly reductive group can take degree one. In positive characteristic, higher degree is essential: a power $p^r$ can separate a fixed vector even when an invariant linear functional cannot.

Choose a torus-invariant functional nonzero on a fixed vector, and map the representation to right-torus-invariant matrix coefficients on $S{L}_2$. These functions lie in a finite filtration by polynomials of bidegree $(n,n)$ in the two matrix columns. For $n=q-1$, all binomial coefficients are nonzero in characteristic $p$, so the invariant tensor given by the determinant to power $n$ induces an invertible map from the symmetric power's dual to itself. The filtration piece becomes its endomorphism space, with the constant function corresponding to the identity. The endomorphism determinant is an invariant degree-$q$ polynomial equal to one there; composing it with the original representation map proves geometric reductivity.