Linear reductivity is equivalent to vanishing of first Hochschild cohomology

Suppose first that $H^1(G,M)=0$ for every finite-dimensional representation $M$. For representations $E,F$, define $g\cdot a=r_F(g)a{r}_E(g{)}^{-1}$ on ${\mathrm{Hom}}_k(E,F)$ after every base change. This is a natural linear action satisfying the group law, hence a representation in the Given convention and a $G$-module of [F2]; its invariant vectors are precisely the equivariant maps. Let $0\to V^{\mathrm{\prime }}\to V\to V^{\mathrm{\prime }\mathrm{\prime }}\to 0$ be an exact sequence of finite-dimensional representations. Applying ${\mathrm{Hom}}_k(V^{\mathrm{\prime }\mathrm{\prime }},-)$ gives an exact sequence of these representations, since a vector-space surjection splits linearly. Applying [F1] gives the exact sequence $H^0(G,\mathrm{Hom}(V^{\mathrm{\prime }\mathrm{\prime }},V))\to H^0(G,\mathrm{Hom}(V^{\mathrm{\prime }\mathrm{\prime }},V^{\mathrm{\prime }\mathrm{\prime }}))\stackrel{\delta }{\to }{H}^1(G,\mathrm{Hom}(V^{\mathrm{\prime }\mathrm{\prime }},V^{\mathrm{\prime }}))$. The identity of $V^{\mathrm{\prime }\mathrm{\prime }}$ is a $G$-fixed element of $\mathrm{Hom}(V^{\mathrm{\prime }\mathrm{\prime }},V^{\mathrm{\prime }\mathrm{\prime }})$, and its image under $\delta$ lies in $H^1(G,\mathrm{Hom}(V^{\mathrm{\prime }\mathrm{\prime }},V^{\mathrm{\prime }}))=0$; hence the identity lifts to a $G$-fixed element of $\mathrm{Hom}(V^{\mathrm{\prime }\mathrm{\prime }},V)$, i.e. to a $G$-equivariant splitting of the sequence. Every short exact sequence of finite-dimensional representations splits, so every such representation is a direct sum of simple representations and $G$ is linearly reductive.

Conversely suppose $G$ is linearly reductive and let $f:G\to V$ be a natural $1$-cocycle. On $W=V\oplus k$ define $g\cdot (v,a)=(gv+af(g),a)$, functorially on every base algebra. The cocycle identity [F3] proves the action law; $f(e)=0$ follows by evaluating the identity at $e$, so the identity acts trivially. The entries are regular because $f$ is natural, hence a scheme morphism by Yoneda. This is a finite-dimensional rational representation and fits into $0\to V\to W\to k\to 0$. Linear reductivity gives a $G$-equivariant splitting, whose value at $1$ is $(m,1)$. Its invariance means $gm+f(g)=m$, so $f(g)=m-gm$ is the coboundary of $-m$. Thus every $H^1$ class vanishes. This argument preserves the full general-group Statement.

Step1.1 proves vanishing implies linear reductivity, and step1.2 proves the converse through an explicit finite-dimensional cocycle representation. Thus the equivalence holds, with no use of affine free-comodule effacement for a nonaffine group.