This question arose from my last question, which I considered answered - from the comments, however, it is obvious that the answer is only complete in characteristic zero, and I am trying to understand why.

Let $A$ be a $\Bbbk$-algebra¹, where $\Bbbk$ is some algebraically closed field. Let $G$ be a reductive algebraic group. If $G$ acts algebraically on $X=\newcommand{\Spec}{\mathrm{Spec}}\Spec(A)$, then it induces an action of $G$ on $A$.

On the other hand, assume that $G$ acts on $A$ by $\Bbbk$-algebra automorphisms. For any maximal ideal $\mathfrak m\subset A$ and any $g\in G$, the image $g.\mathfrak m$ is again a maximal ideal. This defines an action of $G$ on the closed points of $X$ and for any $f\in A$, note that the image of $f$ in $A/g.\mathfrak m=\Bbbk$ is the same as the image of $g^{-1}.f$ in $A/\mathfrak m=\Bbbk$, $$\begin{matrix} g^{-1}.f & \in & A & \xrightarrow{\quad g\quad} & A & \ni & f \\ && \downarrow && \downarrow && \\ g^{-1}.f+\mathfrak m & \in & A/\mathfrak m & \xrightarrow{\quad \sim\quad} & A/g.\mathfrak m & \ni & f+g.\mathfrak m \end{matrix} $$

so if $\mathfrak m$ is viewed as a closed point of $X$, we have the familiar formula $(g^{-1}.f)(\mathfrak m) = f(g.\mathfrak m)$. This is good, but I forgot to ask myself (and now I am asking you):

When is this action algebraic?

By this, I mean that there is a morphism $G\times X\to X$ of $\Bbbk$-schemes (or varieties) which gives the above action on closed points.

The first assumption should probably be that each $f\in A$ is contained in a finite-dimensional $G$-module. because this property holds when the action on $A$ comes from an algebraic action on $X$. On the other hand, I suspect that one will need at least characteristic zero (or more generally, some separability condition). However, I don't know exactly how to put this together.

¹ You may assume $A$ finitely generated over $\Bbbk$ and reduced, or even a domain, but I have a feeling that it won't matter much whether we deal with varieties or $\Bbbk$-schemes.