This question arose from [my last question](https://mathoverflow.net/questions/144880/equivariant-normalization), 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.



When does a group action on a k-algebra induce an algebraic action on the spectrum?
If I understand the question correctly, you have a map of schemes $G \times X \to X$, and the corresponding map of $k$-points is a group action (meaning that the obvious two maps $G(k) \times G(k) \times X(k) \to X(k)$ coincide), but you are not sure that it is a group action in the category of schemes. In other words, you fear that you may have two maps $G \times G \times X \to X$ which coincide on $k$ points but not as maps of schemes.


This certainly can't happen if $G$ and $X$ are reduced. So, if you are talking about varieties, there is no issue. It's not obvious to me what happens when $G$ is reduced (which is automatic in characteristic zero) but $X$ isn't.




---


Based on comments below, and on reading the motivating question, I didn't understand right. The question is, given an action $G(k) \times X(k) \to X(k)$, so that $g \times X(k) \to X(k)$ is algebraic for every $k \in G(k)$, can we conclude that it comes from an algebraic map $G \times X \to X$. But I don't think there is any good way to force this. For example, suppose that $k$ has a nontrivial automorphism $\sigma$ and $G$ is defined over the fixed field of $\sigma$. (Think of complex conjugation.) Then $\sigma$ induces an automorphism of $G(k)$ as an abstract group. Take any algebraic action $G \times X \to X$ and compose with the automorphism of $k$ to get a very nonalgebraic action of $G(k)$ on $X(k)$. 


