I'm reading the paper How to use finite fields for problems concerning infinite fields, by Serre. In Theorem 1.2 on page 1, he says
Let $G$ be a finite $p$-group acting algebraically...
In the context of actions, what is the meaning of the word "algebraically"?
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This means that, under the action, each element $g\in G$ induces an "algebraic" map $\mathbb{A}^n\to\mathbb{A}^n$, i.e. a map that consists of polynomials. See the Wikipedia page Morphism of varieties.
On
Since the group acts on a variety, the action being algebraic means that for each $g\in G$, the map $X\to X$ given by $x\mapsto g.x$ is a morphism of varieties.
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The map $\mathbb{A}^n\to\mathbb{A}^n$ defined by each $g$ should be an algebraic map, i.e. defined by a polynomial equation.
The action is defined by a map $G\to\operatorname{Aut}(\mathbb{A}^n)$, where the automorphism group is taken in the category of algebraic varieties. There would be more ambiguity if one replaced $\mathbb{A}^n$ by $\mathbb{C}^n$; then $\mathbb{C}^n$ could be considered (for example) as either an algebraic variety or as a topological space (with the Euclidean topology), and the group $\operatorname{Aut}(\mathbb{C}^n)$ would be different in each case.
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Saying that a group acts algebraically is analogous to saying that it acts continuously, acts smoothly, etc.
It specifies what type of map the action morphism is intended to be, as a map $G \times X \to X$. (Or equivalently, when $G$ is finite, what type of endomorphism $\varphi(g) \colon X \to X$ each $g \in G$ induces.)
More naturally, the action of an algebraic group $G\times X\rightarrow X$ on an algerbaic variety is called algebraic, if $(g,x)\mapsto g.x$ is a morphism satisfying the two usual axioms $e.x=x$ and $(gh).x=g.(h.x)$. Every finite group is algebraic, so this makes sense.