How does a (finite) group act on a(n affine) scheme? I understand what it means for a group scheme to act on a scheme, but I don't understand what it means for a finite group to act on a scheme. I am sure it doesn't mean that the group acts on the scheme as though it were merely a set.
I have some ideas about what it could be. You can associate to a group the constant group scheme over S (say all these schemes are over a base S). Then you can act on a scheme by the associated constant group scheme. Another thing you can do if the group is finite is build a group scheme (actually, I believe this is the constant group scheme) associated to the trivial action of the fundamental group on the finite group. Now you can act by that. Is my hunch true?
Even if this is the case, I do not know how to interpret some descriptions of group actions. For example, lets say I take the line with two origins. $\mathbb Z/2\mathbb Z$ "acts" on this scheme by sending $y$ to $-y$. Geometrically this seems reasonable enough, but I do not know how to write down an actual group scheme $G$ that acts on the line of two origins through an explicit action. What is this description a shorthand for?
(Now that I have everyone's attention: what is a finite group scheme $G\to S$? Does finite simply mean that $G$ is finite over $S$ in the sense of schemes, or that somehow I can associate some kind of finite group to $G$?)
 A: You can talk about a group acting on an object in any category.  If $X$ is an object in a category, then an action of a group $G$ on $X$ is a homomorphism $G\to \operatorname{Aut}(X)$ where $\operatorname{Aut}(X)$ is the group of automorphisms of $X$.  So if your category is schemes (over some base), this just means that each element of the group "acts" on your scheme by some automorphism of the scheme, and this turns the group operation into composition of automorphisms.  There's no need to involve group schemes.
That said, you can formulate it as a group scheme action if you like.  As you suggest, if $G$ is a group, then there is a constant group scheme $G_S$ associated to $G$ over any base $S$.  Explicitly, this group scheme is just a disjoint union of copies of $S$ indexed by the elements of $G$, with the group operations defined using the group operations of $G$ on the index set of the copies of $S$.  For any scheme $X$ over $S$, the product $G_S\times_S X$ is just a disjoint union of copies of $X$ indexed by $G$, so a morphism $G_S\times_S X\to X$ is the same as giving a morphism $X\to X$ for each element of $G$.  It is then straightforward to verify that such a morphism is an action of $G_S$ on $X$ iff all of the morphisms $X\to X$ are automorphisms and form an action of $G$ on $X$ in the sense of the first paragraph.
If you say $\mathbb{Z}/2\mathbb{Z}$ acts on $X$ by "sending $y$ to $-y$", this means you are considering the homomorphism $\mathbb{Z}/2\mathbb{Z}\to \operatorname{Aut}(X)$ which sends the nontrivial element of $\mathbb{Z}/2\mathbb{Z}$ to the automorphism that "sends $y$ to $-y$".  Presumably in this case that would refer to the automorphism which on each copy of $\mathbb{A}^1_S=\operatorname{Spec}\mathbb{Z}[t]\times S$ in the line with two origins is the morphism corresponding to the automorphism $t\mapsto -t$ of the ring $\mathbb{Z}[t]$.
(A finite group scheme over $S$ is just a group scheme that is also finite in the sense of schemes, i.e. it is a finite morphism $G\to S$.  Sometimes it is also required to be a flat morphism.)
