Suppose G is an algebraic group with an action G×X→X on a scheme. Does the fixed locus (the set of points x∈X fixed by all of G) have a scheme structure? You can obviously define the functor Fix(T)={t∈X(T)|t is fixed by every element of G(T)}. Is this functor always representable?


(This question was "broken off" of [a compound question of mine](https://mathoverflow.net/questions/3124/do-orbits-and-stable-loci-of-group-actions-have-natural-scheme-structures) after Scott Carnahan answered the other part so wonderfully that I had to accept his answer.)



Is the fixed locus of a group action always a scheme?
The question gives the "wrong" definition of Fix(T), hence the resulting confusion. 


A more natural definition of the subfunctor X^G of "G-fixed points in X" is  

(X^G)(T) = {x in X(T) | G\_T-action on X\_T fixes x}  

               = {x in X(T) | G(T')-action on X(T') fixes x for *all* T-schemes T'}.  

(Of course, can just as well restriction to affine T and T' for "practical" purposes.) 


By way of analogy with more classical situations, if the base is a field k then a moment's reflection with the case of finite k shows that  

{x in X(k) | G(k) fixes x}  

is the "wrong" notion of (X^G)(k), whereas  

{x in X(k) | G-action on X fixes x}  

is a "better" notion, and is what the above definition of (X^G)(k) says. 


From this point of view, if (for simplicity of notation) the base scheme is an affine Spec(k) for a commutative ring k then the "scheme of G-fixed points" exists whenever G is affine and X is separated provided that k[G] is k-free (or becomes so after faithfully flat extension on k). So this works when k is a field, or any k if G is a k-torus (or "of multiplicative type"). See Proposition A.8.10(1) in the book [Pseudo-reductive groups](http://math.stanford.edu/~conrad/papers/predfinal.pdf).


