ag.algebraic geometry - Why do automorphism groups of algebraic varieties have natural algebraic group structure?

It is not always true that the automorphism group of an algebraic variety has a natural algebraic group structure. For example, the automorphism group of $mathbb{A}^2$ includes all the maps of the form $(x,y) mapsto (x, y+f(x))$ where $f$ is any polynomial. I haven't thought through how to say this precisely in terms of functors, but this subgroup morally should be a connected infinite dimensional object, and is thus not a subobject of an algebraic group.

On the other hand, I believe that the automorphism group of a projective algebraic variety, $X$, can be given the structure of algebraic group in a fairly natural way. This is something I've thought about myself, but not written down a careful proof nor found a reference for: For any automorphism $f$ of $X$, consider the graph of $f$ as a subscheme of $X times X$, and thus a point of the Hilbert scheme of $Xtimes X$. In this way, we get an embedding of point sets from $mathrm{Aut}(X)$ into $mathrm{Hilb}(Xtimes X)$.

I believe that it should be easy to show that (1) $mathrm{Aut}(X)$ is open in $mathrm{Hilb}(Xtimes X)$, and thus acquires a natural scheme structure and (2) composition of automorphisms is a map of schemes.

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