I have a probably stupid question on schemes ...
 Let $S$ be a scheme, and let $A = \mathsf{Aut}(S)$ be its automorphism group. Does $A$ carry 
a scheme structure itself, that is, can one see $A$ as a group scheme ? 
 Thanks !



Automorphism group of a scheme
The answer is yes when the scheme is flat and projective over the base. This follows from the existence of the Hom scheme, which in turn is proven via the existence of the Hilbert scheme. 


A readable reference is Nitin Nitsure's part of the book *Fundamental algebraic geometry*. In particular Theorem 5.23 in his notes states that when *S* is noetherian, *X* projective and flat over *S*, and *Y* quasi-projective over *S*, then there exists a Hom-scheme parametrizing morphisms from *X* to *Y* over *S*. (The precise definition of the functor that this scheme represents is given in the text). The automorphism group scheme of *X* is then an open subscheme of $\mathrm{Hom}\_S(X,X)$. 


**Addendum**: The proof that Isom is open in Hom is very similar to the proof that Hom is open in Hilb. The map from the Hom scheme to $\mathrm{Hilb}\_{X \times\_S Y / S}$ is given by associating to a morphism $f : X \to Y$ its graph. Now the image consists of those $Z \subset X \times\_S Y$ such that projection onto *X* induces an isomorphism $Z \cong X$, and the crucial part of the proof is showing that this is an open condition. (In Nitsure's notes this is Theorem 5.22.(b).) But then the condition that *Z* maps isomorphically onto *Y* is also open, and this is exactly the condition that defines the Isom scheme inside the Hom scheme.


