TITLE: Why do automorphism groups of algebraic varieties have natural algebraic group structure?
QUESTION [17 upvotes]: I am not sure that all automorphism groups of algebraic varieties have natrual algebraic group structure. 
But if the automorphism group of a variety has algebraic group structure, how do I know the automorphism group is an algebraic group.
 For example, the automorphism group of an elliptic curve $A$ is an extension of the group $G$ of automorphisms which preserve the structure of the elliptic curve, by the group $A(k)$  of translations in the points of $A$, i.e. the sequence of groups
$0\to A(k)\to \text{Aut}(A)\to G \to 0$ is exact, see  Springer online ref - automorphism group of algebraic variaties. 
In this example, how do I know $\text{Aut}(A)$ is an algebraic group.

REPLY [25 votes]: This is really a comment on Pete's comment for Mikhail's answer, but I am making it an answer because it raises a question which I think should be more widely known.
The construction of Aut-scheme uses the entire Hilbert scheme, which has countably many components (due to varying Hilbert polynomials), and it is not obvious how many of these intervene in the construction. Mikhail's answer illustrates the basic example showing it can be infinite.  But this leads to the following problem (suggested by what is known about Neron-Severi groups, whose construction encounters the same issue via construction of Pic schemes in terms of Hilbert schemes, at least in the original Grothendieck construction):  
Q. Does the Aut scheme of a proper scheme over an alg. closed field have finitely generated component group? (This is a more basic question than finiteness, which is really Pete's comment to Mikhail's answer.) 
Incredibly, even for smooth projective varieties over  C this appears to be a wide open problem!! I have mentioned this to several experts in alg. geom. (including Oort), and nobody has an idea. If anyone has an idea or a solution, please let me know right away.