Let $X$ denote an algebraic scheme over $\operatorname{Spec} k$ such that its deformation functor $\operatorname{Def}\_X$ has a semi-universal couple $(R,u)$, where $R$ is an Artinian $k$-algebra and $u \in \operatorname{Def}\_X(R)$.


On pg. 91 of the book *"Deformations of algebraic schemes"* by E. Sernesi the following automorphism functor is introduced


$$ \operatorname{Aut}\_u: \mathcal{A}\_R \to \operatorname{Set} $$


$$ \operatorname{Aut}\_u(A) = \text{the group of automorphisms of the deformations} \ \mathcal{X}\_A$$


My question concerns the following proposition:



> 
> $\textbf{Proposition 2.6.2} $ *If $X$ is projective, then
> $\operatorname{Aut}\_u$ has $H^0(X,T\_X)$ as tangent space.*
> 
> 
> 


The proof of the proposition concludes (Eqn 2.29, pg. 92) with the claim that $$Aut\_u(k[\epsilon]) \cong H^0(X, T\_X).$$


However, $k[\epsilon]$ does not have a natural $R$-algebra structure so writing $Aut\_u(k[\epsilon])$ doesn't really make sense.


What did Sernesi mean when he wrote this?



Question about automorphism functor in Sernesi's "Deformations of algebraic schemes"
In fact, there is a natural such structure, namely
$$ R \to k \to k[\varepsilon], $$
corresponding to the constant deformation $X\otimes k[\varepsilon]$, and indeed the automorphisms of $X\otimes k[\varepsilon]$ over $k[\varepsilon]$ restricting to the identity on $X$ correspond to derivations $\mathcal{O}\_X \to \mathcal{O}\_X$.


