# Question about automorphism functor in Sernesi's "Deformations of algebraic schemes"

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?

• It seems to me that a $k$-algebra homomorphism $R \longrightarrow k[\varepsilon]$ exist by semi-universality, or am I missing something? Jul 1, 2020 at 8:13
• It appear that the same problem would occur at the beginning of the proof, when Sernesi shows that $\mathrm{Aut}_u$ satisfies condition $H_0$ (as in pag45), that is $\mathrm{Aut}_u(k)$ is the identity. Jul 1, 2020 at 8:16
• In my mind, geometrically, in the case of complex projective varieties this corresponds to a morphism of schemes $\mathrm{Spec}(k[\varepsilon]) \longrightarrow \mathrm{Def}(X)$, where $\mathrm{Def}(X)$ is (the germ of) the Kuranishi family of $X$. Jul 1, 2020 at 8:21

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 $$\mathcal{O}_X \to \mathcal{O}_X$$.