Lemma 93.9.3. In Example 93.9.1 let $X$ be a scheme over $k$. Then
and
Lemma 93.9.3. In Example 93.9.1 let $X$ be a scheme over $k$. Then
and
Proof. Recall that $\text{Inf}(\mathcal{D}\! \mathit{ef}_ X)$ is the set of automorphisms of the trivial deformation $X' = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k[\epsilon ])$ of $X$ to $k[\epsilon ]$ equal to the identity modulo $\epsilon $. By Deformation Theory, Lemma 91.8.1 this is equal to $\text{Ext}^0_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/k}, \mathcal{O}_ X)$. The equality $\text{Ext}^0_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/k}, \mathcal{O}_ X) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\Omega _{X/k}, \mathcal{O}_ X)$ follows from More on Morphisms, Lemma 37.13.3. The equality $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\Omega _{X/k}, \mathcal{O}_ X) = \text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X)$ follows from Morphisms, Lemma 29.32.2.
Recall that $T_{x_0}\mathcal{D}\! \mathit{ef}_ X$ is the set of isomorphism classes of flat deformations $X'$ of $X$ to $k[\epsilon ]$, more precisely, the set of isomorphism classes of $\mathcal{D}\! \mathit{ef}_ X(k[\epsilon ])$. Thus the second statement of the lemma follows from Deformation Theory, Lemma 91.8.1. $\square$
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