Lemma 92.4.3. In Example 92.4.1 let $V$ be a finite dimensional $k$-vector space. Then

are finite dimensional.

Lemma 92.4.3. In Example 92.4.1 let $V$ be a finite dimensional $k$-vector space. Then

\[ T\mathcal{D}\! \mathit{ef}_ V = (0) \quad \text{and}\quad \text{Inf}(\mathcal{D}\! \mathit{ef}_ V) = \text{End}_ k(V) \]

are finite dimensional.

**Proof.**
With $\mathcal{F}$ as in Example 92.4.1 set $x_0 = (k, V) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$. Recall that $T\mathcal{D}\! \mathit{ef}_ V = T_{x_0}\mathcal{F}$ is the set of isomorphism classes of pairs $(x, \alpha )$ consisting of an object $x$ of $\mathcal{F} $ over the dual numbers $k[\epsilon ]$ and a morphism $\alpha : x \to x_0$ of $\mathcal{F}$ lying over $k[\epsilon ] \to k$.

Up to isomorphism, there is a unique pair $(M, \alpha )$ consisting of a finite projective module $M$ over $k[\epsilon ]$ and $k[\epsilon ]$-linear map $\alpha : M \to V$ which induces an isomorphism $M \otimes _{k[\epsilon ]} k \to V$. For example, if $V = k^{\oplus n}$, then we take $M = k[\epsilon ]^{\oplus n}$ with the obvious map $\alpha $.

Similarly, $\text{Inf}(\mathcal{D}\! \mathit{ef}_ V) = \text{Inf}_{x_0}(\mathcal{F})$ is the set of automorphisms of the trivial deformation $x'_0$ of $x_0$ over $k[\epsilon ]$. See Formal Deformation Theory, Definition 89.19.2 for details.

Given $(M, \alpha )$ as in the second paragraph, we see that an element of $\text{Inf}_{x_0}(\mathcal{F})$ is an automorphism $\gamma : M \to M$ with $\gamma \bmod \epsilon = \text{id}$. Then we can write $\gamma = \text{id}_ M + \epsilon \psi $ where $\psi : M/\epsilon M \to M/\epsilon M$ is $k$-linear. Using $\alpha $ we can think of $\psi $ as an element of $\text{End}_ k(V)$ and this finishes the proof. $\square$

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