Lemma 89.15.4. Let $\mathcal{F}$ be a predeformation category satisfying (S2) which has a versal formal object. Then its minimal versal formal object satisfies (89.15.0.2).

Proof. Let $\xi$ be a minimal versal formal object for $\mathcal{F}$, see Lemma 89.14.5. Say $\xi$ lies over $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. In order to parse (89.15.0.2) we point out that $T\mathcal{F}$ has a natural $k$-vector space structure (see Lemma 89.12.2), that $d\underline{\xi } : \text{Der}_\Lambda (R, k) \to T\mathcal{F}$ is linear (see Lemma 89.12.4), and that the action of $\text{Der}_\Lambda (k, k)$ is given by addition (see Lemma 89.12.6). Consider the diagram

$\xymatrix{ & \mathop{\mathrm{Hom}}\nolimits _ k(\mathfrak m_ R/\mathfrak m_ R^2, k) \\ K \ar[r] & \text{Der}_\Lambda (R, k) \ar[r]^{d\underline{\xi }} \ar[u] & T\mathcal{F} \\ & \text{Der}_\Lambda (k, k) \ar[u] \ar[ru] }$

The vector space $K$ is the kernel of $d\underline{\xi }$. Note that the middle column is exact in the middle as it is dual to the sequence (89.3.10.1). If (89.15.0.2) fails, then we can find a nonzero element $D \in K$ which does not map to zero in $\mathop{\mathrm{Hom}}\nolimits _ k(\mathfrak m_ R/\mathfrak m_ R^2, k)$. This means there exists an $t \in \mathfrak m_ R$ such that $D(t) = 1$. Set $R' = \{ a \in R \mid D(a) = 0\}$. As $D$ is a derivation this is a subring of $R$. Since $D(t) = 1$ we see that $R' \to k$ is surjective (compare with the proof of Lemma 89.3.12). Note that $\mathfrak m_{R'} = \mathop{\mathrm{Ker}}(D : \mathfrak m_ R \to k)$ is an ideal of $R$ and $\mathfrak m_ R^2 \subset \mathfrak m_{R'}$. Hence

$\mathfrak m_ R/\mathfrak m_ R^2 = \mathfrak m_{R'}/\mathfrak m_ R^2 + k\overline{t}$

which implies that the map

$R'/\mathfrak m_ R^2 \times _ k k[\epsilon ] \to R/\mathfrak m_ R^2$

sending $\epsilon$ to $\overline{t}$ is an isomorphism. In particular there is a map $R/\mathfrak m_ R^2 \to R'/\mathfrak m_ R^2$.

Let $\xi \to y$ be a morphism lying over $R \to R/\mathfrak m_ R^2$. Let $y \to x$ be a morphism lying over $R/\mathfrak m_ R^2 \to R'/\mathfrak m_ R^2$. Let $y \to x_\epsilon$ be a morphism lying over $R/\mathfrak m_ R^2 \to k[\epsilon ]$. Let $x_0$ be the unique (up to unique isomorphism) object of $\mathcal{F}$ over $k$. By the axioms of a category cofibred in groupoids we obtain a commutative diagram

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & x_\epsilon \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ R'/\mathfrak m_ R^2 \times _ k k[\epsilon ] \ar[r] \ar[d] & k[\epsilon ] \ar[d] \\ R'/\mathfrak m_ R^2 \ar[r] & k. } }$

Because $D \in K$ we see that $x_\epsilon$ is isomorphic to $0 \in \mathcal{F}(k[\epsilon ])$, i.e., $x_\epsilon$ is the pushforward of $x_0$ via $k \to k[\epsilon ], a \mapsto a$. Hence by Lemma 89.10.7 we see that there exists a morphism $x \to y$. Since $\text{length}_\Lambda (R'/\mathfrak m_ R^2) < \text{length}_\Lambda (R/\mathfrak m_ R^2)$ the corresponding ring map $R'/\mathfrak m_ R^2 \to R/\mathfrak m_ R^2$ is not surjective. This contradicts the minimality of $\xi /R$, see part (1) of Lemma 89.14.2. This contradiction shows that such a $D$ cannot exist, hence we win. $\square$

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