Lemma 89.15.1. Let $\mathcal{F}$ be a predeformation category. Let $\xi $ be a versal formal object of $\mathcal{F}$ such that (89.15.0.2) holds. Then $\xi $ is a minimal versal formal object. In particular, such $\xi $ are unique up to isomorphism.

## 89.15 Miniversal formal objects and tangent spaces

The general notion of minimality introduced in Definition 89.14.4 can sometimes be deduced from the behaviour on tangent spaces. Let $\xi $ be a formal object of the predeformation category $\mathcal{F}$ and let $\underline{\xi } : \underline{R}|_{\mathcal{C}_\Lambda } \to \mathcal{F}$ be the corresponding morphism. Then we can consider the following the condition

and the condition

Here we are using the identification $T\underline{R}|_{\mathcal{C}_\Lambda } = \text{Der}_\Lambda (R, k)$ of Example 89.11.11 and the action (89.12.6.2) of derivations on the tangent spaces. If $k' \subset k$ is separable, then $\text{Der}_\Lambda (k, k) = 0$ and the two conditions are equivalent. It turns out that, in the presence of condition (S2) a versal formal object is minimal if and only if $\underline{\xi }$ satisfies (89.15.0.2). Moreover, if $\underline{\xi }$ satisfies (89.15.0.1), then $\mathcal{F}$ satisfies (S2).

**Proof.**
If $\xi $ is not minimal, then there exists a morphism $\xi ' \to \xi $ lying over $R' \to R$ such that $R = R'[[x_1, \ldots , x_ n]]$ with $n > 0$, see Lemma 89.14.5. Thus $d\underline{\xi }$ factors as

and we see that (89.15.0.2) cannot hold because $D : f \mapsto \partial /\partial x_1(f) \bmod \mathfrak m_ R$ is an element of the kernel of the first arrow which is not in the image of $\text{Der}_\Lambda (k, k) \to \text{Der}_\Lambda (R, k)$. $\square$

Lemma 89.15.2. Let $\mathcal{F}$ be a predeformation category. Let $\xi $ be a versal formal object of $\mathcal{F}$ such that (89.15.0.1) holds. Then

$\mathcal{F}$ satisfies (S1).

$\mathcal{F}$ satisfies (S2).

$\dim _ k T\mathcal{F}$ is finite.

**Proof.**
Condition (S1) holds by Lemma 89.13.1. The first part of (S2) holds since (S1) holds. Let

be diagrams as in the second part of (S2). As above we can find morphisms $b : \xi \to y$ and $b' : \xi \to y'$ such that

commutes. Let $p : \mathcal{F} \to \mathcal{C}_\Lambda $ denote the structure morphism. Say $\widehat{p}(\xi ) = R$, i.e., $\xi $ lies over $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. We see that the pushforward of $\xi $ via $p(c) \circ p(b)$ is $x_\epsilon $ and that the pushforward of $\xi $ via $p(c') \circ p(b')$ is $x_\epsilon $. Since $\xi $ satisfies (89.15.0.1), we see that $p(c) \circ p(b) = p(c') \circ p(b')$ as maps $R \to k[\epsilon ]$. Hence $p(b) = p(b')$ as maps from $R \to A \times _ k k[\epsilon ]$. Thus we see that $y$ and $y'$ are isomorphic to the pushforward of $\xi $ along this map and we get a unique morphism $y \to y'$ over $A \times _ k k[\epsilon ]$ compatible with $b$ and $b'$ as desired.

Finally, by Example 89.11.11 we see $\dim _ k T\mathcal{F} = \dim _ k T\underline{R}|_{\mathcal{C}_\Lambda }$ is finite. $\square$

Example 89.15.3. There exist predeformation categories which have a versal formal object satisfying (89.15.0.2) but which do not satisfy (S2). A quick example is to take $F = \underline{k[\epsilon ]}/G$ where $G \subset \text{Aut}_{\mathcal{C}_\Lambda }(k[\epsilon ])$ is a finite nontrivial subgroup. Namely, the map $\underline{k[\epsilon ]} \to F$ is smooth, but the tangent space of $F$ does not have a natural $k$-vector space structure (as it is a quotient of a $k$-vector space by a finite group).

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

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

which implies that the map

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

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. $\square$

Theorem 89.15.5. Let $\mathcal{F}$ be a predeformation category. Consider the following conditions

$\mathcal{F}$ has a minimal versal formal object satisfying (89.15.0.1),

$\mathcal{F}$ has a minimal versal formal object satisfying (89.15.0.2),

the following conditions hold:

$\mathcal{F}$ satisfies (S1).

$\mathcal{F}$ satisfies (S2).

$\dim _ k T\mathcal{F}$ is finite.

We always have

If $k' \subset k$ is separable, then all three are equivalent.

**Proof.**
Lemma 89.15.2 shows that (1) $\Rightarrow $ (3). Lemmas 89.13.4 and 89.15.4 show that (3) $\Rightarrow $ (2). If $k' \subset k$ is separable then $\text{Der}_\Lambda (k, k) = 0$ and we see that (89.15.0.1) $=$ (89.15.0.2), i.e., (1) is the same as (2).

An alternative proof of (3) $\Rightarrow $ (1) in the classical case is to add a few words to the proof of Lemma 89.13.4 to see that one can right away construct a versal object which satisfies (89.15.0.1) in this case. This avoids the use of Lemma 89.13.4 in the classical case. Details omitted. $\square$

Remark 89.15.6. Let $F : \mathcal{C}_\Lambda \to \textit{Sets}$ be a predeformation functor satisfying (S1) and (S2) and $\dim _ k TF < \infty $. Recall that these conditions correspond to the conditions (H1), (H2), and (H3) from Schlessinger's paper, see Remark 89.13.5. Now, in the classical case (or if $k' \subset k$ is separable) following Schlessinger we introduce the notion of a hull: a *hull* is a versal formal object $\xi \in \widehat{F}(R)$ such that $d\underline{\xi } : T\underline{R}|_{\mathcal{C}_\Lambda } \to TF$ is an isomorphism, i.e., (89.15.0.1) holds. Thus Theorem 89.15.5 tells us

in the classical case. In other words, our theorem recovers Schlessinger's theorem on the existence of hulls.

Remark 89.15.7. Let $\mathcal{F}$ be a predeformation category. Recall that $\mathcal{F} \to \overline{\mathcal{F}}$ is smooth, see Remark 89.8.5. Hence if $\xi \in \widehat{\mathcal{F}}(R)$ is a versal formal object, then the composition

is smooth (Lemma 89.8.7) and we conclude that the image $\overline{\xi }$ of $\xi $ in $\overline{\mathcal{F}}$ is a versal formal object. If (89.15.0.1) holds, then $\overline{\xi }$ induces an isomorphism $T\underline{R}|_{\mathcal{C}_\Lambda } \to T\overline{\mathcal{F}}$ because $\mathcal{F} \to \overline{\mathcal{F}}$ identifies tangent spaces. Hence in this case $\overline{\xi }$ is a hull for $\overline{\mathcal{F}}$, see Remark 89.15.6. By Theorem 89.15.5 we can always find such a $\xi $ if $k' \subset k$ is separable and $\mathcal{F}$ is a predeformation category satisfying (S1), (S2), and $\dim _ k T\mathcal{F} < \infty $.

Example 89.15.8. In Lemma 89.9.5 we constructed objects $R \in \widehat{\mathcal{C}}_\Lambda $ such that $\underline{R}|_{\mathcal{C}_\Lambda }$ is smooth and such that

Let us reinterpret this using the theorem above. Namely, consider $\mathcal{F} = \mathcal{C}_\Lambda $ as a category cofibred in groupoids over itself (using the identity functor). Then $\mathcal{F}$ is a predeformation category, satisfies (S1) and (S2), and we have $T\mathcal{F} = 0$. Thus $\mathcal{F}$ satisfies condition (3) of Theorem 89.15.5. The theorem implies that (2) holds, i.e., we can find a minimal versal formal object $\xi \in \widehat{\mathcal{F}}(S)$ over some $S \in \widehat{\mathcal{C}}_\Lambda $ satisfying (89.15.0.2). Lemma 89.9.3 shows that $\Lambda \to S$ is formally smooth in the $\mathfrak m_ S$-adic topology (because $\underline{\xi } : \underline{R}|_{\mathcal{C}_\Lambda } \to \mathcal{F} = \mathcal{C}_\Lambda $ is smooth). Now condition (89.15.0.2) tells us that $\text{Der}_\Lambda (S, k) \to 0$ is bijective on $\text{Der}_\Lambda (k, k)$-orbits. This means the injection $\text{Der}_\Lambda (k, k) \to \text{Der}_\Lambda (S, k)$ is also surjective. In other words, we have $\Omega _{S/\Lambda } \otimes _ S k = \Omega _{k/\Lambda }$. Since $\Lambda \to S$ is formally smooth in the $\mathfrak m_ S$-adic topology, we can apply More on Algebra, Lemma 15.40.4 to conclude the exact sequence (89.3.10.2) turns into a pair of identifications

Reading the argument backwards, we find that the $R$ constructed in Lemma 89.9.5 carries a minimal versal object. By the uniqueness of minimal versal objects (Lemma 89.14.5) we also conclude $R \cong S$, i.e., the two constructions give the same answer.

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