We do a little bit of work to try and understand (non)uniqueness of versal formal objects. It turns out that if a predeformation category has a versal formal object, then it has a minimal versal formal object and any two such are isomorphic. Moreover, all versal formal objects are “more or less” the same up to replacing the base ring by a power series extension.

Let $\mathcal{F}$ be a category cofibred in groupoids over $\mathcal{C}_\Lambda $. For every object $x$ of $\mathcal{F}$ lying over $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_\Lambda )$ consider the category $\mathcal{S}_ x$ with objects

\[ \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ x) = \{ x' \to x \mid x' \to x\text{ lies over }A' \subset A\} \]

and morphisms are morphisms over $x$. For every $y \to x$ in $\mathcal{F}$ lying over $f : B \to A$ in $\mathcal{C}_\Lambda $ there is a functor $f_* : \mathcal{S}_ y \to \mathcal{S}_ x$ defined as follows: Given $y' \to y$ lying over $B' \subset B$ set $A' = f(B')$ and let $y' \to x'$ be over $B' \to f(B')$ be the pushforward of $y'$. By the axioms of a category cofibred in groupoids we obtain a unique morphism $x' \to x$ lying over $f(B') \to A$ such that

\[ \xymatrix{ y' \ar[d] \ar[r] & x' \ar[d] \\ y \ar[r] & x } \]

commutes. Then $x' \to x$ is an object of $\mathcal{S}_ x$. We say an object $x' \to x$ of $\mathcal{S}_ x$ is *minimal* if any morphism $(x'_1 \to x) \to (x' \to x)$ in $\mathcal{S}_ x$ is an isomorphism, i.e., $x'$ and $x'_1$ are defined over the same subring of $A$. Since $A$ has finite length as a $\Lambda $-module we see that minimal objects always exist.

Lemma 89.14.1. Let $\mathcal{F}$ be a category cofibred in groupoids over $\mathcal{C}_\Lambda $ which has (S1).

For $y \to x$ in $\mathcal{F}$ a minimal object in $\mathcal{S}_ y$ maps to a minimal object of $\mathcal{S}_ x$.

For $y \to x$ in $\mathcal{F}$ lying over a surjection $f : B \to A$ in $\mathcal{C}_\Lambda $ every minimal object of $\mathcal{S}_ x$ is the image of a minimal object of $\mathcal{S}_ y$.

**Proof.**
Proof of (1). Say $y \to x$ lies over $f : B \to A$. Let $y' \to y$ lying over $B' \subset B$ be a minimal object of $\mathcal{S}_ y$. Let

\[ \vcenter { \xymatrix{ y' \ar[d] \ar[r] & x' \ar[d] \\ y \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ B' \ar[d] \ar[r] & f(B') \ar[d] \\ B \ar[r] & A } } \]

be as in the construction of $f_*$ above. Suppose that $(x'' \to x) \to (x' \to x)$ is a morphism of $\mathcal{S}_ x$ with $x'' \to x'$ lying over $A'' \subset f(B')$. By (S1) there exists $y'' \to y'$ lying over $B' \times _{f(B')} A'' \to B'$. Since $y' \to y$ is minimal we conclude that $B' \times _{f(B')} A'' \to B'$ is an isomorphism, which implies that $A'' = f(B')$, i.e., $x' \to x$ is minimal.

Proof of (2). Suppose $f : B \to A$ is surjective and $y \to x$ lies over $f$. Let $x' \to x$ be a minimal object of $\mathcal{S}_ x$ lying over $A' \subset A$. By (S1) there exists $y' \to y$ lying over $B' = f^{-1}(A') = B \times _ A A' \to B$ whose image in $\mathcal{S}_ x$ is $x' \to x$. So $f_*(y' \to y) = x' \to x$. Choose a morphism $(y'' \to y) \to (y' \to y)$ in $\mathcal{S}_ y$ with $y'' \to y$ a minimal object (this is possible by the remark on lengths above the lemma). Then $f_*(y'' \to y)$ is an object of $\mathcal{S}_ x$ which maps to $x' \to x$ (by functoriality of $f_*$) hence is isomorphic to $x' \to x$ by minimality of $x' \to x$.
$\square$

Lemma 89.14.2. Let $\mathcal{F}$ be a category cofibred in groupoids over $\mathcal{C}_\Lambda $ which has (S1). Let $\xi $ be a versal formal object of $\mathcal{F}$ lying over $R$. There exists a morphism $\xi ' \to \xi $ lying over $R' \subset R$ with the following minimality properties

for every $f : R \to A$ with $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_\Lambda )$ the pushforwards

\[ \vcenter { \xymatrix{ \xi ' \ar[d] \ar[r] & x' \ar[d] \\ \xi \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ R' \ar[d] \ar[r] & f(R') \ar[d] \\ R \ar[r] & A } } \]

produce a minimal object $x' \to x$ of $\mathcal{S}_ x$, and

for any morphism of formal objects $\xi '' \to \xi '$ the corresponding morphism $R'' \to R'$ is surjective.

**Proof.**
Write $\xi = (R, \xi _ n, f_ n)$. Set $R'_1 = k$ and $\xi '_1 = \xi _1$. Suppose that we have constructed minimal objects $\xi '_ m \to \xi _ m$ of $\mathcal{S}_{\xi _ m}$ lying over $R'_ m \subset R/\mathfrak m_ R^ m$ for $m \leq n$ and morphisms $f'_ m : \xi '_{m + 1} \to \xi '_ m$ compatible with $f_ m$ for $m \leq n - 1$. By Lemma 89.14.1 (2) there exists a minimal object $\xi '_{n + 1} \to \xi _{n + 1}$ lying over $R'_{n + 1} \subset R/\mathfrak m_ R^{n + 1}$ whose image is $\xi '_ n \to \xi _ n$ over $R'_ n \subset R/\mathfrak m_ R^ n$. This produces the commutative diagram

\[ \xymatrix{ \xi '_{n + 1} \ar[r]_{f'_ n} \ar[d] & \xi '_ n \ar[d] \\ \xi _{n + 1} \ar[r]^{f_ n} & \xi _ n } \]

by construction. Moreover the ring map $R'_{n + 1} \to R'_ n$ is surjective. Set $R' = \mathop{\mathrm{lim}}\nolimits _ n R'_ n$. Then $R' \to R$ is injective.

However, it isn't a priori clear that $R'$ is Noetherian. To prove this we use that $\xi $ is versal. Namely, versality implies that there exists a morphism $\xi \to \xi '_ n$ in $\widehat{\mathcal{F}}$, see Lemma 89.8.11. The corresponding map $R \to R'_ n$ has to be surjective (as $\xi '_ n \to \xi _ n$ is minimal in $\mathcal{S}_{\xi _ n}$). Thus the dimensions of the cotangent spaces are bounded and Lemma 89.4.8 implies $R'$ is Noetherian, i.e., an object of $\widehat{\mathcal{C}}_\Lambda $. By Lemma 89.7.4 (plus the result on filtrations of Lemma 89.4.8) the sequence of elements $\xi '_ n$ defines a formal object $\xi '$ over $R'$ and we have a map $\xi ' \to \xi $.

By construction (1) holds for $R \to R/\mathfrak m_ R^ n$ for each $n$. Since each $R \to A$ as in (1) factors through $R \to R/\mathfrak m_ R^ n \to A$ we see that (1) for $x' \to x$ over $f(R) \subset A$ follows from the minimality of $\xi '_ n \to \xi _ n$ over $R'_ n \to R/\mathfrak m_ R^ n$ by Lemma 89.14.1 (1).

If $R'' \to R'$ as in (2) is not surjective, then $R'' \to R' \to R'_ n$ would not be surjective for some $n$ and $\xi '_ n \to \xi _ n$ wouldn't be minimal, a contradiction. This contradiction proves (2).
$\square$

Lemma 89.14.3. Let $\mathcal{F}$ be a category cofibred in groupoids over $\mathcal{C}_\Lambda $ which has (S1). Let $\xi $ be a versal formal object of $\mathcal{F}$ lying over $R$. Let $\xi ' \to \xi $ be a morphism of formal objects lying over $R' \subset R$ as constructed in Lemma 89.14.2. Then

\[ R \cong R'[[x_1, \ldots , x_ r]] \]

is a power series ring over $R'$. Moreover, $\xi '$ is a versal formal object too.

**Proof.**
By Lemma 89.8.11 there exists a morphism $\xi \to \xi '$. By Lemma 89.14.2 the corresponding map $f : R \to R'$ induces a surjection $f|_{R'} : R' \to R'$. This is an isomorphism by Algebra, Lemma 10.31.10. Hence $I = \mathop{\mathrm{Ker}}(f)$ is an ideal of $R$ such that $R = R' \oplus I$. Let $x_1, \ldots , x_ n \in I$ be elements which form a basis for $I/\mathfrak m_ RI$. Consider the map $S = R'[[X_1, \ldots , X_ r]] \to R$ mapping $X_ i$ to $x_ i$. For every $n \geq 1$ we get a surjection of Artinian $R'$-algebras $B = S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n = A$. Denote $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(B)$, resp. $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(A))$ the pushforward of $\xi '$ along $R' \to S \to B$, resp. $R' \to S \to A$. Note that $x$ is also the pushforward of $\xi $ along $R \to A$ as $\xi $ is the pushforward of $\xi '$ along $R' \to R$. Thus we have a solid diagram

\[ \vcenter { \xymatrix{ & y \ar[d] \\ \xi \ar[r] \ar@{..>}[ru] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & S/\mathfrak m_ S^ n \ar[d] \\ R \ar[r] \ar@{..>}[ru] & R/\mathfrak m_ R^ n } } \]

Because $\xi $ is versal, using Remark 89.8.10 we obtain the dotted arrows fitting into these diagrams. In particular, the maps $S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n$ have sections $h_ n : R/\mathfrak m_ R^ n \to S/\mathfrak m_ S^ n$. It follows from Lemma 89.4.9 that $S \to R$ is an isomorphism.

As $\xi $ is a pushforward of $\xi '$ along $R' \to R$ we obtain from Remark 89.7.13 a commutative diagram

\[ \xymatrix{ \underline{R}|_{\mathcal{C}_\Lambda } \ar[rr] \ar[rd]_{\underline{\xi }} & & \underline{R'}|_{\mathcal{C}_\Lambda } \ar[ld]^{\underline{\xi '}} \\ & \mathcal{F} } \]

Since $R' \to R$ has a left inverse (namely $R \to R/I = R'$) we see that $\underline{R}|_{\mathcal{C}_\Lambda } \to \underline{R'}|_{\mathcal{C}_\Lambda }$ is essentially surjective. Hence by Lemma 89.8.7 we see that $\underline{\xi '}$ is smooth, i.e., $\xi '$ is a versal formal object.
$\square$

Motivated by the preceding lemmas we make the following definition.

Definition 89.14.4. Let $\mathcal{F}$ be a predeformation category. We say a versal formal object $\xi $ of $\mathcal{F}$ is *minimal*^{1} if for any morphism of formal objects $\xi ' \to \xi $ the underlying map on rings is surjective. Sometimes a minimal versal formal object is called *miniversal*.

The work in this section shows this definition is reasonable. First of all, the existence of a versal formal object implies that $\mathcal{F}$ has (S1). Then the preceding lemmas show there exists a minimal versal formal object. Finally, any two minimal versal formal objects are isomorphic. Here is a summary of our results (with detailed proofs).

Lemma 89.14.5. Let $\mathcal{F}$ be a predeformation category which has a versal formal object. Then

$\mathcal{F}$ has a minimal versal formal object,

minimal versal objects are unique up to isomorphism, and

any versal object is the pushforward of a minimal versal object along a power series ring extension.

**Proof.**
Suppose $\mathcal{F}$ has a versal formal object $\xi $ over $R$. Then it satisfies (S1), see Lemma 89.13.1. Let $\xi ' \to \xi $ over $R' \subset R$ be any of the morphisms constructed in Lemma 89.14.2. By Lemma 89.14.3 we see that $\xi '$ is versal, hence it is a minimal versal formal object (by construction). This proves (1). Also, $R \cong R'[[x_1, \ldots , x_ n]]$ which proves (3).

Suppose that $\xi _ i/R_ i$ are two minimal versal formal objects. By Lemma 89.8.11 there exist morphisms $\xi _1 \to \xi _2$ and $\xi _2 \to \xi _1$. The corresponding ring maps $f : R_1 \to R_2$ and $g : R_2 \to R_1$ are surjective by minimality. Hence the compositions $g \circ f : R_1 \to R_1$ and $f \circ g : R_2 \to R_2$ are isomorphisms by Algebra, Lemma 10.31.10. Thus $f$ and $g$ are isomorphisms whence the maps $\xi _1 \to \xi _2$ and $\xi _2 \to \xi _1$ are isomorphisms (because $\widehat{\mathcal{F}}$ is cofibred in groupoids by Lemma 89.7.2). This proves (2) and finishes the proof of the lemma.
$\square$

## Comments (0)