## 90.14 Minimal versal formal objects

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

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

2. 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 90.14.2. 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

1. 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

2. 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 90.14.1 (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 90.8.11. 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 90.4.8 implies $R'$ is Noetherian, i.e., an object of $\widehat{\mathcal{C}}_\Lambda$. By Lemma 90.7.4 (plus the result on filtrations of Lemma 90.4.8) 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 90.14.1 (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 90.14.3. 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 90.14.2. 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 90.8.11 there exists a morphism $\xi \to \xi '$. By Lemma 90.14.2 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 90.8.10 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 90.4.9 that $S \to R$ is an isomorphism.

As $\xi$ is a pushforward of $\xi '$ along $R' \to R$ we obtain from Remark 90.7.13 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 90.8.7 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 90.14.4. Let $\mathcal{F}$ be a predeformation category. We say a versal formal object $\xi$ of $\mathcal{F}$ is minimal1 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 90.14.5. Let $\mathcal{F}$ be a predeformation category which has a versal formal object. Then

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

2. minimal versal objects are unique up to isomorphism, and

3. 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 90.13.1. Let $\xi ' \to \xi$ over $R' \subset R$ be any of the morphisms constructed in Lemma 90.14.2. By Lemma 90.14.3 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 90.8.11 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 90.7.2). This proves (2) and finishes the proof of the lemma. $\square$

 This may be nonstandard terminology. Many authors tie this notion in with properties of tangent spaces. We will make the link in Section 90.15.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).