The Stacks project

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

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 $x' \to x$. $\square$

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

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). $\square$

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

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

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. $\square$

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. $\square$