The Stacks project

Lemma 90.13.4. Let $\mathcal{F}$ be a category cofibred in groupoids over $\mathcal{C}_\Lambda $. Assume the following conditions hold:

  1. $\mathcal{F}$ is a predeformation category.

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

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

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

Then $\mathcal{F}$ has a versal formal object.

Proof. Assume (1), (2), (3), and (4) hold. Choose an object $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$ such that $\underline{R}|_{\mathcal{C}_\Lambda }$ is smooth. See Lemma 90.9.5. Let $r = \dim _ k T\mathcal{F}$ and put $S = R[[X_1, \ldots , X_ r]]$.

We are going to inductively construct for $n \geq 2$ pairs $(J_ n, f_{n - 1} : \xi _ n \to \xi _{n - 1})$ where $J_ n \subset S$ is an decreasing sequence of ideals and $f_{n - 1} : \xi _ n \to \xi _{n - 1}$ is a morphism of $\mathcal{F}$ lying over the projection $S/J_ n \to S/J_{n - 1}$.

Step 1. Let $J_1 = \mathfrak m_ S$. Let $\xi _1$ be the unique (up to unique isomorphism) object of $\mathcal{F}$ over $k = S/J_1 = S/\mathfrak m_ S$

Step 2. Let $J_2 = \mathfrak m_ S^2 + \mathfrak {m}_ R S$. Then $S/J_2 = k[V]$ with $V = kX_1 \oplus \ldots \oplus kX_ r$ By (S2) for $\overline{\mathcal{F}}$ we get a bijection

\[ \overline{\mathcal{F}}(S/J_2) \longrightarrow T\mathcal{F} \otimes _ k V, \]

see Lemmas 90.10.5 and 90.12.2. Choose a basis $\theta _1, \ldots , \theta _ r$ for $T\mathcal{F}$ and set $\xi _2 = \sum \theta _ i \otimes X_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(S/J_2))$. The point of this choice is that

\[ d\xi _2 : \mathop{\mathrm{Mor}}\nolimits _{\mathcal{C}_\Lambda }(S/J_2, k[\epsilon ]) \longrightarrow T\mathcal{F} \]

is surjective. Let $f_1 : \xi _2 \to \xi _1$ be the unique morphism.

Induction step. Assume $(J_ n, f_{n - 1} : \xi _ n \to \xi _{n - 1})$ has been constructed for some $n \geq 2$. There is a minimal element $J_{n + 1}$ of the set of ideals $J \subset S$ satisfying: (a) $\mathfrak m_ S J_ n \subset J \subset J_ n$ and (b) there exists a morphism $\xi _{n + 1} \to \xi _ n$ lying over $S/J \to S/J_ n$, see Lemma 90.13.3. Let $f_ n : \xi _{n + 1} \to \xi _ n$ be any morphism of $\mathcal{F}$ lying over $S/J_{n + 1} \to S/J_ n$.

Set $J = \bigcap J_ n$. Set $\overline{S} = S/J$. Set $\overline{J}_ n = J_ n/J$. By Lemma 90.4.7 the sequence of ideals $(\overline{J}_ n)$ induces the $\mathfrak m_{\overline{S}}$-adic topology on $\overline{S}$. Since $(\xi _ n, f_ n)$ is an object of $\widehat{\mathcal{F}}_\mathcal {I}(\overline{S})$, where $\mathcal{I}$ is the filtration $(\overline{J}_ n)$ of $\overline{S}$, we see that $(\xi _ n, f_ n)$ induces an object $\xi $ of $\widehat{\mathcal{F}}(\overline{S})$. see Lemma 90.7.4.

We prove $\xi $ is versal. For versality it suffices to check conditions (1) and (2) of Lemma 90.13.2. Condition (1) follows from our choice of $\xi _2$ in Step 2 above. Suppose given a diagram in $\widehat{\mathcal{F}}$

\[ \vcenter { \xymatrix{ & y \ar[d] \\ \xi \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & B \ar[d]^{f} \\ \overline{S} \ar[r] & A } } \]

in $\widehat{\mathcal{C}}_\Lambda $ with $f: B \to A$ a small extension of Artinian rings. We have to show there is a map $\overline{S} \to B$ fitting into the diagram on the right. Choose $n$ such that $\overline{S} \to A$ factors through $\overline{S} \to S/J_ n$. This is possible as the sequence $(\overline{J}_ n)$ induces the $\mathfrak m_{\overline{S}}$-adic topology as we saw above. The pushforward of $\xi $ along $\overline{S} \to S/J_ n$ is $\xi _ n$. We may factor $\xi \to x$ as $\xi \to \xi _ n \to x$ hence we get a diagram in $\mathcal{F}$

\[ \vcenter { \xymatrix{ & y \ar[d] \\ \xi _ n \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & B \ar[d]^{f} \\ S/J_ n \ar[r] & A . } } \]

To check condition (2) of Lemma 90.13.2 it suffices to complete the diagram

\[ \xymatrix{ S/J_{n + 1} \ar[d] \ar@{-->}[r] & B \ar[d]^{f} \\ S/J_ n \ar[r] & A } \]

or equivalently, to complete the diagram

\[ \xymatrix{ & S/J_ n \times _ A B \ar[d]^{p_1} \\ S/J_{n + 1} \ar@{-->}[ur] \ar[r] & S/J_ n. } \]

If $p_1$ has a section we are done. If not, by Lemma 90.3.8 (2) $p_1$ is a small extension, so by Lemma 90.3.12 (4) $p_1$ is an essential surjection. Recall that $S = R[[X_1, \ldots , X_ r]]$ and that we chose $R$ such that $\underline{R}|_{\mathcal{C}_\Lambda }$ is smooth. Hence there exists a map $h : R \to B$ lifting the map $R \to S \to S/J_ n \to A$. By the universal property of a power series ring there is an $R$-algebra map $h : S = R[[X_1, \ldots , X_ r]] \to B$ lifting the given map $S \to S/J_ n \to A$. This induces a map $g: S \to S/J_ n \times _ A B$ making the solid square in the diagram

\[ \xymatrix{ S \ar[d] \ar[r]_-g & S/J_ n \times _ A B \ar[d]^{p_1} \\ S/J_{n + 1} \ar@{-->}[ur] \ar[r] & S/J_ n } \]

commute. Then $g$ is a surjection since $p_1$ is an essential surjection. We claim the ideal $K = \mathop{\mathrm{Ker}}(g)$ of $S$ satisfies conditions (a) and (b) of the construction of $J_{n + 1}$ in the induction step above. Namely, $K \subset J_ n$ is clear and $\mathfrak m_ SJ_ n \subset K$ as $p_1$ is a small extension; this proves (a). By (S1) applied to

\[ \xymatrix{ & y \ar[d] \\ \xi _ n \ar[r] & x, } \]

there exists a lifting of $\xi _ n$ to $S/K \cong S/J_ n \times _ A B$, so (b) holds. Since $J_{n + 1}$ was the minimal ideal with properties (a) and (b) this implies $J_{n + 1} \subset K$. Thus the desired map $S/J_{n+1} \to S/K \cong S/J_ n \times _ A B$ exists. $\square$

Comments (3)

Comment #7733 by Mingchen on

In the proof, eta in the diagram seems to be a typo, I think you mean xi

Comment #7734 by Mingchen on

In the definition of h, 2 should be replaced by r

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06IW. Beware of the difference between the letter 'O' and the digit '0'.