The Stacks project

89.13 Versal formal objects

The existence of a versal formal object forces $\mathcal{F}$ to have property (S1).

Lemma 89.13.1. Let $\mathcal{F}$ be a predeformation category. Assume $\mathcal{F}$ has a versal formal object. Then $\mathcal{F}$ satisfies (S1).

Proof. Let $\xi $ be a versal formal object of $\mathcal{F}$. Let

\[ \xymatrix{ & x_2 \ar[d] \\ x_1 \ar[r] & x } \]

be a diagram in $\mathcal{F}$ such that $x_2 \to x$ lies over a surjective ring map. Since the natural morphism $\widehat{\mathcal{F}}|_{\mathcal{C}_\Lambda } \xrightarrow {\sim } \mathcal{F}$ is an equivalence (see Remark 89.7.7), we can consider this diagram also as a diagram in $\widehat{\mathcal{F}}$. By Lemma 89.8.11 there exists a morphism $\xi \to x_1$, so by Remark 89.8.10 we also get a morphism $\xi \to x_2$ making the diagram

\[ \xymatrix{ \xi \ar[r] \ar[d] & x_2 \ar[d] \\ x_1 \ar[r] & x } \]

commute. If $x_1 \to x$ and $x_2 \to x$ lie above ring maps $A_1 \to A$ and $A_2 \to A$ then taking the pushforward of $\xi $ to $A_1 \times _ A A_2$ gives an object $y$ as required by (S1). $\square$

In the case that our cofibred category satisfies (S1) and (S2) we can characterize the versal formal objects as follows.

Lemma 89.13.2. Let $\mathcal{F}$ be a predeformation category satisfying (S1) and (S2). Let $\xi $ be a formal object of $\mathcal{F}$ corresponding to $\underline{\xi } : \underline{R}|_{\mathcal{C}_\Lambda } \to \mathcal{F}$, see Remark 89.7.12. Then $\xi $ is versal if and only if the following two conditions hold:

  1. the map $d\underline{\xi } : T\underline{R}|_{\mathcal{C}_\Lambda } \to T\mathcal{F}$ on tangent spaces is surjective, and

  2. given a diagram in $\widehat{\mathcal{F}}$

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

    in $\widehat{\mathcal{C}}_\Lambda $ with $B \to A$ a small extension of Artinian rings, then there exists a ring map $R \to B$ such that

    \[ \xymatrix{ & B \ar[d]^{f} \\ R \ar[ur] \ar[r] & A } \]


Proof. If $\xi $ is versal then (1) holds by Lemma 89.8.8 and (2) holds by Remark 89.8.10. Assume (1) and (2) hold. By Remark 89.8.10 we must show that given a diagram in $\widehat{\mathcal{F}}$ as in (2), there exists $\xi \to y$ such that

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

commutes. Let $b : R \to B$ be the map guaranteed by (2). Denote $y' = b_*\xi $ and choose a factorization $\xi \to y' \to x$ lying over $R \to B \to A$ of the given morphism $\xi \to x$. By (S1) we obtain a commutative diagram

\[ \vcenter { \xymatrix{ z \ar[r] \ar[d] & y \ar[d] \\ y' \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ B \times _ A B \ar[d] \ar[r] & B \ar[d]^{f} \\ B \ar[r]^{f} & A . } } \]

Set $I = \mathop{\mathrm{Ker}}(f)$. Let $\overline{g} : B \times _ A B \to k[I]$ be the ring map $(u, v) \mapsto \overline{u} \oplus (v - u)$, cf. Lemma 89.10.8. By (1) there exists a morphism $\xi \to \overline{g}_*z$ which lies over a ring map $i : R \to k[\epsilon ]$. Choose an Artinian quotient $b_1 : R \to B_1$ such that both $b : R \to B$ and $i : R \to k[\epsilon ]$ factor through $R \to B_1$, i.e., giving $h : B_1 \to B$ and $i' : B_1 \to k[\epsilon ]$. Choose a pushforward $y_1 = b_{1, *}\xi $, a factorization $\xi \to y_1 \to y'$ lying over $R \to B_1 \to B$ of $\xi \to y'$, and a factorization $\xi \to y_1 \to \overline{g}_*z$ lying over $R \to B_1 \to k[\epsilon ]$ of $\xi \to \overline{g}_*z$. Applying (S1) once more we obtain

\[ \vcenter { \xymatrix{ z_1 \ar[r] \ar[d] & z \ar[r] \ar[d] & y \ar[d] \\ y_1 \ar[r] & y' \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ B_1 \times _ A B \ar[d] \ar[r] & B \times _ A B \ar[r] \ar[d] & B \ar[d]^{f} \\ B_1 \ar[r] & B \ar[r] & A . } } \]

Note that the map $g : B_1 \times _ A B \to k[I]$ of Lemma 89.10.8 (defined using $h$) is the composition of $B_1 \times _ A B \to B \times _ A B$ and the map $\overline{g}$ above. By construction there exists a morphism $y_1 \to g_*z_1 \cong \overline{g}_*z$! Hence Lemma 89.10.8 applies (to the outer rectangles in the diagrams above) to give a morphism $y_1 \to y$ and precomposing with $\xi \to y_1$ gives the desired morphism $\xi \to y$. $\square$

If $\mathcal{F}$ has property (S1) then the “largest quotient where a lift exists” exists. Here is a precise statement.

Lemma 89.13.3. Let $\mathcal{F}$ be a category cofibred in groupoids over $\mathcal{C}_\Lambda $ which has (S1). Let $B \to A$ be a surjection in $\mathcal{C}_\Lambda $ with kernel $I$ annihilated by $\mathfrak m_ B$. Let $x \in \mathcal{F}(A)$. The set of ideals

\[ \mathcal{J} = \{ J \subset I \mid \text{there exists an }y \to x\text{ lying over }B/J \to A\} \]

has a smallest element.

Proof. Note that $\mathcal{J}$ is nonempty as $I \in \mathcal{J}$. Also, if $J \in \mathcal{J}$ and $J \subset J' \subset I$ then $J' \in \mathcal{J}$ because we can pushforward the object $y$ to an object $y'$ over $B/J'$. Let $J$ and $K$ be elements of the displayed set. We claim that $J \cap K \in \mathcal{J}$ which will prove the lemma. Since $I$ is a $k$-vector space we can find an ideal $J \subset J' \subset I$ such that $J \cap K = J' \cap K$ and such that $J' + K = I$. By the above we may replace $J$ by $J'$ and assume that $J + K = I$. In this case

\[ A/(J \cap K) = A/J \times _{A/I} A/K. \]

Hence the existence of an element $z \in \mathcal{F}(A/(J \cap K))$ mapping to $x$ follows, via (S1), from the existence of the elements we have assumed exist over $A/J$ and $A/K$. $\square$

We will improve on the following result later.

Lemma 89.13.4. 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 89.9.5. 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 89.10.5 and 89.12.2. 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 89.13.3. 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 89.4.7 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 89.7.4.

We prove $\xi $ is versal. For versality it suffices to check conditions (1) and (2) of Lemma 89.13.2. Condition (1) follows from our choice of $\xi _2$ in Step 2 above. Suppose given a diagram in $\widehat{\mathcal{F}}$

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

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 89.13.2 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 89.3.8 (2) $p_1$ is a small extension, so by Lemma 89.3.12 (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 $h : S = R[[X_1, \ldots , X_2]] \to B$ 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. $\square$

Remark 89.13.5. Let $F : \mathcal{C}_\Lambda \to \textit{Sets}$ be a predeformation functor satisfying (S1) and (S2). The condition $\dim _ k TF < \infty $ is precisely condition (H3) from Schlessinger's paper. Recall that (S1) and (S2) correspond to conditions (H1) and (H2), see Remark 89.10.3. Thus Lemma 89.13.4 tells us

\[ (H1) + (H2) + (H3) \Rightarrow \text{ there exists a versal formal object} \]

for predeformation functors. We will make the link with hulls in Remark 89.15.6.

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