The Stacks project

106.2 Versal rings

In this section we elucidate the relationship between deformation rings and local rings on algebraic stacks of finite type over a locally Noetherian base.

Situation 106.2.1. Here $\mathcal{X}$ is an algebraic stack locally of finite type over a locally Noetherian scheme $S$.

Here is the definition.

Definition 106.2.2. In Situation 106.2.1 let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism, where $k$ is a finite type field over $S$. A versal ring to $\mathcal{X}$ at $x_0$ is a complete Noetherian local $S$-algebra $A$ with residue field $k$ such that there exists a versal formal object $(A, \xi _ n, f_ n)$ as in Artin's Axioms, Definition 97.12.1 with $\xi _1 \cong x_0$ (a $2$-isomorphism).

We want to prove that versal rings exist and are unique up to smooth factors. To do this, we will use the predeformation categories of Artin's Axioms, Section 97.3. These are always deformation categories in our situation.

Lemma 106.2.3. In Situation 106.2.1 let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism, where $k$ is a finite type field over $S$. Then $\mathcal{F}_{\mathcal{X}, k, x_0}$ is a deformation category and $T\mathcal{F}_{\mathcal{X}, k, x_0}$ and $\text{Inf}(\mathcal{F}_{\mathcal{X}, k, x_0})$ are finite dimensional $k$-vector spaces.

Proof. Choose an affine open $\mathop{\mathrm{Spec}}(\Lambda ) \subset S$ such that $\mathop{\mathrm{Spec}}(k) \to S$ factors through it. By Artin's Axioms, Section 97.3 we obtain a predeformation category $\mathcal{F}_{\mathcal{X}, k, x_0}$ over the category $\mathcal{C}_\Lambda $. (As pointed out in locus citatus this category only depends on the morphism $\mathop{\mathrm{Spec}}(k) \to S$ and not on the choice of $\Lambda $.) By Artin's Axioms, Lemmas 97.6.1 and 97.5.2 $\mathcal{F}_{\mathcal{X}, k, x_0}$ is actually a deformation category. By Artin's Axioms, Lemma 97.8.1 we find that $T\mathcal{F}_{\mathcal{X}, k, x_0}$ and $\text{Inf}(\mathcal{F}_{\mathcal{X}, k, x_0})$ are finite dimensional $k$-vector spaces. $\square$

Lemma 106.2.4. In Situation 106.2.1 let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism, where $k$ is a finite type field over $S$. Then a versal ring to $\mathcal{X}$ at $x_0$ exists. Given a pair $A$, $A'$ of these, then $A \cong A'[[t_1, \ldots , t_ r]]$ or $A' \cong A[[t_1, \ldots , t_ r]]$ as $S$-algebras for some $r$.

Proof. By Lemma 106.2.3 and Formal Deformation Theory, Lemma 89.13.4 (note that the assumptions of this lemma hold by Formal Deformation Theory, Lemmas 89.16.6 and Definition 89.16.8). By the uniquness result of Formal Deformation Theory, Lemma 89.14.5 there exists a “minimal” versal ring $A$ of $\mathcal{X}$ at $x_0$ such that any other versal ring of $\mathcal{X}$ at $x_0$ is isomorphic to $A[[t_1, \ldots , t_ r]]$ for some $r$. This clearly implies the second statement. $\square$

Lemma 106.2.5. In Situation 106.2.1 let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism, where $k$ is a finite type field over $S$. Let $l/k$ be a finite extension of fields and denote $x_{l, 0} : \mathop{\mathrm{Spec}}(l) \to \mathcal{X}$ the induced morphism. Given a versal ring $A$ to $\mathcal{X}$ at $x_0$ there exists a versal ring $A'$ to $\mathcal{X}$ at $x_{l, 0}$ such that there is a $S$-algebra map $A \to A'$ which induces the given field extension $l/k$ and is formally smooth in the $\mathfrak m_{A'}$-adic topology.

Proof. Follows immediately from Artin's Axioms, Lemma 97.7.1 and Formal Deformation Theory, Lemma 89.29.6. (We also use that $\mathcal{X}$ satisfies (RS) by Artin's Axioms, Lemma 97.5.2.) $\square$

Lemma 106.2.6. In Situation 106.2.1 let $x : U \to \mathcal{X}$ be a morphism where $U$ is a scheme locally of finite type over $S$. Let $u_0 \in U$ be a finite type point. Set $k = \kappa (u_0)$ and denote $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ the induced map. The following are equivalent

  1. $x$ is versal at $u_0$ (Artin's Axioms, Definition 97.12.2),

  2. $\hat x : \mathcal{F}_{U, k, u_0} \to \mathcal{F}_{\mathcal{X}, k, x_0}$ is smooth,

  3. the formal object associated to $x|_{\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u_0}^\wedge )}$ is versal, and

  4. there is an open neighbourhood $U' \subset U$ of $x$ such that $x|_{U'} : U' \to \mathcal{X}$ is smooth.

Moreover, in this case the completion $\mathcal{O}_{U, u_0}^\wedge $ is a versal ring to $\mathcal{X}$ at $x_0$.

Proof. Since $U \to S$ is locally of finite type (as a composition of such morphisms), we see that $\mathop{\mathrm{Spec}}(k) \to S$ is of finite type (again as a composition). Thus the statement makes sense. The equivalence of (1) and (2) is the definition of $x$ being versal at $u_0$. The equivalence of (1) and (3) is Artin's Axioms, Lemma 97.12.3. Thus (1), (2), and (3) are equivalent.

If $x|_{U'}$ is smooth, then the functor $\hat x : \mathcal{F}_{U, k, u_0} \to \mathcal{F}_{\mathcal{X}, k, x_0}$ is smooth by Artin's Axioms, Lemma 97.3.2. Thus (4) implies (1), (2), and (3). For the converse, assume $x$ is versal at $u_0$. Choose a surjective smooth morphism $y : V \to \mathcal{X}$ where $V$ is a scheme. Set $Z = V \times _\mathcal {X} U$ and pick a finite type point $z_0 \in |Z|$ lying over $u_0$ (this is possible by Morphisms of Spaces, Lemma 66.25.5). By Artin's Axioms, Lemma 97.12.6 the morphism $Z \to V$ is smooth at $z_0$. By definition we can find an open neighbourhood $W \subset Z$ of $z_0$ such that $W \to V$ is smooth. Since $Z \to U$ is open, let $U' \subset U$ be the image of $W$. Then we see that $U' \to \mathcal{X}$ is smooth by our definition of smooth morphisms of stacks.

The final statement follows from the definitions as $\mathcal{O}_{U, u_0}^\wedge $ prorepresents $\mathcal{F}_{U, k, u_0}$. $\square$

Lemma 106.2.7. In Situation 106.2.1. Let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism such that $\mathop{\mathrm{Spec}}(k) \to S$ is of finite type with image $s$. Let $A$ be a versal ring to $\mathcal{X}$ at $x_0$. The following are equivalent

  1. $x_0$ is in the smooth locus of $\mathcal{X} \to S$ (Morphisms of Stacks, Lemma 100.33.6),

  2. $\mathcal{O}_{S, s} \to A$ is formally smooth in the $\mathfrak m_ A$-adic topology, and

  3. $\mathcal{F}_{\mathcal{X}, k, x_0}$ is unobstructed.

Proof. The equivalence of (2) and (3) follows immediately from Formal Deformation Theory, Lemma 89.9.4.

Note that $\mathcal{O}_{S, s} \to A$ is formally smooth in the $\mathfrak m_ A$-adic topology if and only if $\mathcal{O}_{S, s} \to A' = A[[t_1, \ldots , t_ r]]$ is formally smooth in the $\mathfrak m_{A'}$-adic topology. Hence (2) does not depend on the choice of our versal ring by Lemma 106.2.4. Next, let $l/k$ be a finite extension and choose $A \to A'$ as in Lemma 106.2.5. If $\mathcal{O}_{S, s} \to A$ is formally smooth in the $\mathfrak m_ A$-adic topology, then $\mathcal{O}_{S, s} \to A'$ is formally smooth in the $\mathfrak m_{A'}$-adic topology, see More on Algebra, Lemma 15.37.7. Conversely, if $\mathcal{O}_{S, s} \to A'$ is formally smooth in the $\mathfrak m_{A'}$-adic topology, then $\mathcal{O}_{S, s}^\wedge \to A'$ and $A \to A'$ are regular (More on Algebra, Proposition 15.49.2) and hence $\mathcal{O}_{S, s}^\wedge \to A$ is regular (More on Algebra, Lemma 15.41.7), hence $\mathcal{O}_{S, s} \to A$ is formally smooth in the $\mathfrak m_ A$-adic topology (same lemma as before). Thus the equivalence of (2) and (1) holds for $k$ and $x_0$ if and only if it holds for $l$ and $x_{0, l}$.

Choose a scheme $U$ and a smooth morphism $U \to \mathcal{X}$ such that $\mathop{\mathrm{Spec}}(k) \times _\mathcal {X} U$ is nonempty. Choose a finite extension $l/k$ and a point $w_0 : \mathop{\mathrm{Spec}}(l) \to \mathop{\mathrm{Spec}}(k) \times _\mathcal {X} U$. Let $u_0 \in U$ be the image of $w_0$. We may apply the above to $l/k$ and to $l/\kappa (u_0)$ to see that we can reduce to $u_0$. Thus we may assume $A = \mathcal{O}_{U, u_0}^\wedge $, see Lemma 106.2.6. Observe that $x_0$ is in the smooth locus of $\mathcal{X} \to S$ if and only if $u_0$ is in the smooth locus of $U \to S$, see for example Morphisms of Stacks, Lemma 100.33.6. Thus the equivalence of (1) and (2) follows from More on Algebra, Lemma 15.38.6. $\square$

We recall a consequence of Artin approximation.

Lemma 106.2.8. In Situation 106.2.1. Let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism such that $\mathop{\mathrm{Spec}}(k) \to S$ is of finite type with image $s$. Let $A$ be a versal ring to $\mathcal{X}$ at $x_0$. If $\mathcal{O}_{S, s}$ is a G-ring, then we may find a smooth morphism $U \to \mathcal{X}$ whose source is a scheme and a point $u_0 \in U$ with residue field $k$, such that

  1. $\mathop{\mathrm{Spec}}(k) \to U \to \mathcal{X}$ coincides with the given morphism $x_0$,

  2. there is an isomorphism $\mathcal{O}_{U, u_0}^\wedge \cong A$.

Proof. Let $(\xi _ n, f_ n)$ be the versal formal object over $A$. By Artin's Axioms, Lemma 97.9.5 we know that $\xi = (A, \xi _ n, f_ n)$ is effective. By assumption $\mathcal{X}$ is locally of finite presentation over $S$ (use Morphisms of Stacks, Lemma 100.27.5), and hence limit preserving by Limits of Stacks, Proposition 101.3.8. Thus Artin approximation as in Artin's Axioms, Lemma 97.12.7 shows that we may find a morphism $U \to \mathcal{X}$ with source a finite type $S$-scheme, containing a point $u_0 \in U$ of residue field $k$ satisfying (1) and (2) such that $U \to \mathcal{X}$ is versal at $u_0$. By Lemma 106.2.6 after shrinking $U$ we may assume $U \to \mathcal{X}$ is smooth. $\square$

Remark 106.2.9 (Upgrading versal rings). In Situation 106.2.1 let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism, where $k$ is a finite type field over $S$. Let $A$ be a versal ring to $\mathcal{X}$ at $x_0$. By Artin's Axioms, Lemma 97.9.5 our versal formal object in fact comes from a morphism

\[ \mathop{\mathrm{Spec}}(A) \longrightarrow \mathcal{X} \]

over $S$. Moreover, the results above each can be upgraded to be compatible with this morphism. Here is a list:

  1. in Lemma 106.2.4 the isomorphism $A \cong A'[[t_1, \ldots , t_ r]]$ or $A' \cong A[[t_1, \ldots , t_ r]]$ may be chosen compatible with these morphisms,

  2. in Lemma 106.2.5 the homomorphism $A \to A'$ may be chosen compatible with these morphisms,

  3. in Lemma 106.2.6 the morphism $\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u_0}^\wedge ) \to \mathcal{X}$ is the composition of the canonical map $\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u_0}^\wedge ) \to U$ and the given map $U \to \mathcal{X}$,

  4. in Lemma 106.2.8 the isomorphism $\mathcal{O}_{U, u_0}^\wedge \cong A$ may be chosen so $\mathop{\mathrm{Spec}}(A) \to \mathcal{X}$ corresponds to the canonical map in the item above.

In each case the statement follows from the fact that our maps are compatible with versal formal elements; we note however that the implied diagrams are $2$-commutative only up to a (noncanonical) choice of a $2$-arrow. Still, this means that the implied map $A' \to A$ or $A \to A'$ in (1) is well defined up to formal homotopy, see Formal Deformation Theory, Lemma 89.28.3.

Lemma 106.2.10. In Situation 106.2.1 let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism, where $k$ is a finite type field over $S$. Let $A$ be a versal ring to $\mathcal{X}$ at $x_0$. Then the morphism $\mathop{\mathrm{Spec}}(A) \to \mathcal{X}$ of Remark 106.2.9 is flat.

Proof. If the local ring of $S$ at the image point is a G-ring, then this follows immediately from Lemma 106.2.8 and the fact that the map from a Noetherian local ring to its completion is flat. In general we prove it as follows.

Step I. If $A$ and $A'$ are two versal rings to $\mathcal{X}$ at $x_0$, then the result is true for $A$ if and only if it is true for $A'$. Namely, after possible swapping $A$ and $A'$, we may assume there is a formally smooth map $\varphi : A \to A'$ such that the composition

\[ \mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A) \to \mathcal{X} \]

is the morphism $\mathop{\mathrm{Spec}}(A') \to \mathcal{X}$, see Lemma 106.2.4 and Remark 106.2.9. Since $A \to A'$ is faithfully flat we obtain the equivalence from Morphisms of Stacks, Lemmas 100.25.2 and 100.25.5.

Step II. Let $l/k$ be a finite extension of fields. Let $x_{l, 0} : \mathop{\mathrm{Spec}}(l) \to \mathcal{X}$ be the induced morphism. Let $A$ be a versal ring to $\mathcal{X}$ at $x_0$ and let $A \to A'$ be as in Lemma 106.2.5. Then again the composition

\[ \mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A) \to \mathcal{X} \]

is the morphism $\mathop{\mathrm{Spec}}(A') \to \mathcal{X}$, see Remark 106.2.9. Arguing as before and using step I to see choice of versal rings is irrelevant, we see that the lemma holds for $x_0$ if and only if it holds for $x_{l, 0}$.

Step III. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then we can choose a finite type point $z_0$ on $Z = U \times _\mathcal {X} x_0$ (this is a nonempty algebraic space). Let $u_0 \in U$ be the image of $z_0$ in $U$. Choose a scheme and a surjective étale map $W \to Z$ such that $z_0$ is the image of a closed point $w_0 \in W$ (see Morphisms of Spaces, Section 66.25). Since $W \to \mathop{\mathrm{Spec}}(k)$ and $W \to U$ are of finite type, we see that $\kappa (w_0)/k$ and $\kappa (w_0)/\kappa (u_0)$ are finite extensions of fields (see Morphisms, Section 29.16). Applying Step II twice we may replace $x_0$ by $u_0 \to U \to \mathcal{X}$. Then we see our morphism is the composition

\[ \mathop{\mathrm{Spec}}(\mathcal{O}_{U, u_0}^\wedge ) \to U \to \mathcal{X} \]

The first arrow is flat because completion of Noetherian local rings are flat (Algebra, Lemma 10.97.2) and the second arrow is flat as a smooth morphism is flat. The composition is flat as composition preserves flatness. $\square$

Remark 106.2.11. In Situation 106.2.1 let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism, where $k$ is a finite type field over $S$. By Lemma 106.2.3 and Formal Deformation Theory, Theorem 89.26.4 we know that $\mathcal{F}_{\mathcal{X}, k, x_0}$ has a presentation by a smooth prorepresentable groupoid in functors on $\mathcal{C}_\Lambda $. Unwinding the definitions, this means we can choose

  1. a Noetherian complete local $\Lambda $-algebra $A$ with residue field $k$ and a versal formal object $\xi $ of $\mathcal{F}_{\mathcal{X}, k, x_0}$ over $A$,

  2. a Noetherian complete local $\Lambda $-algebra $B$ with residue field $k$ and an isomorphism

    \[ \underline{B}|_{\mathcal{C}_\Lambda } \longrightarrow \underline{A}|_{\mathcal{C}_\Lambda } \times _{\underline{\xi }, \mathcal{F}_{\mathcal{X}, k, x_0}, \underline{\xi }} \underline{A}|_{\mathcal{C}_\Lambda } \]

The projections correspond to formally smooth maps $t : A \to B$ and $s : A \to B$ (because $\xi $ is versal). There is a map $c : B \to B \widehat{\otimes }_{s, A, t} B$ which turns $(A, B, s, t, c)$ into a cogroupoid in the category of Noetherian complete local $\Lambda $-algebras with residue field $k$ (on prorepresentable functors this map is constructed in Formal Deformation Theory, Lemma 89.25.2). Finally, the cited theorem tells us that $\xi $ induces an equivalence

\[ [\underline{A}|_{\mathcal{C}_\Lambda } / \underline{B}|_{\mathcal{C}_\Lambda }] \longrightarrow \mathcal{F}_{\mathcal{X}, k, x_0} \]

of groupoids cofibred over $\mathcal{C}_\Lambda $. In fact, we also get an equivalence

\[ [\underline{A}/\underline{B}] \longrightarrow \widehat{\mathcal{F}}_{\mathcal{X}, k, x_0} \]

of groupoids cofibred over the completed category $\widehat{\mathcal{C}}_\Lambda $ (see discussion in Formal Deformation Theory, Section 89.22 as to why this works). Of course $A$ is a versal ring to $\mathcal{X}$ at $x_0$.


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