The Stacks project

105.2 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 105.2.1. Here $\mathcal{X}$ is an algebraic stack locally of finite type over a locally Noetherian scheme $S$.

Here is the definition.

Definition 105.2.2. In Situation 105.2.1 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 96.12.1 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 96.3. These are always deformation categories in our situation.

Lemma 105.2.3. In Situation 105.2.1 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 96.3 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 96.6.1 and 96.5.2 $\mathcal{F}_{\mathcal{X}, k, x_0}$ is actually a deformation category. By Artin's Axioms, Lemma 96.8.1 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 105.2.4. In Situation 105.2.1 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 105.2.3 and Formal Deformation Theory, Lemma 88.13.4 (note that the assumptions of this lemma hold by Formal Deformation Theory, Lemmas 88.16.6 and Definition 88.16.8). By the uniquness result of Formal Deformation Theory, Lemma 88.14.5 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 105.2.5. In Situation 105.2.1 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 96.7.1 and Formal Deformation Theory, Lemma 88.29.6. (We also use that $\mathcal{X}$ satisfies (RS) by Artin's Axioms, Lemma 96.5.2.) $\square$

Lemma 105.2.6. In Situation 105.2.1 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 96.12.2),

  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 96.12.3. 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 96.3.2. 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 65.25.5). By Artin's Axioms, Lemma 96.12.6 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 105.2.7. In Situation 105.2.1. 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 99.33.6),

  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 88.9.4.

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 105.2.4. Next, let $l/k$ be a finite extension and choose $A \to A'$ as in Lemma 105.2.5. 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 105.2.6. 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 99.33.6. 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 105.2.8. In Situation 105.2.1. 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 96.9.5 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 99.27.5), and hence limit preserving by Limits of Stacks, Proposition 100.3.8. Thus Artin approximation as in Artin's Axioms, Lemma 96.12.7 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 105.2.6 after shrinking $U$ we may assume $U \to \mathcal{X}$ is smooth. $\square$

Remark 105.2.9 (Upgrading versal rings). In Situation 105.2.1 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 96.9.5 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 105.2.4 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 105.2.5 the homomorphism $A \to A'$ may be chosen compatible with these morphisms,

  3. in Lemma 105.2.6 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 105.2.8 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 88.28.3.

Lemma 105.2.10. In Situation 105.2.1 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 105.2.9 is flat.

Proof. If the local ring of $S$ at the image point is a G-ring, then this follows immediately from Lemma 105.2.8 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 105.2.4 and Remark 105.2.9. Since $A \to A'$ is faithfully flat we obtain the equivalence from Morphisms of Stacks, Lemmas 99.25.2 and 99.25.5.

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 105.2.5. 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 105.2.9. 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 65.25). 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 105.2.11. In Situation 105.2.1 let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism, where $k$ is a finite type field over $S$. By Lemma 105.2.3 and Formal Deformation Theory, Theorem 88.26.4 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 88.25.2). 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 88.22 as to why this works). Of course $A$ is a versal ring to $\mathcal{X}$ at $x_0$.


Comments (0)


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 0DQT. Beware of the difference between the letter 'O' and the digit '0'.