Lemma 106.7.1. Let $\mathcal{X}$ be an algebraic stack over a scheme $S$. Assume $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is locally of finite presentation. Let $A \to B$ be a flat $S$-algebra homomorphism. Let $x$ be an object of $\mathcal{X}$ over $A$ and set $y = x|_ B$. Then $\text{Inf}_ x(M) \otimes _ A B = \text{Inf}_ y(M \otimes _ A B)$.

## 106.7 Infinitesimal deformations

We continue the discussion from Artin's Axioms, Section 98.21.

**Proof.**
Recall that $\text{Inf}_ x(M)$ is the set of automorphisms of the trivial deformation of $x$ to $A[M]$ which induce the identity automorphism of $x$ over $A$. The trivial deformation is the pullback of $x$ to $\mathop{\mathrm{Spec}}(A[M])$ via $\mathop{\mathrm{Spec}}(A[M]) \to \mathop{\mathrm{Spec}}(A)$. Let $G \to \mathop{\mathrm{Spec}}(A)$ be the automorphism group algebraic space of $x$ (this exists because $\mathcal{X}$ is an algebraic space). Let $e : \mathop{\mathrm{Spec}}(A) \to G$ be the neutral element. The discussion in More on Morphisms of Spaces, Section 76.17 gives

By the same token

Since $G \to \mathop{\mathrm{Spec}}(A)$ is locally of finite presentation by assumption, we see that $\Omega _{G/A}$ is locally of finite presentation, see More on Morphisms of Spaces, Lemma 76.7.15. Hence $e^*\Omega _{G/A}$ is a finitely presented $A$-module. Moreover, $\Omega _{G_ B/B}$ is the pullback of $\Omega _{G/A}$ by More on Morphisms of Spaces, Lemma 76.7.12. Therefore $e_ B^*\Omega _{G_ B/B} = e^*\Omega _{G/A} \otimes _ A B$. we conclude by More on Algebra, Lemma 15.65.4. $\square$

Lemma 106.7.2. Let $\mathcal{X}$ be an algebraic stack over a base scheme $S$. Assume $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is locally of finite presentation. Let $(A' \to A, x)$ be a deformation situation. Then the functor

is a sheaf on the site $(\textit{Aff}/\mathop{\mathrm{Spec}}(A'))_{fppf}$ of Topologies, Definition 34.7.8.

**Proof.**
Let $\{ T'_ i \to T'\} _{i = 1, \ldots n}$ be a standard fppf covering of affine schemes over $A'$. Write $T' = \mathop{\mathrm{Spec}}(B')$. As usual denote

where the ring is a suitable tensor product. Set $B = B' \otimes _{A'} A$ and $B_{i_0 \ldots i_ p} = B'_{i_0 \ldots i_ p} \otimes _{A'} A$. Denote $y = x|_ B$ and $y_{i_0 \ldots i_ p} = x|_{B_{i_0 \ldots i_ p}}$. Let $\gamma _ i \in F(B'_ i)$ such that $\gamma _{i_0}$ and $\gamma _{i_1}$ map to the same element of $F(B'_{i_0i_1})$. We have to find a unique $\gamma \in F(B')$ mapping to $\gamma _ i$ in $F(B'_ i)$.

Choose an actual object $y'_ i$ of $\textit{Lift}(y_ i, B'_ i)$ in the isomorphism class $\gamma _ i$. Choose isomorphisms $\varphi _{i_0i_1} : y'_{i_0}|_{B'_{i_0i_1}} \to y'_{i_1}|_{B'_{i_0i_1}}$ in the category $\textit{Lift}(y_{i_0i_1}, B'_{i_0i_1})$. If the maps $\varphi _{i_0i_1}$ satisfy the cocycle condition, then we obtain our object $\gamma $ because $\mathcal{X}$ is a stack in the fppf topology. The cocycle condition is that the composition

is the identity. If not, then these maps give elements

Here $J = \mathop{\mathrm{Ker}}(B' \to B)$ and $J_{i_0 \ldots i_ p} = \mathop{\mathrm{Ker}}(B'_{i_0 \ldots i_ p} \to B_{i_0 \ldots i_ p})$. The equality in the displayed equation holds by Lemma 106.7.1 applied to $B' \to B'_{i_0 \ldots i_ p}$ and $y$ and $y_{i_0 \ldots i_ p}$, the flatness of the maps $B' \to B'_{i_0 \ldots i_ p}$ which also guarantees that $J_{i_0 \ldots i_ p} = J \otimes _{B'} B'_{i_0 \ldots i_ p}$. A computation (omitted) shows that $\delta _{i_0i_1i_2}$ gives a $2$-cocycle in the Čech complex

By Descent, Lemma 35.9.2 this complex is acyclic in positive degrees and has $H^0 = \text{Inf}_ y(J)$. Since $\text{Inf}_{y_{i_0i_1}}(J_{i_0i_1})$ acts on morphisms (Artin's Axioms, Remark 98.21.4) this means we can modify our choice of $\varphi _{i_0i_1}$ to get to the case where $\delta _{i_0i_1i_2} = 0$.

Uniqueness. We still have to show there is at most one $\gamma $ restricting to $\gamma _ i$ for all $i$. Suppose we have objects $y', z'$ of $\textit{Lift}(y, B')$ and isomorphisms $\psi _ i : y'|_{B'_ i} \to z'|_{B'_ i}$ in $\textit{Lift}(y_ i, B'_ i)$. Then we can consider

Arguing as before, the obstruction to finding an isomorphism between $y'$ and $z'$ over $B'$ is an element in the $H^1$ of the Čech complex displayed above which is zero. $\square$

Lemma 106.7.3. Let $\mathcal{X}$ be an algebraic stack over a scheme $S$ whose structure morphism $\mathcal{X} \to S$ is locally of finite presentation. Let $A \to B$ be a flat $S$-algebra homomorphism. Let $x$ be an object of $\mathcal{X}$ over $A$. Then $T_ x(M) \otimes _ A B = T_ y(M \otimes _ A B)$.

**Proof.**
Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. We first reduce the lemma to the case where $x$ lifts to $U$. Recall that $T_ x(M)$ is the set of isomorphism classes of lifts of $x$ to $A[M]$. Therefore Lemma 106.7.2^{1} says that the rule

is a sheaf on the small étale site of $\mathop{\mathrm{Spec}}(A)$; the tensor product is needed to make $A[M] \to A_1[M \otimes _ A A_1]$ a flat ring map. We may choose a faithfully flat étale ring map $A \to A_1$ such that $x|_{A_1}$ lifts to a morphism $u_1 : \mathop{\mathrm{Spec}}(A_1) \to U$, see for example Sheaves on Stacks, Lemma 96.19.10. Write $A_2 = A_1 \otimes _ A A_1$ and set $B_1 = B \otimes _ A A_1$ and $B_2 = B \otimes _ A A_2$. Consider the diagram

The rows are exact by the sheaf condition. We have $M \otimes _ A B_ i = (M \otimes _ A A_ i) \otimes _{A_ i} B_ i$. Thus if we prove the result for the middle and right vertical arrow, then the result follows. This reduces us to the case discussed in the next paragraph.

Assume that $x$ is the image of a morphism $u : \mathop{\mathrm{Spec}}(A) \to U$. Observe that $T_ u(M) \to T_ x(M)$ is surjective since $U \to \mathcal{X}$ is smooth and representable by algebraic spaces, see Criteria for Representability, Lemma 97.6.3 (see discussion preceding it for explanation) and More on Morphisms of Spaces, Lemma 76.19.6. Set $R = U \times _\mathcal {X} U$. Recall that we obtain a groupoid $(U, R, s, t, c, e, i)$ in algebraic spaces with $\mathcal{X} = [U/R]$. By Artin's Axioms, Lemma 98.21.6 we have an exact sequence

where the zero on the right was shown above. A similar sequence holds for the base change to $B$. Thus the result we want follows if we can prove the result of the lemma for $T_ u(M)$ and $T_{e \circ u}(M)$. This reduces us to the case discussed in the next paragraph.

Assume that $\mathcal{X} = X$ is an algebraic space locally of finite presentation over $S$. Then we have

by the discussion in More on Morphisms of Spaces, Section 76.17. By the same token

Since $X \to S$ is locally of finite presentation, we see that $\Omega _{X/S}$ is locally of finite presentation, see More on Morphisms of Spaces, Lemma 76.7.15. Hence $x^*\Omega _{X/S}$ is a finitely presented $A$-module. Clearly, we have $y^*\Omega _{X/S} = x^*\Omega _{X/S} \otimes _ A B$. we conclude by More on Algebra, Lemma 15.65.4. $\square$

Lemma 106.7.4. Let $\mathcal{X}$ be an algebraic stack over a scheme $S$ whose structure morphism $\mathcal{X} \to S$ is locally of finite presentation. Let $(A' \to A, x)$ be a deformation situation. If there exists a faithfully flat finitely presented $A'$-algebra $B'$ and an object $y'$ of $\mathcal{X}$ over $B'$ lifting $x|_{B' \otimes _{A'} A}$, then there exists an object $x'$ over $A'$ lifting $x$.

**Proof.**
Let $I = \mathop{\mathrm{Ker}}(A' \to A)$. Set $B'_1 = B' \otimes _{A'} B'$ and $B'_2 = B' \otimes _{A'} B' \otimes _{A'} B'$. Let $J = IB'$, $J_1 = IB'_1$, $J_2 = IB'_2$ and $B = B'/J$, $B_1 = B'_1/J_1$, $B_2 = B'_2/J_2$. Set $y = x|_ B$, $y_1 = x|_{B_1}$, $y_2 = x|_{B_2}$. Let $F$ be the fppf sheaf of Lemma 106.7.2 (which applies, see footnote in the proof of Lemma 106.7.3). Thus we have an equalizer diagram

On the other hand, we have $F(B') = \text{Lift}(y, B')$, $F(B'_1) = \text{Lift}(y_1, B'_1)$, and $F(B'_2) = \text{Lift}(y_2, B'_2)$ in the terminology from Artin's Axioms, Section 98.21. These sets are nonempty and are (canonically) principal homogeneous spaces for $T_ y(J)$, $T_{y_1}(J_1)$, $T_{y_2}(J_2)$, see Artin's Axioms, Lemma 98.21.2. Thus the difference of the two images of $y'$ in $F(B'_1)$ is an element

The equality in the displayed equation holds by Lemma 106.7.3 applied to $A' \to B'_1$ and $x$ and $y_1$, the flatness of $A' \to B'_1$ which also guarantees that $J_1 = I \otimes _{A'} B'_1$. We have similar equalities for $B'$ and $B'_2$. A computation (omitted) shows that $\delta _1$ gives a $1$-cocycle in the Čech complex

By Descent, Lemma 35.9.2 this complex is acyclic in positive degrees and has $H^0 = T_ x(I)$. Thus we may choose an element in $T_ x(I) \otimes _ A B = T_ y(J)$ whose boundary is $\delta _1$. Replacing $y'$ by the result of this element acting on it, we find a new choice $y'$ with $\delta _1 = 0$. Thus $y'$ maps to the same element under the two maps $F(B') \to F(B'_1)$ and we obtain an element o $F(A')$ by the sheaf condition. $\square$

## 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.

## Comments (0)