The Stacks project

Proof. Let $A_1 \to A$ and $A_2 \to A$ be morphisms of $\mathcal{C}_\Lambda $. Assume $A_2 \to A$ is surjective. According to Formal Deformation Theory, Lemma 90.16.4 it suffices to show that the functor $\mathcal{F}(A_1 \times _ A A_2) \to \mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$ is an equivalence of categories.

Consider an object

of the category $\mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$. Then we consider $P_1 \times _\varphi P_2$. Since $\varphi : P_1 \otimes _{A_1} A \to P_2 \otimes _{A_2} A$ is an isomorphism of graded algebras, we see that the graded pieces of $P_1 \times _\varphi P_2$ are finite projective $A_1 \times _ A A_2$-modules, see proof of Lemma 93.4.2. Thus $P_1 \times _\varphi P_2$ is an object of $\mathcal{F}(A_1 \times _ A A_2)$. This construction determines a quasi-inverse to our functor and the proof is complete. $\square$

Proof. We first deal with the infinitesimal automorphisms. Let $Q = P \otimes _ k k[\epsilon ]$. Then an element of $\text{Inf}(\mathcal{D}\! \mathit{ef}_ P)$ is given by an automorphism $\gamma = \text{id} + \epsilon \delta : Q \to Q$ as above where now $\delta : P \to P$. The fact that $\gamma $ is graded implies that $\delta $ is homogeneous of degree $0$. The fact that $\gamma $ is $k$-linear implies that $\delta $ is $k$-linear. The fact that $\gamma $ is multiplicative implies that $\delta $ is a $k$-derivation. Conversely, given a $k$-derivation $\delta : P \to P$ homogeneous of degree $0$, we obtain an automorphism $\gamma = \text{id} + \epsilon \delta $ as above. Thus we see that

as predicted in the lemma. Clearly, if $P$ is generated in degrees $P_ i$, $0 \leq i \leq N$, then $\delta $ is determined by the linear maps $\delta _ i : P_ i \to P_ i$ for $0 \leq i \leq N$ and we see that

as desired.

To finish the proof of the lemma we show that there is a finite dimensional deformation space. To do this we choose a presentation

of graded $k$-algebras where $\deg (X_ i) = d_ i$ and $F_ j$ is homogeneous of degree $e_ j$. Let $Q$ be any graded $k[\epsilon ]$-algebra finite free in each degree which comes with an isomorphsm $\alpha : Q/\epsilon Q \to P$ so that $(Q, \alpha )$ defines an element of $T\mathcal{D}\! \mathit{ef}_ P$. Choose a homogeneous element $q_ i \in Q$ of degree $d_ i$ mapping to the image of $X_ i$ in $P$. Then we obtain

and since $P = Q/\epsilon Q$ this map is surjective by Nakayama's lemma. A small diagram chase shows we can choose homogeneous elements $F_{\epsilon , j} \in k[\epsilon ][X_1, \ldots , X_ n]$ of degree $e_ j$ mapping to zero in $Q$ and mapping to $F_ j$ in $k[X_1, \ldots , X_ n]$. Then

is a presentation of $Q$ by flatness of $Q$ over $k[\epsilon ]$. Write

There is some ambiguity in the vector $(G_1, \ldots , G_ m)$. First, using different choices of $F_{\epsilon , j}$ we can modify $G_ j$ by an arbitrary element of degree $e_ j$ in the kernel of $k[X_1, \ldots , X_ n] \to P$. Hence, instead of $(G_1, \ldots , G_ m)$, we remember the element

where $g_ j$ is the image of $G_ j$ in $P_{e_ j}$. Moreover, if we change our choice of $q_ i$ into $q_ i + \epsilon p_ i$ with $p_ i$ of degree $d_ i$ then a computation (omitted) shows that $g_ j$ changes into

We conclude that the isomorphism class of $Q$ is determined by the image of the vector $(G_1, \ldots , G_ m)$ in the $k$-vector space

In this way we see that we obtain an injection

Since $W$ visibly has finite dimension, we conclude that the lemma is true. $\square$

Proof. This follows immediately from Lemmas 93.7.2 and 93.7.3 and Formal Deformation Theory, Lemma 90.16.6 and Remark 90.15.7. $\square$