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18.35 The naive cotangent complex

This section is the analogue of Algebra, Section 10.134 and Modules, Section 17.30. We advise the reader to read those sections first.

Let $\mathcal{C}$ be a site. Let $\mathcal{A} \to \mathcal{B}$ be a homomorphism of sheaves of rings on $\mathcal{C}$. In this section, for any sheaf of sets $\mathcal{E}$ on $\mathcal{C}$ we denote $\mathcal{A}[\mathcal{E}]$ the sheafification of the presheaf $U \mapsto \mathcal{A}(U)[\mathcal{E}(U)]$. Here $\mathcal{A}(U)[\mathcal{E}(U)]$ denotes the polynomial algebra over $\mathcal{A}(U)$ whose variables correspond to the elements of $\mathcal{E}(U)$. We denote $[e] \in \mathcal{A}(U)[\mathcal{E}(U)]$ the variable corresponding to $e \in \mathcal{E}(U)$. There is a canonical surjection of $\mathcal{A}$-algebras
\begin{equation} \label{sites-modules-equation-canonical-presentation} \mathcal{A}[\mathcal{B}] \longrightarrow \mathcal{B},\quad [b] \longmapsto b \end{equation}

whose kernel we denote $\mathcal{I} \subset \mathcal{A}[\mathcal{B}]$. It is a simple observation that $\mathcal{I}$ is generated by the local sections $[b][b'] - [bb']$ and $[a] - a$. According to Lemma 18.33.8 there is a canonical map
\begin{equation} \label{sites-modules-equation-naive-cotangent-complex} \mathcal{I}/\mathcal{I}^2 \longrightarrow \Omega _{\mathcal{A}[\mathcal{B}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{B}]} \mathcal{B} \end{equation}

whose cokernel is canonically isomorphic to $\Omega _{\mathcal{B}/\mathcal{A}}$.

Definition 18.35.1. Let $\mathcal{C}$ be a site. Let $\mathcal{A} \to \mathcal{B}$ be a homomorphism of sheaves of rings on $\mathcal{C}$. The naive cotangent complex $\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}}$ is the chain complex (

\[ \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} = \left(\mathcal{I}/\mathcal{I}^2 \longrightarrow \Omega _{\mathcal{A}[\mathcal{B}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{B}]} \mathcal{B}\right) \]

with $\mathcal{I}/\mathcal{I}^2$ placed in degree $-1$ and $\Omega _{\mathcal{A}[\mathcal{B}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{B}]} \mathcal{B}$ placed in degree $0$.

This construction satisfies a functoriality similar to that discussed in Lemma 18.33.7 for modules of differentials. Namely, given a commutative diagram
\begin{equation} \label{sites-modules-equation-commutative-square-sheaves} \vcenter { \xymatrix{ \mathcal{B} \ar[r] & \mathcal{B}' \\ \mathcal{A} \ar[u] \ar[r] & \mathcal{A}' \ar[u] } } \end{equation}

of sheaves of rings on $\mathcal{C}$ there is a canonical $\mathcal{B}$-linear map of complexes

\[ \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \longrightarrow \mathop{N\! L}\nolimits _{\mathcal{B}'/\mathcal{A}'} \]

Namely, the maps in the commutative diagram give rise to a canonical map $\mathcal{A}[\mathcal{B}] \to \mathcal{A}'[\mathcal{B}']$ which maps $\mathcal{I}$ into $\mathcal{I}' = \mathop{\mathrm{Ker}}(\mathcal{A}'[\mathcal{B}'] \to \mathcal{B}')$. Thus a map $\mathcal{I}/\mathcal{I}^2 \to \mathcal{I}'/(\mathcal{I}')^2$ and a map between modules of differentials, which together give the desired map between the naive cotangent complexes.

We can choose a different presentation of $\mathcal{B}$ as a quotient of a polynomial algebra over $\mathcal{A}$ and still obtain the same object of $D(\mathcal{B})$. To explain this, suppose that $\mathcal{E}$ is a sheaves of sets on $\mathcal{C}$ and $\alpha : \mathcal{E} \to \mathcal{B}$ a map of sheaves of sets. Then we obtain an $\mathcal{A}$-algebra homomorphism $\mathcal{A}[\mathcal{E}] \to \mathcal{B}$. Assume this map is surjective, and let $\mathcal{J} \subset \mathcal{A}[\mathcal{E}]$ be the kernel. Set

\[ \mathop{N\! L}\nolimits (\alpha ) = \left( \mathcal{J}/\mathcal{J}^2 \longrightarrow \Omega _{\mathcal{A}[\mathcal{E}]/\mathcal{A}} \otimes _{\mathcal{A}[\mathcal{E}]} \mathcal{B}\right) \]

Here is the result.

Lemma 18.35.2. In the situation above there is a canonical isomorphism $\mathop{N\! L}\nolimits (\alpha ) = \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}}$ in $D(\mathcal{B})$.

Proof. Observe that $\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} = \mathop{N\! L}\nolimits (\text{id}_\mathcal {B})$. Thus it suffices to show that given two maps $\alpha _ i : \mathcal{E}_ i \to \mathcal{B}$ as above, there is a canonical quasi-isomorphism $\mathop{N\! L}\nolimits (\alpha _1) = \mathop{N\! L}\nolimits (\alpha _2)$ in $D(\mathcal{B})$. To see this set $\mathcal{E} = \mathcal{E}_1 \amalg \mathcal{E}_2$ and $\alpha = \alpha _1 \amalg \alpha _2 : \mathcal{E} \to \mathcal{B}$. Set $\mathcal{J}_ i = \mathop{\mathrm{Ker}}(\mathcal{A}[\mathcal{E}_ i] \to \mathcal{B})$ and $\mathcal{J} = \mathop{\mathrm{Ker}}(\mathcal{A}[\mathcal{E}] \to \mathcal{B})$. We obtain maps $\mathcal{A}[\mathcal{E}_ i] \to \mathcal{A}[\mathcal{E}]$ which send $\mathcal{J}_ i$ into $\mathcal{J}$. Thus we obtain canonical maps of complexes

\[ \mathop{N\! L}\nolimits (\alpha _ i) \longrightarrow \mathop{N\! L}\nolimits (\alpha ) \]

and it suffices to show these maps are quasi-isomorphism. To see this we argue as follows. First, observe that $H^0(\mathop{N\! L}\nolimits (\alpha _ i)) = \Omega _{\mathcal{B}/\mathcal{A}}$ and $H^0(\mathop{N\! L}\nolimits (\alpha )) = \Omega _{\mathcal{B}/\mathcal{A}}$ by Lemma 18.33.8 hence the map is an isomorphism on cohomology sheaves in degree $0$. Similarly, we claim that $H^{-1}(\mathop{N\! L}\nolimits (\alpha _ i))$ and $H^{-1}(\mathop{N\! L}\nolimits (\alpha ))$ are the sheaves associated to the presheaf $U \mapsto H_1(L_{\mathcal{B}(U)/\mathcal{A}(U)})$ where $H_1(L_{-/-})$ is as in Algebra, Definition 10.134.1. If the claim holds, then the proof is finished.

Proof of the claim. Let $\alpha : \mathcal{E} \to \mathcal{B}$ be as above. Let $\mathcal{B}' \subset \mathcal{B}$ be the subpresheaf of $\mathcal{A}$-algebras whose value on $U$ is the image of $\mathcal{A}(U)[\mathcal{E}(U)] \to \mathcal{B}(U)$. Let $\mathcal{I}'$ be the presheaf whose value on $U$ is the kernel of $\mathcal{A}(U)[\mathcal{E}(U)] \to \mathcal{B}(U)$. Then $\mathcal{I}$ is the sheafification of $\mathcal{I}'$ and $\mathcal{B}$ is the sheafification of $\mathcal{B}'$. Similarly, $H^{-1}(\mathop{N\! L}\nolimits (\alpha ))$ is the sheafification of the presheaf

\[ U \longmapsto \mathop{\mathrm{Ker}}(\mathcal{I}'(U)/\mathcal{I}'(U)^2 \to \Omega _{\mathcal{A}(U)[\mathcal{E}(U)]/\mathcal{A}(U)} \otimes _{\mathcal{A}(U)[\mathcal{E}(U)]} \mathcal{B}'(U)) \]

by Lemma 18.33.4. By Algebra, Lemma 10.134.2 we conclude $H^{-1}(\mathop{N\! L}\nolimits (\alpha ))$ is the sheaf associated to the presheaf $U \mapsto H_1(L_{\mathcal{B}'(U)/\mathcal{A}(U)})$. Thus we have to show that the maps $H_1(L_{\mathcal{B}'(U)/\mathcal{A}(U)}) \to H_1(L_{\mathcal{B}(U)/\mathcal{A}(U)})$ induce an isomorphism $\mathcal{H}'_1 \to \mathcal{H}_1$ of sheafifications.

Injectivity of $\mathcal{H}'_1 \to \mathcal{H}_1$. Let $f \in H_1(L_{\mathcal{B}'(U)/\mathcal{A}(U)})$ map to zero in $\mathcal{H}_1(U)$. To show: $f$ maps to zero in $\mathcal{H}'_1(U)$. The assumption means there is a covering $\{ U_ i \to U\} $ such that $f$ maps to zero in $H_1(L_{\mathcal{B}(U_ i)/\mathcal{A}(U_ i)})$ for all $i$. Replace $U$ by $U_ i$ to get to the point where $f$ maps to zero in $H_1(L_{\mathcal{B}(U)/\mathcal{A}(U)})$. By Algebra, Lemma 10.134.9 we can find a finitely generated subalgebra $\mathcal{B}'(U) \subset B \subset \mathcal{B}(U)$ such that $f$ maps to zero in $H_1(L_{B/\mathcal{A}(U)})$. Since $\mathcal{B} = (\mathcal{B}')^\# $ we can find a covering $\{ U_ i \to U\} $ such that $B \to \mathcal{B}(U_ i)$ factors through $\mathcal{B}'(U_ i)$. Hence $f$ maps to zero in $H_1(L_{\mathcal{B}'(U_ i)/\mathcal{A}(U_ i)})$ as desired.

The surjectivity of $\mathcal{H}'_1 \to \mathcal{H}_1$ is proved in exactly the same way. $\square$

Lemma 18.35.3. Let $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ be morphism of topoi. Let $\mathcal{A} \to \mathcal{B}$ be a homomorphism of sheaves of rings on $\mathcal{D}$. Then $f^{-1}\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} = \mathop{N\! L}\nolimits _{f^{-1}\mathcal{B}/f^{-1}\mathcal{A}}$.

Proof. Omitted. Hint: Use Lemma 18.33.5. $\square$

The cotangent complex of a morphism of ringed topoi is defined in terms of the cotangent complex we defined above.

Definition 18.35.4. Let $X = (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O})$ and $Y = (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ be ringed topoi. Let $(f, f^\sharp ) : X \to Y$ be a morphism of ringed topoi. The naive cotangent complex $\mathop{N\! L}\nolimits _ f = \mathop{N\! L}\nolimits _{X/Y}$ of the given morphism of ringed topoi is $\mathop{N\! L}\nolimits _{\mathcal{O}/f^{-1}\mathcal{O}'}$. We sometimes write $\mathop{N\! L}\nolimits _{X/Y} = \mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}'}$.

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