## 91.24 The cotangent complex of a morphism of schemes

As promised above we define the cotangent complex of a morphism of schemes as follows.

Definition 91.24.1. Let $f : X \to Y$ be a morphism of schemes. The cotangent complex $L_{X/Y}$ of $X$ over $Y$ is the cotangent complex of $f$ as a morphism of ringed spaces (Definition 91.20.1).

In particular, the results of Section 91.20 apply to cotangent complexes of morphisms of schemes. The next lemma shows this definition is compatible with the definition for ring maps and it also implies that $L_{X/Y}$ is an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$.

Lemma 91.24.2. Let $f : X \to Y$ be a morphism of schemes. Let $U = \mathop{\mathrm{Spec}}(A) \subset X$ and $V = \mathop{\mathrm{Spec}}(B) \subset Y$ be affine opens such that $f(U) \subset V$. There is a canonical map

$\widetilde{L_{B/A}} \longrightarrow L_{X/Y}|_ U$

of complexes which is an isomorphism in $D(\mathcal{O}_ U)$. This map is compatible with restricting to smaller affine opens of $X$ and $Y$.

Proof. By Remark 91.18.5 there is a canonical map of complexes $L_{\mathcal{O}_ X(U)/f^{-1}\mathcal{O}_ Y(U)} \to L_{X/Y}(U)$ of $B = \mathcal{O}_ X(U)$-modules, which is compatible with further restrictions. Using the canonical map $A \to f^{-1}\mathcal{O}_ Y(U)$ we obtain a canonical map $L_{B/A} \to L_{\mathcal{O}_ X(U)/f^{-1}\mathcal{O}_ Y(U)}$ of complexes of $B$-modules. Using the universal property of the $\widetilde{\ }$ functor (see Schemes, Lemma 26.7.1) we obtain a map as in the statement of the lemma. We may check this map is an isomorphism on cohomology sheaves by checking it induces isomorphisms on stalks. This follows immediately from Lemmas 91.18.9 and 91.8.6 (and the description of the stalks of $\mathcal{O}_ X$ and $f^{-1}\mathcal{O}_ Y$ at a point $\mathfrak p \in \mathop{\mathrm{Spec}}(B)$ as $B_\mathfrak p$ and $A_\mathfrak q$ where $\mathfrak q = A \cap \mathfrak p$; references used are Schemes, Lemma 26.5.4 and Sheaves, Lemma 6.21.5). $\square$

Lemma 91.24.3. Let $\Lambda$ be a ring. Let $X$ be a scheme over $\Lambda$. Then

$L_{X/\mathop{\mathrm{Spec}}(\Lambda )} = L_{\mathcal{O}_ X/\underline{\Lambda }}$

where $\underline{\Lambda }$ is the constant sheaf with value $\Lambda$ on $X$.

Proof. Let $p : X \to \mathop{\mathrm{Spec}}(\Lambda )$ be the structure morphism. Let $q : \mathop{\mathrm{Spec}}(\Lambda ) \to (*, \Lambda )$ be the obvious morphism. By the distinguished triangle of Lemma 91.20.3 it suffices to show that $L_ q = 0$. To see this it suffices to show for $\mathfrak p \in \mathop{\mathrm{Spec}}(\Lambda )$ that

$(L_ q)_\mathfrak p = L_{\mathcal{O}_{\mathop{\mathrm{Spec}}(\Lambda ), \mathfrak p}/\Lambda } = L_{\Lambda _\mathfrak p/\Lambda }$

(Lemma 91.18.9) is zero which follows from Lemma 91.8.4. $\square$

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