Definition 37.13.1. Let $f : X \to Y$ be a morphism of schemes. The *naive cotangent complex of $f$* is the complex defined in Modules, Definition 17.28.6. Notation: $\mathop{N\! L}\nolimits _ f$ or $\mathop{N\! L}\nolimits _{X/Y}$.

## 37.13 The naive cotangent complex

This section is the continuation of Modules, Section 17.28 which in turn continues the discussion in Algebra, Section 10.133.

Lemma 37.13.2. Let $f : X \to Y$ be a morphism of schemes. Let $\mathop{\mathrm{Spec}}(A) = U \subset X$ and $\mathop{\mathrm{Spec}}(R) = V \subset S$ be affine opens with $f(U) \subset V$. There is a canonical map

of complexes which is an isomorphism in $D(\mathcal{O}_ U)$.

**Proof.**
From the construction of $\mathop{N\! L}\nolimits _{X/Y}$ in Modules, Section 17.28 we see there is a canonical map of complexes $\mathop{N\! L}\nolimits _{\mathcal{O}_ X(U)/f^{-1}\mathcal{O}_ Y(U)} \to \mathop{N\! L}\nolimits _{X/Y}(U)$ of $A = \mathcal{O}_ X(U)$-modules, which is compatible with further restrictions. Using the canonical map $R \to f^{-1}\mathcal{O}_ Y(U)$ we obtain a canonical map $\mathop{N\! L}\nolimits _{A/R} \to \mathop{N\! L}\nolimits _{\mathcal{O}_ X(U)/f^{-1}\mathcal{O}_ Y(U)}$ of complexes of $A$-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 from Algebra, Lemma 10.133.11 and 10.133.13 and Modules, Lemma 17.28.4 (and the description of the stalks of $\mathcal{O}_ X$ and $f^{-1}\mathcal{O}_ Y$ at a point $\mathfrak p \in \mathop{\mathrm{Spec}}(A)$ as $A_\mathfrak p$ and $R_\mathfrak q$ where $\mathfrak q = R \cap \mathfrak p$; references used are Schemes, Lemma 26.5.4 and Sheaves, Lemma 6.21.5).
$\square$

Lemma 37.13.3. Let $f : X \to Y$ be a morphism of schemes. The cohomology sheaves of the complex $\mathop{N\! L}\nolimits _{X/Y}$ are quasi-coherent, zero outside degrees $-1$, $0$ and equal to $\Omega _{X/Y}$ in degree $0$.

**Proof.**
By construction of the naive cotangent complex in Modules, Section 17.28 we have that $\mathop{N\! L}\nolimits _{X/Y}$ is a complex sitting in degrees $-1$, $0$ and that its cohomology in degree $0$ is $\Omega _{X/Y}$. The sheaf of differentials is quasi-coherent (by Morphisms, Lemma 29.32.7). To finish the proof it suffices to show that $H^{-1}(\mathop{N\! L}\nolimits _{X/Y})$ is quasi-coherent. This follows by checking over affines using Lemma 37.13.2.
$\square$

Lemma 37.13.4. Let $f : X \to Y$ be a morphism of schemes. If $f$ is locally of finite presentation, then $\mathop{N\! L}\nolimits _{X/Y}$ is locally on $X$ quasi-isomorphic to a complex

of quasi-coherent $\mathcal{O}_ X$-modules with $\mathcal{F}^0$ of finite presentation and $\mathcal{F}^{-1}$ of finite type.

**Proof.**
By Lemma 37.13.2 it suffices to show that $\mathop{N\! L}\nolimits _{A/R}$ has this shape if $R \to A$ is a finitely presented ring map. Write $A = R[x_1, \ldots , x_ n]/I$ with $I$ finitely generated. Then $I/I^2$ is a finite $A$-module and $\mathop{N\! L}\nolimits _{A/R}$ is quasi-isomorphic to

by Algebra, Section 10.133 and in particular Algebra, Lemma 10.133.2. $\square$

Lemma 37.13.5. Let $f : X \to Y$ be a morphism of schemes. The following are equivalent

$f$ is formally smooth,

$H^{-1}(\mathop{N\! L}\nolimits _{X/Y}) = 0$ and $H^0(\mathop{N\! L}\nolimits _{X/Y}) = \Omega _{X/Y}$ is locally projective.

**Proof.**
This follows from Algebra, Proposition 10.137.8 and Lemma 37.11.10.
$\square$

Lemma 37.13.6. Let $f : X \to Y$ be a morphism of schemes. The following are equivalent

$f$ is formally étale,

$H^{-1}(\mathop{N\! L}\nolimits _{X/Y}) = H^0(\mathop{N\! L}\nolimits _{X/Y}) = 0$.

**Proof.**
A formally étale morphism is formally smooth and hence we have $H^{-1}(\mathop{N\! L}\nolimits _{X/Y}) = 0$ by Lemma 37.13.5. On the other hand, we have $\Omega _{X/Y} = 0$ by Lemma 37.8.6. Conversely, if (2) holds, then $f$ is formally smooth by Lemma 37.13.5 and formally unramified by Lemma 37.6.7 and hence formally étale by Lemmas 37.11.4.
$\square$

Lemma 37.13.7. Let $f : X \to Y$ be a morphism of schemes. The following are equivalent

$f$ is smooth, and

$f$ is locally of finite presentation, $H^{-1}(\mathop{N\! L}\nolimits _{X/Y}) = 0$, and $H^0(\mathop{N\! L}\nolimits _{X/Y}) = \Omega _{X/Y}$ is finite locally free.

**Proof.**
This follows from the definition of a smooth ring homomorphism (Algebra, Definition 10.136.1), Lemma 37.13.2, and the definition of a smooth morphism of schemes (Morphisms, Definition 29.33.1). We also use that finite locally free is the same as finite projective for modules over rings (Algebra, Lemma 10.77.2).
$\square$

Lemma 37.13.8. Let $i : Z \to X$ be an immersion of schemes. Then $\mathop{N\! L}\nolimits _{Z/X}$ is isomorphic to $\mathcal{C}_{Z/X}[1]$ in $D(\mathcal{O}_ Z)$ where $\mathcal{C}_{Z/X}$ is the conormal sheaf of $Z$ in $X$.

**Proof.**
This follows from Algebra, Lemma 10.133.6, Morphisms, Lemma 29.31.2, and Lemma 37.13.2.
$\square$

Lemma 37.13.9. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes. There is a canonical six term exact sequence

of cohomology sheaves.

**Proof.**
Special case of Modules, Lemma 17.28.7.
$\square$

Lemma 37.13.10. Let $f : X \to Y$ and $Y \to Z$ be morphisms of schemes. Assume $X \to Y$ is a complete intersection morphism. Then there is a canonical distinguished triangle

in $D(\mathcal{O}_ X)$ which recovers the $6$-term exact sequence of Lemma 37.13.9.

**Proof.**
It suffices to show the canonical map

of Modules, Lemma 17.28.7 is an isomorphism in $D(\mathcal{O}_ X)$. In order to show this, it suffices to show that the $6$-term sequence has a zero on the left, i.e., that $H^{-1}(f^*\mathop{N\! L}\nolimits _{Y/Z}) \to H^{-1}(\mathop{N\! L}\nolimits _{X/Z})$ is injective. Affine locally this follows from the corresponding algebra result in More on Algebra, Lemma 15.32.6. To translate into algebra use Lemma 37.13.2. $\square$

Lemma 37.13.11. Let $f : X \to Y$ be a morphism of schemes which factors as $f = g \circ i$ with $i$ an immersion and $g : P \to Y$ formally smooth (for example smooth). Then there is a canonical isomorphism

in $D(\mathcal{O}_ X)$ where the conormal sheaf $\mathcal{C}_{X/P}$ is placed in degree $-1$.

**Proof.**
(For the parenthetical statement see Lemma 37.11.7.) By Lemmas 37.13.8 and 37.13.5 we have $\mathop{N\! L}\nolimits _{X/P} = \mathcal{C}_{X/P}[1]$ and $\mathop{N\! L}\nolimits _{P/Y} = \Omega _{P/Y}$ with $\Omega _{P/Y}$ locally projective. This implies that $i^*\mathop{N\! L}\nolimits _{P/Y} \to i^*\Omega _{P/Y}$ is a quasi-isomorphism too (small detail omitted; the reason is that $i^*\mathop{N\! L}\nolimits _{P/Y}$ is the same thing as $\tau _{\geq -1}Li^*\mathop{N\! L}\nolimits _{P/Y}$, see More on Algebra, Lemma 15.79.1). Thus the canonical map

of Modules, Lemma 17.28.7 is an isomorphism in $D(\mathcal{O}_ X)$ because the cohomology group $H^{-1}(i^*\mathop{N\! L}\nolimits _{P/Y})$ is zero by what we said above. In other words, we have a distinguished triangle

Clearly, this means that $\mathop{N\! L}\nolimits _{X/Y}$ is the cone on the map $\mathop{N\! L}\nolimits _{X/P}[-1] \to i^*\mathop{N\! L}\nolimits _{P/Y}$ which is equivalent to the statement of the lemma by our computation of the cohomology sheaves of these objects in the derived category given above. $\square$

Lemma 37.13.12. Consider a cartesian diagram of schemes

The canonical map $(g')^*\mathop{N\! L}\nolimits _{X/Y} \to \mathop{N\! L}\nolimits _{X'/Y'}$ induces an isomorphism on $H^0$ and a surjection on $H^{-1}$.

**Proof.**
Translated into algebra this is More on Algebra, Lemma 15.79.2. To do the translation use Lemma 37.13.2.
$\square$

Lemma 37.13.13. Consider a cartesian diagram of schemes

If $Y' \to Y$ is flat, then the canonical map $(g')^*\mathop{N\! L}\nolimits _{X/Y} \to \mathop{N\! L}\nolimits _{X'/Y'}$ is a quasi-isomorphism.

**Proof.**
By Lemma 37.13.2 this follows from Algebra, Lemma 10.133.8.
$\square$

Lemma 37.13.14. Consider a cartesian diagram of schemes

If $X \to Y$ is flat, then the canonical map $(g')^*\mathop{N\! L}\nolimits _{X/Y} \to \mathop{N\! L}\nolimits _{X'/Y'}$ is a quasi-isomorphism. If in addition $\mathop{N\! L}\nolimits _{X/Y}$ has tor-amplitude in $[-1, 0]$ then $L(g')^*\mathop{N\! L}\nolimits _{X/Y} \to \mathop{N\! L}\nolimits _{X'/Y'}$ is a quasi-isomorphism too.

**Proof.**
Translated into algebra this is More on Algebra, Lemma 15.79.3. To do the translation use Lemma 37.13.2 and Derived Categories of Schemes, Lemmas 36.3.5 and 36.9.4.
$\square$

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