## 90.11 Comparison with the naive cotangent complex

The naive cotangent complex was introduced in Algebra, Section 10.134.

Remark 90.11.1. Let $A \to B$ be a ring map. Working on $\mathcal{C}_{B/A}$ as in Section 90.4 let $\mathcal{J} \subset \mathcal{O}$ be the kernel of $\mathcal{O} \to \underline{B}$. Note that $L\pi _!(\mathcal{J}) = 0$ by Lemma 90.5.7. Set $\Omega = \Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}$ so that $L_{B/A} = L\pi _!(\Omega )$ by Lemma 90.4.3. It follows that $L\pi _!(\mathcal{J} \to \Omega ) = L\pi _!(\Omega ) = L_{B/A}$. Thus, for any object $U = (P \to B)$ of $\mathcal{C}_{B/A}$ we obtain a map

90.11.1.1
$$\label{cotangent-equation-comparison-map-A} (J \to \Omega _{P/A} \otimes _ P B) \longrightarrow L_{B/A}$$

where $J = \mathop{\mathrm{Ker}}(P \to B)$ in $D(A)$, see Cohomology on Sites, Remark 21.38.4. Continuing in this manner, note that $L\pi _!(\mathcal{J} \otimes _\mathcal {O}^\mathbf {L} \underline{B}) = L\pi _!(\mathcal{J}) = 0$ by Lemma 90.5.6. Since $\text{Tor}_0^\mathcal {O}(\mathcal{J}, \underline{B}) = \mathcal{J}/\mathcal{J}^2$ the spectral sequence

$H_ p(\mathcal{C}_{B/A}, \text{Tor}_ q^\mathcal {O}(\mathcal{J}, \underline{B})) \Rightarrow H_{p + q}(\mathcal{C}_{B/A}, \mathcal{J} \otimes _\mathcal {O}^\mathbf {L} \underline{B}) = 0$

(dual of Derived Categories, Lemma 13.21.3) implies that $H_0(\mathcal{C}_{B/A}, \mathcal{J}/\mathcal{J}^2) = 0$ and $H_1(\mathcal{C}_{B/A}, \mathcal{J}/\mathcal{J}^2) = 0$. It follows that the complex of $\underline{B}$-modules $\mathcal{J}/\mathcal{J}^2 \to \Omega$ satisfies $\tau _{\geq -1}L\pi _!(\mathcal{J}/\mathcal{J}^2 \to \Omega ) = \tau _{\geq -1}L_{B/A}$. Thus, for any object $U = (P \to B)$ of $\mathcal{C}_{B/A}$ we obtain a map

90.11.1.2
$$\label{cotangent-equation-comparison-map} (J/J^2 \to \Omega _{P/A} \otimes _ P B) \longrightarrow \tau _{\geq -1}L_{B/A}$$

in $D(B)$, see Cohomology on Sites, Remark 21.38.4.

The first case is where we have a surjection of rings.

Lemma 90.11.2. Let $A \to B$ be a surjective ring map with kernel $I$. Then $H^0(L_{B/A}) = 0$ and $H^{-1}(L_{B/A}) = I/I^2$. This isomorphism comes from the map (90.11.1.2) for the object $(A \to B)$ of $\mathcal{C}_{B/A}$.

Proof. We will show below (using the surjectivity of $A \to B$) that there exists a short exact sequence

$0 \to \pi ^{-1}(I/I^2) \to \mathcal{J}/\mathcal{J}^2 \to \Omega \to 0$

of sheaves on $\mathcal{C}_{B/A}$. Taking $L\pi _!$ and the associated long exact sequence of homology, and using the vanishing of $H_1(\mathcal{C}_{B/A}, \mathcal{J}/\mathcal{J}^2)$ and $H_0(\mathcal{C}_{B/A}, \mathcal{J}/\mathcal{J}^2)$ shown in Remark 90.11.1 we obtain what we want using Lemma 90.4.4.

What is left is to verify the local statement mentioned above. For every object $U = (P \to B)$ of $\mathcal{C}_{B/A}$ we can choose an isomorphism $P = A[E]$ such that the map $P \to B$ maps each $e \in E$ to zero. Then $J = \mathcal{J}(U) \subset P = \mathcal{O}(U)$ is equal to $J = IP + (e; e \in E)$. The value on $U$ of the short sequence of sheaves above is the sequence

$0 \to I/I^2 \to J/J^2 \to \Omega _{P/A} \otimes _ P B \to 0$

Verification omitted (hint: the only tricky point is that $IP \cap J^2 = IJ$; which follows for example from More on Algebra, Lemma 15.30.9). $\square$

Lemma 90.11.3. Let $A \to B$ be a ring map. Then $\tau _{\geq -1}L_{B/A}$ is canonically quasi-isomorphic to the naive cotangent complex.

Proof. Consider $P = A[B] \to B$ with kernel $I$. The naive cotangent complex $\mathop{N\! L}\nolimits _{B/A}$ of $B$ over $A$ is the complex $I/I^2 \to \Omega _{P/A} \otimes _ P B$, see Algebra, Definition 10.134.1. Observe that in (90.11.1.2) we have already constructed a canonical map

$c : \mathop{N\! L}\nolimits _{B/A} \longrightarrow \tau _{\geq -1}L_{B/A}$

Consider the distinguished triangle (90.7.0.1)

$L_{P/A} \otimes _ P^\mathbf {L} B \to L_{B/A} \to L_{B/P} \to (L_{P/A} \otimes _ P^\mathbf {L} B)[1]$

associated to the ring maps $A \to A[B] \to B$. We know that $L_{P/A} = \Omega _{P/A}[0] = \mathop{N\! L}\nolimits _{P/A}$ in $D(P)$ (Lemma 90.4.7 and Algebra, Lemma 10.134.3) and that $\tau _{\geq -1}L_{B/P} = I/I^2[1] = \mathop{N\! L}\nolimits _{B/P}$ in $D(B)$ (Lemma 90.11.2 and Algebra, Lemma 10.134.6). To show $c$ is a quasi-isomorphism it suffices by Algebra, Lemma 10.134.4 and the long exact cohomology sequence associated to the distinguished triangle to show that the maps $L_{P/A} \to L_{B/A} \to L_{B/P}$ are compatible on cohomology groups with the corresponding maps $\mathop{N\! L}\nolimits _{P/A} \to \mathop{N\! L}\nolimits _{B/A} \to \mathop{N\! L}\nolimits _{B/P}$ of the naive cotangent complex. We omit the verification. $\square$

Remark 90.11.4. We can make the comparison map of Lemma 90.11.3 explicit in the following way. Let $P_\bullet$ be the standard resolution of $B$ over $A$. Let $I = \mathop{\mathrm{Ker}}(A[B] \to B)$. Recall that $P_0 = A[B]$. The map of the lemma is given by the commutative diagram

$\xymatrix{ L_{B/A} \ar[d] & \ldots \ar[r] & \Omega _{P_2/A} \otimes _{P_2} B \ar[r] \ar[d] & \Omega _{P_1/A} \otimes _{P_1} B \ar[r] \ar[d] & \Omega _{P_0/A} \otimes _{P_0} B \ar[d] \\ \mathop{N\! L}\nolimits _{B/A} & \ldots \ar[r] & 0 \ar[r] & I/I^2 \ar[r] & \Omega _{P_0/A} \otimes _{P_0} B }$

We construct the downward arrow with target $I/I^2$ by sending $\text{d}f \otimes b$ to the class of $(d_0(f) - d_1(f))b$ in $I/I^2$. Here $d_ i : P_1 \to P_0$, $i = 0, 1$ are the two face maps of the simplicial structure. This makes sense as $d_0 - d_1$ maps $P_1$ into $I = \mathop{\mathrm{Ker}}(P_0 \to B)$. We omit the verification that this rule is well defined. Our map is compatible with the differential $\Omega _{P_1/A} \otimes _{P_1} B \to \Omega _{P_0/A} \otimes _{P_0} B$ as this differential maps $\text{d}f \otimes b$ to $\text{d}(d_0(f) - d_1(f)) \otimes b$. Moreover, the differential $\Omega _{P_2/A} \otimes _{P_2} B \to \Omega _{P_1/A} \otimes _{P_1} B$ maps $\text{d}f \otimes b$ to $\text{d}(d_0(f) - d_1(f) + d_2(f)) \otimes b$ which are annihilated by our downward arrow. Hence a map of complexes. We omit the verification that this is the same as the map of Lemma 90.11.3.

Remark 90.11.5. Adopt notation as in Remark 90.11.1. The arguments given there show that the differential

$H_2(\mathcal{C}_{B/A}, \mathcal{J}/\mathcal{J}^2) \longrightarrow H_0(\mathcal{C}_{B/A}, \text{Tor}_1^\mathcal {O}(\mathcal{J}, \underline{B}))$

of the spectral sequence is an isomorphism. Let $\mathcal{C}'_{B/A}$ denote the full subcategory of $\mathcal{C}_{B/A}$ consisting of surjective maps $P \to B$. The agreement of the cotangent complex with the naive cotangent complex (Lemma 90.11.3) shows that we have an exact sequence of sheaves

$0 \to \underline{H_1(L_{B/A})} \to \mathcal{J}/\mathcal{J}^2 \xrightarrow {\text{d}} \Omega \to \underline{H_2(L_{B/A})} \to 0$

on $\mathcal{C}'_{B/A}$. It follows that $\mathop{\mathrm{Ker}}(d)$ and $\mathop{\mathrm{Coker}}(d)$ on the whole category $\mathcal{C}_{B/A}$ have vanishing higher homology groups, since these are computed by the homology groups of constant simplicial abelian groups by Lemma 90.4.1. Hence we conclude that

$H_ n(\mathcal{C}_{B/A}, \mathcal{J}/\mathcal{J}^2) \to H_ n(L_{B/A})$

is an isomorphism for all $n \geq 2$. Combined with the remark above we obtain the formula $H_2(L_{B/A}) = H_0(\mathcal{C}_{B/A}, \text{Tor}_1^\mathcal {O}(\mathcal{J}, \underline{B}))$.

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