In this section we prove some results on two term complexes of modules which will help us understand conditions on the naive cotangent complex.
Lemma 15.84.1. Let $R$ be a ring. Let $K \in D(R)$ with $H^ i(K) = 0$ for $i \not\in \{ -1, 0\} $. The following are equivalent
$H^{-1}(K) = 0$ and $H^0(K)$ is a projective module and
$\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, M) = 0$ for every $R$-module $M$.
If $R$ is Noetherian and $H^ i(K)$ is a finite $R$-module for $i = -1, 0$, then these are also equivalent to
$\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, M) = 0$ for every finite $R$-module $M$.
Proof. The equivalence of (1) and (2) follows from Lemma 15.68.2. If $R$ is Noetherian and $H^ i(K)$ is a finite $R$-module for $i = -1, 0$, then $K$ is pseudo-coherent, see Lemma 15.64.17. Thus the equivalence of (1) and (3) follows from Lemma 15.77.4. $\square$
Remark 15.84.2. The following two statements follow from Lemma 15.84.1, Algebra, Definition 10.137.1, and Algebra, Proposition 10.138.8.
A ring map $A \to B$ is smooth if and only if $A \to B$ is of finite presentation and $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N) = 0$ for every $B$-module $N$.
A ring map $A \to B$ is formally smooth if and only if $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N) = 0$ for every $B$-module $N$.
Lemma 15.84.3. Let $R$ be a ring. Let $K$ be an object of $D(R)$ with $H^ i(K) = 0$ for $i \not\in \{ -1, 0\} $. Then
$K$ can be represented by a two term complex $K^{-1} \to K^0$ with $K^0$ a free module, and
if $R$ is Noetherian and $H^ i(K)$ is a finite $R$-module for $i = -1, 0$, then $K$ can be represented by a two term complex $K^{-1} \to K^0$ with $K^0$ a finite free module and $K^{-1}$ finite.
Proof. Proof of (1). Suppose $K$ is given by the complex of modules $M^\bullet $. We may first replace $M^\bullet $ by $\tau _{\leq 0}M^\bullet $. Thus we may assume $M^ i = 0$ for $i > 0$, Next, we may choose a free resolution $P^\bullet \to M^\bullet $ with $P^ i = 0$ for $i > 0$, see Derived Categories, Lemma 13.15.4. Finally, we can set $K^\bullet = \tau _{\geq -1}P^\bullet $.
Proof of (2). Assume $R$ is Noetherian and $H^ i(K)$ is a finite $R$-module for $i = -1, 0$. By Lemma 15.64.5 we can choose a quasi-isomorphism $F^\bullet \to M^\bullet $ with $F^ i = 0$ for $i > 0$ and $F^ i$ finite free. Then we can set $K^\bullet = \tau _{\geq -1}F^\bullet $. $\square$
Maps in the derived category out of the naive cotangent complex $\mathop{N\! L}\nolimits _{B/A}$ or $\mathop{N\! L}\nolimits (\alpha )$ (see Algebra, Section 10.134) are easy to understand by the result of the following lemma.
Lemma 15.84.4. Let $R$ be a ring. Let $M^\bullet $ be a complex of modules over $R$ with $M^ i = 0$ for $i > 0$ and $M^0$ a projective $R$-module. Let $K^\bullet $ be a second complex.
Assume $K^ i = 0$ for $i \leq -2$. Then $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(M^\bullet , K^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(M^\bullet , K^\bullet )$.
Assume $K^ i = 0$ for $i \not\in [-1, 0]$ and $K^0$ a projective $R$-module. Then for a map of complexes $a^\bullet : M^\bullet \to K^\bullet $, the following are equivalent
$a^\bullet $ induces the zero map $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K^\bullet , N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M^\bullet , N)$ for all $R$-modules $N$, and
there is a map $h^0 : M^0 \to K^{-1}$ such that $a^{-1} + h^0 \circ d^{-1}_ K = 0$.
Assume $K^ i = 0$ for $i \leq -3$. Let $\alpha \in \mathop{\mathrm{Hom}}\nolimits _{D(R)}(M^\bullet , K^\bullet )$. If the composition of $\alpha $ with $K^\bullet \to K^{-2}[2]$ comes from an $R$-module map $a : M^{-2} \to K^{-2}$ with $a \circ d_ M^{-3} = 0$, then $\alpha $ can be represented by a map of complexes $a^\bullet : M^\bullet \to K^\bullet $ with $a^{-2} = a$.
In (2) for any second map of complexes $(a')^\bullet : M^\bullet \to K^\bullet $ representing $\alpha $ with $a = (a')^{-2}$ there exist $h^ i : M^ i \to K^{i - 1}$ for $i = 0, -1$ such that
\[ h^{-1} \circ d_ M^{-2} = 0, \quad (a')^{-1} = a^{-1} + d_ K^{-2} \circ h^{-1} + h^0 \circ d_ M^{-1},\quad (a')^0 = a^0 + d_ K^{-1} \circ h^0 \]
Proof. Set $F^0 = M^0$. Choose a free $R$-module $F^{-1}$ and a surjection $F^{-1} \to M^{-1}$. Choose a free $R$-module $F^{-2}$ and a surjection $F^{-2} \to M^{-2} \times _{M^{-1}} F^{-1}$. Continuing in this way we obtain a quasi-isomorphism $p^\bullet : F^\bullet \to M^\bullet $ which is termwise surjective and with $F^ i$ projective for all $i$.
Proof of (1). By Derived Categories, Lemma 13.19.8 we have
If $K^ i = 0$ for $i \leq -2$, then any morphism of complexes $F^\bullet \to K^\bullet $ factors through $p^\bullet $. Similarly, any homotopy $\{ h^ i : F^ i \to K^{i - 1}\} $ factors through $p^\bullet $. Thus (1) holds.
Proof of (2). If (2)(b) holds, then $a^\bullet $ is homotopic to a map of complexes $(a')^\bullet : M^\bullet \to K^\bullet $ which is zero in degree $-1$. On the other hand, let $N \to I^\bullet $ be an injective resolution. We have
by Derived Categories, Lemma 13.18.8. Let $b^\bullet : K^\bullet \to I^\bullet [1]$ be a map of complexes. Since $K^1 = 0$ the map $b^0 : K^0 \to I^1$ maps into the kernel of $I^1 \to I^2$ which is the image of $I^0 \to I^1$. Since $K^0$ is projective we can lift $b^0$ to a map $h : K^0 \to I^0$. Thus we see that $b^\bullet $ is homotopic to a map of complexes $(b')^\bullet $ with $(b')^0 = 0$. Since $K^ i = 0$ for $i \not\in [-1, 0]$ it follows that $(b')^\bullet \circ (a')^\bullet = 0$ as a map of complexes. Hence the map $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K^\bullet , N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M^\bullet , N)$ is zero. In this way we see that (2)(b) implies (2)(a). Conversely, assume (2)(a). We see that the canonical element in $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K^\bullet , K^{-1})$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^1_ R(M^\bullet , K^{-1})$. Using (1) we see immediately that we get a map $h^0$ as in (2)(b).
Proof of (3). Choose $b^\bullet : F^\bullet \to K^\bullet $ representing $\alpha $. The composition of $\alpha $ with $K^\bullet \to K^{-2}[2]$ is represented by $b^{-2} : F^{-2} \to K^{-2}$. As this is homotopic to $a \circ p^{-2} : F^{-2} \to M^{-2} \to K^{-2}$, there is a map $h : F^{-1} \to K^{-2}$ such that $b^{-2} = a \circ p^{-2} + h \circ d_ F^{-2}$. Adjusting $b^\bullet $ by $h$ viewed as a homotopy from $F^\bullet $ to $K^\bullet $, we find that $b^{-2} = a \circ p^{-2}$. Hence $b^{-2}$ factors through $p^{-2}$. Since $F^0 = M^0$ the kernel of $p^{-2}$ surjects onto the kernel of $p^{-1}$ (for example because the kernel of $p^\bullet $ is an acyclic complex or by a diagram chase). Hence $b^{-1}$ necessarily factors through $p^{-1}$ as well and we see that (3) holds for these factorizations and $a^0 = b^0$.
Proof of (4) is omitted. Hint: There is a homotopy between $a^\bullet \circ p^\bullet $ and $(a')^\bullet \circ p^\bullet $ and we argue as before that this homotopy factors through $p^\bullet $. $\square$
Let $A \to B$ be a finitely presented ring map. Given an ideal $I \subset B$ we can consider the condition
$\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ is annihilated by $I$ for all $B$-modules $N$.
This condition is one possible precise mathematical formulation of the notion “the singular locus of $A \to B$ is scheme theoretically contained in $V(I)$”. Please compare with Remark 15.84.2 and the following lemmas.
Lemma 15.84.5. Let $R$ be a ring and let $I \subset R$ be an ideal. Let $K \in D(R)$. Assume $H^ i(K) = 0$ for $i \not\in \{ -1, 0\} $. The following are equivalent
$\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$ is annihilated by $I$ for all $R$-modules $N$,
$K$ can be represented by a complex $K^{-1} \to K^0$ with $K^0$ free such that for any $a \in I$ the map $a : K^{-1} \to K^{-1}$ factors through $d_ K^{-1} : K^{-1} \to K^0$,
whenever $K$ is represented by a two term complex $K^{-1} \to K^0$ with $K^0$ projective, then for any $a \in I$ the map $a : K^{-1} \to K^{-1}$ factors through $d_ K^{-1} : K^{-1} \to K^0$.
If $R$ is Noetherian and $H^ i(K)$ is a finite $R$-module for $i = -1, 0$, then these are also equivalent to
$\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$ is annihilated by $I$ for every finite $R$-module $N$,
$K$ can be represented by a complex $K^{-1} \to K^0$ with $K^0$ finite free and $K^{-1}$ finite such that for any $a \in I$ the map $a : K^{-1} \to K^{-1}$ factors through $d_ K^{-1} : K^{-1} \to K^0$.
Proof. Assume (1) and let $K^{-1} \to K^0$ be a two term complex representing $K$ with $K^0$ projective. We will use the description of maps in $D(R)$ out of $K^\bullet $ given in Lemma 15.84.4 without further mention. Choosing $N = K^{-1}$ consider the element $\xi $ of $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$ given by $\text{id}_{K^{-1}} : K^{-1} \to K^{-1}$. Since is annihilated by $a \in I$ we see that we get the dotted arrow fitting into the following commutative diagram
This proves that (3) holds. Part (3) implies (2) in view of Lemma 15.84.3 part (1). Assume $K^\bullet $ is as in (2) and $N$ is an arbitrary $R$-module. Any element $\xi $ of $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$ is given as the class of a map $\varphi : K^{-1} \to N$. Then for $a \in I$ by assumption we may choose a map $h$ as in the diagram above and we see that $a\varphi = \varphi \circ a = \varphi \circ h \circ \text{d}_ K^{-1}$ which proves that $a \xi $ is zero in $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$. Thus (1), (2), and (3) are equivalent.
Assume $R$ is Noetherian and $H^ i(K)$ is a finite $R$-module for $i = -1, 0$. Part (3) implies (5) in view of Lemma 15.84.3 part (2). It is clear that (5) implies (2). Trivially (1) implies (4). Thus to finish the proof it suffices to show that (4) implies any of the other conditions. Let $K^{-1} \to K^0$ be a complex representing $K$ with $K^0$ finite free and $K^{-1}$ finite as in Lemma 15.84.3 part (2). The argument given in the proof of (2) $\Rightarrow $ (1) shows that if $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, K^{-1})$ is annihilated by $I$, then (1) holds. In this way we see that (4) implies (1) and the proof is complete. $\square$
Lemma 15.84.6. Let $R$ be a ring. Let $K$ be an object of $D(R)$ with $H^ i(K) = 0$ for $i \not\in \{ -1, 0\} $. Let $K^{-1} \to K^0$ be a two term complex of $R$-modules representing $K$ such that $K^0$ is a flat $R$-module (for example projective or free). Let $R \to R'$ be a ring map. Then the complex $K^\bullet \otimes _ R R'$ represents $\tau _{\geq -1}(K \otimes _ R^\mathbf {L} R')$.
Proof. We have a distinguished triangle
in $D(R)$. This determines a map of distinguished triangles
The left and right vertical arrows are isomorphisms as $K^0$ is flat. Since $K^{-1} \otimes _ R^\mathbf {L} R' \to K^{-1} \otimes _ R R'$ is an isomorphism on cohomology in degree $0$ we conclude. $\square$
Lemma 15.84.7. Let $I$ be an ideal of a ring $R$. Let $K$ be an object of $D(R)$ with $H^ i(K) = 0$ for $i \not\in \{ -1, 0\} $. Let $R \to R'$ be a ring map. If $K$ satisfies the equivalent conditions (1), (2), and (3) of Lemma 15.84.5 with respect to $(R, I)$, then $\tau _{\geq -1}(K \otimes _ R^\mathbf {L} R')$ satisfies the equivalent conditions (1), (2), and (3) of Lemma 15.84.5 with respect to $(R', IR')$
Proof. We may assume $K$ is represented by a two term complex $K^{-1} \to K^0$ with $K^0$ free such that for any $a \in I$ the map $a : K^{-1} \to K^{-1}$ is equal to $h_ a \circ d_ K^{-1}$ for some map $h_ a : K^0 \to K^{-1}$. By Lemma 15.84.6 we see that $\tau _{\geq -1}(K \otimes _ R^\mathbf {L} R')$ is represented by $K^\bullet \otimes _ R R'$. Then of course for every $a \in I$ we see that $a \otimes 1 : K^{-1} \otimes _ R R' \to K^{-1} \otimes _ R R'$ is equal to $(h_ a \otimes 1) \circ (d_ K^{-1} \otimes 1)$. Since the collection of maps $K^{-1} \otimes _ R R' \to K^{-1} \otimes _ R R'$ which factor through $\text{d}_ K^{-1} \otimes 1$ forms an $R'$-module we conclude. $\square$
Lemma 15.84.8. Let $R$ be a ring. Let $\alpha : K \to K'$ be a morphism of $D(R)$. Assume
$H^ i(K) = H^ i(K') = 0$ for $i \not\in \{ -1, 0\} $
$H^0(\alpha )$ is an isomorphism and $H^{-1}(\alpha )$ is surjective.
For any $f \in R$ if $f : K \to K$ is $0$, then $f : K' \to K'$ is $0$.
Proof. Set $M = \mathop{\mathrm{Ker}}(H^{-1}(\alpha ))$. Then $\alpha $ fits into a distinguished triangle
Since $K \to K' \xrightarrow {f} K'$ is zero by our assumption, we see that $f : K' \to K'$ factors over a map $M[2] \to K'$. However $\mathop{\mathrm{Hom}}\nolimits (M[2], K') = 0$ for example by Derived Categories, Lemma 13.27.3. $\square$
Lemma 15.84.9. Let $I$ be an ideal of a ring $R$. Let $\alpha : K \to K'$ be a morphism of $D(R)$. Assume
$H^ i(K) = H^ i(K') = 0$ for $i \not\in \{ -1, 0\} $
$H^0(\alpha )$ is an isomorphism and $H^{-1}(\alpha )$ is surjective.
If $K$ satisfies the equivalent conditions (1), (2), and (3) of Lemma 15.84.5, then $K'$ does too.
Proof. Set $M = \mathop{\mathrm{Ker}}(H^{-1}(\alpha ))$. Then $\alpha $ fits into a distinguished trangle
For any $R$-module $N$ this determines an exact sequence
Since $\mathop{\mathrm{Ext}}\nolimits ^0_ R(M[1], N) = \mathop{\mathrm{Ext}}\nolimits ^{-1}_ R(M, N) = 0$ we see that $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K', N)$ is a submodule of $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$. Hence if $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$ is annihilated by $I$ so is $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K', N)$. $\square$
Lemma 15.84.10. Let $R$ be ring and let $I \subset R$ be an ideal. Let $K \in D(R)$ with $H^ i(K) = 0$ for $i \not\in \{ -1, 0\} $. The following are equivalent
If $R$ is Noetherian and $H^ i(K)$ is a finite $R$-module for $i = -1, 0$, then these are also equivalent to
there exists a $c \geq 0$ such that the equivalent conditions (4), (5) of Lemma 15.84.5 hold for $K$ and the ideal $I^ c$,
$H^{-1}(K)$ is $I$-power torsion and there exist $f_1, \ldots , f_ s \in R$ with $V(f_1, \ldots , f_ s) \subset V(I)$ such that the localizations $H^0(K)_{f_ i}$ are projective $R_{f_ i}$-modules,
$H^{-1}(K)$ is $I$-power torsion and there exist $f_1, \ldots , f_ s \in I$ with $V(f_1, \ldots , f_ s) = V(I)$ such that the localizations $H^0(K)_{f_ i}$ are projective $R_{f_ i}$-modules.
Proof. The distinguished triangle $H^{-1}(K)[1] \to K \to H^0(K)[0] \to H^{-1}(K)[2]$ determines an exact sequence
Thus (2) implies that $I^{2c}$ annihilates $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$ for every $R$-module $N$. Assuming (1) we immediately see that $H^0(K)$ is $I^ c$-projective. On the other hand, we may choose an injective map $H^{-1}(K) \to N$ for some injective $R$-module $N$. Then this map is the image of an element of $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$ by the vanishing of the $\mathop{\mathrm{Ext}}\nolimits ^2$ in the sequence and we conclude $H^{-1}(K)$ is annihilated by $I^ c$.
Assume $R$ is Noetherian and $H^ i(K)$ is a finite $R$-module for $i = -1, 0$. By Lemma 15.84.5 we see that (3) is equivalent to (1) and (2). Also, if (3) holds then for $f \in I$ the multiplication by $f$ on $H^0(K)$ factors through a projective module, which implies that $H^0(K)_ f$ is a summand of a projective $R_ f$-module and hence itself a projective $R_ f$-module. Choosing $f_1, \ldots , f_ s$ to be generators of $I$ we find the equivalent conditions (1), (2), and (3) imply (5). Of course (5) trivially implies (4).
Assume (4). Since $H^{-1}(K)$ is a finite $R$-module and $I$-power torsion we see that $I^{c_1}$ annihilates $H^{-1}(K)$ for some $c_1 \geq 0$. Choose a short exact sequence
which determines an element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_ R(H^0(K), M)$. For any $f \in I$ we have $\mathop{\mathrm{Ext}}\nolimits ^1_ R(H^0(K), M)_ f = \mathop{\mathrm{Ext}}\nolimits ^1_{R_ f}(H^0(K)_ f, M_ f)$ by Lemma 15.65.4. Hence if $H^0(K)_ f$ is projective, then a power of $f$ annihilates $\xi $. We conclude that $\xi $ is annihilated by $(f_1, \ldots , f_ s)^{c_2}$ for some $c_2 \geq 0$. Since $V(f_1, \ldots , f_ s) \subset V(I)$ we have $\sqrt{I} \subset (f_1, \ldots , f_ s)$ (Algebra, Lemma 10.17.2). Since $R$ is Noetherian we find $I^{c_3} \subset (f_1, \ldots , f_ s)$ for some $c_3 \geq 0$ (Algebra, Lemma 10.32.5). Hence $I^{c2c3}$ annihilates $\xi $. This in turn says that $H^0(K)$ is $I^{c_2c_3}$-projective (as multiplication by $a \in I$ which annihilate $\xi $ factor through $R^{\oplus r}$). Hence taking $c = \max (c_1, c_2c_3)$ we see that (2) holds. $\square$
Lemma 15.84.11. Let $R$ be a ring. Let $K_ j \in D(R)$, $j = 1, 2, 3$ with $H^ i(K_ j) = 0$ for $i \not\in \{ -1, 0\} $. Let $\varphi : K_1 \to K_2$ and $\psi : K_2 \to K_3$ be maps in $D(R)$. If $H^0(\varphi ) = 0$ and $H^{-1}(\psi ) = 0$, then $\varphi \circ \psi = 0$.
Proof. Apply Derived Categories, Lemma 13.12.5 to see that $\varphi \circ \psi $ factors through $\tau _{\leq -2}K_2 = 0$. $\square$
Lemma 15.84.12. Let $R$ be a ring. Let $K \in D(R)$ be given by a two term complex of the form $R^{\oplus n} \to R^{\oplus n}$. Denote $A \in \text{Mat}(n \times n, R)$ the matrix of the differential. Then $\det (a) : K \to K$ is zero in $D(R)$.
Proof. Omitted. Good exercise. $\square$
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