Lemma 86.3.1. Let $A$ be a Noetherian ring and let $I \subset A$ be a ideal. Let $B$ be an object of (86.2.0.2). Then $\mathop{N\! L}\nolimits ^\wedge _{B/A} = R\mathop{\mathrm{lim}}\nolimits \mathop{N\! L}\nolimits _{B_ n/A_ n}$ in $D(B)$.

## 86.3 A naive cotangent complex

Let $A$ be a Noetherian ring and let $I \subset A$ be a ideal. Let $B$ be an $A$-algebra which is $I$-adically complete such that $A/I \to B/IB$ is of finite type, i.e., an object of (86.2.0.2). By Lemma 86.2.2 we can write

for some finitely generated ideal $J$. For a choice of presentation as above we define the naive cotangent complex in this setting by the formula

with terms sitting in degrees $-1$ and $0$ where the map sends the residue class of $g \in J$ to the differential $\text{d}g = \sum (\partial g/\partial x_ i) \text{d}x_ i$. Here the partial derivative is taken by thinking of $g$ as a power series. The following lemma shows that $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ is well defined in $D(B)$, i.e., independent of the chosen presentation, although this could be shown directly by comparing presentations as in Algebra, Section 10.133.

**Proof.**
In fact, the presentation $B = A[x_1, \ldots , x_ r]^\wedge / J$ defines presentations

where

By Artin-Rees (Algebra, Lemma 10.50.2) in the Noetherian ring $A[x_1, \ldots , x_ r]^\wedge $ (Lemma 86.2.2) we see that we have canonical surjections

for some $c \geq 0$. It follows that $\mathop{\mathrm{lim}}\nolimits J_ n/J_ n^2 = J/J^2$ as any finite $A[x_1, \ldots , x_ r]^\wedge $-module is $I$-adically complete (Algebra, Lemma 10.96.1). Thus

(termwise limit) and the transition maps in the system are termwise surjective. The two term complex $J_ n/J_ n^2 \longrightarrow \bigoplus B_ n \text{d}x_ i$ represents $\mathop{N\! L}\nolimits _{B_ n/A_ n}$ by Algebra, Section 10.133. It follows that $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ represents $R\mathop{\mathrm{lim}}\nolimits \mathop{N\! L}\nolimits _{B_ n/A_ n}$ in the derived category by More on Algebra, Lemma 15.82.1. $\square$

Lemma 86.3.2. Let $A$ be a Noetherian ring and let $I \subset A$ be a ideal. Let $B \to C$ be morphism of (86.2.0.2). Then there is an exact sequence

**Proof.**
Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. Note that $(B, IB)$ is a pair consisting of a Noetherian ring and an ideal, and $C$ is in the corresponding category (86.2.0.2) for this pair. Hence we can choose a presentation $C = B[y_1, \ldots , y_ s]^\wedge /J'$. Combinging these presentations gives a presentation

Then the reader verifies that we obtain a commutative diagram

with exact rows. Note that the vertical arrow on the left hand side is the tensor product of the arrow defining $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ with $\text{id}_ C$. The lemma follows by applying the snake lemma (Algebra, Lemma 10.4.1). $\square$

Lemma 86.3.3. With assumptions as in Lemma 86.3.2 assume that $B/I^ nB \to C/I^ nC$ is a local complete intersection homomorphism for all $n$. Then $H^{-1}(\mathop{N\! L}\nolimits ^\wedge _{B/A} \otimes _ B C) \to H^{-1}(\mathop{N\! L}\nolimits ^\wedge _{C/A})$ is injective.

**Proof.**
By More on Algebra, Lemma 15.33.6 we see that this holds for the map between naive cotangent complexes of the situation modulo $I^ n$ for all $n$. In other words, we obtain a distinguished triangle in $D(C/I^ nC)$ for every $n$. Using Lemma 86.3.1 this implies the lemma; details omitted.
$\square$

Maps in the derived category out of a complex such as (86.3.0.1) are easy to understand by the result of the following lemma.

Lemma 86.3.4. 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.

If $K^ i = 0$ for $i \leq -2$, then $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(M^\bullet , K^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(M^\bullet , K^\bullet )$,

If $K^ i = 0$ for $i \leq -3$ and $\alpha \in \mathop{\mathrm{Hom}}\nolimits _{D(R)}(M^\bullet , K^\bullet )$ composed 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 $h' : M^0 \to K^{-1}$ and $h : M^{-1} \to K^{-2}$ such that

\[ h \circ d_ M^{-2} = 0, \quad (a')^{-1} = a^{-1} + d_ K^{-2} \circ h + h' \circ d_ M^{-1},\quad (a')^0 = a^0 + d_ K^{-1} \circ h' \]

**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 $F^ i$ free 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). 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 (2) holds for these factorizations and $a^0 = b^0$.

Proof of (3) 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 $. $\square$

Lemma 86.3.5. Let $R$ be a ring. Let $M^\bullet $ be a two term complex $M^{-1} \to M^0$ over $R$. If $\varphi , \psi \in \text{End}_{D(R)}(M^\bullet )$ are zero on $H^ i(M^\bullet )$, then $\varphi \circ \psi = 0$.

**Proof.**
Apply Derived Categories, Lemma 13.12.5 to see that $\varphi \circ \psi $ factors through $\tau _{\leq -2}M^\bullet = 0$.
$\square$

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