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). The naive cotangent complex $\mathop{N\! L}\nolimits _{B/A}^\wedge $ is well defined in $K(B)$.

## 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 _{B/A}^\wedge $ is well defined up to homotopy.

**Proof.**
The lemma signifies that given a second presentation $B = A[y_1, \ldots , y_ s]^\wedge / K$ the complexes of $B$-modules

are homotopy equivalent. To see this, we can argue exactly as in the proof of Algebra, Lemma 10.134.2.

Step 1. If we choose $g_ i(y_1, \ldots , y_ s) \in A[y_1, \ldots , y_ s]^\wedge $ mapping to the image of $x_ i$ in $B$, then we obtain a (unique) continuous $A$-algebra homomorphism

compatible with the given surjections to $B$. Such a map is called a morphism of presentations. It induces a map from $J$ into $K$ and hence induces a $B$-module map $J/J^2 \to K/K^2$. Sending $\text{d}x_ i$ to $\sum (\partial g_ i/\partial y_ j)\text{d}y_ j$ we obtain a map of complexes

Of course we can do the same thing with the roles of the two presentations exchanged to get a map of complexes in the other direction.

Step 2. The construction above is compatible with compositions of morphsms of presentations. Hence to finish the proof it suffices to show: given $g_ i(x_1, \ldots , x_ r) \in A[x_1, \ldots , x_ n]^\wedge $ mapping to the image of $x_ i$ in $B$, the induced map of complexes

is homotopic to the identity map. To see this consider the map $h : \bigoplus B \text{d}x_ i \to J/J^2$ given by the rule $\text{d}x_ i \mapsto g_ i(x_1, \ldots , x_ n) - x_ i$ and compute. $\square$

Lemma 86.3.2. Let $A$ be a Noetherian ring and let $I \subset A$ be a ideal. Let $A \to B$ be a finite type ring map. Choose a presentation $\alpha : A[x_1, \ldots , x_ n] \to B$. Then $\mathop{N\! L}\nolimits _{B^\wedge /A}^\wedge = \mathop{\mathrm{lim}}\nolimits \mathop{N\! L}\nolimits (\alpha ) \otimes _ B B^\wedge $ as complexes and $\mathop{N\! L}\nolimits _{B^\wedge /A}^\wedge = \mathop{N\! L}\nolimits _{B/A} \otimes _ B^\mathbf {L} B^\wedge $ in $D(B^\wedge )$.

**Proof.**
The statement makes sense as $B^\wedge $ is an object of (86.2.0.2) by Lemma 86.2.2. Let $J = \mathop{\mathrm{Ker}}(\alpha )$. The functor of taking $I$-adic completion is exact on finite modules over $A[x_1, \ldots , x_ n]$ and agrees with the functor $M \mapsto M \otimes _{A[x_1, \ldots , x_ n]} A[x_1, \ldots , x_ n]^\wedge $, see Algebra, Lemmas 10.97.1 and 10.97.2. Moreover, the ring maps $A[x_1, \ldots , x_ n] \to A[x_1, \ldots , x_ n]^\wedge $ and $B \to B^\wedge $ are flat. Hence $B^\wedge = A[x_1, \ldots , x_ n]^\wedge / J^\wedge $ and

Since $\mathop{N\! L}\nolimits (\alpha ) = (J/J^2 \to \bigoplus B\text{d}x_ i)$, see Algebra, Section 10.134, we conclude the complex $\mathop{N\! L}\nolimits _{B^\wedge /A}^\wedge $ is equal to $\mathop{N\! L}\nolimits (\alpha ) \otimes _ B B^\wedge $. The final statement follows as $\mathop{N\! L}\nolimits _{B/A}$ is homotopy equivalent to $\mathop{N\! L}\nolimits (\alpha )$ and because the ring map $B \to B^\wedge $ is flat (so derived base change along $B \to B^\wedge $ is just base change). $\square$

Lemma 86.3.3. 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

the pro-objects $\{ \mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B B/I^ nB\} $ and $\{ \mathop{N\! L}\nolimits _{B_ n/A_ n}\} $ of $D(B)$ are strictly isomorphic (see proof for elucidation),

$\mathop{N\! L}\nolimits _{B/A}^\wedge = R\mathop{\mathrm{lim}}\nolimits \mathop{N\! L}\nolimits _{B_ n/A_ n}$ in $D(B)$.

Here $B_ n$ and $A_ n$ are as in Section 86.2.

**Proof.**
The statement means the following: for every $n$ we have a well defined complex $\mathop{N\! L}\nolimits _{B_ n/A_ n}$ of $B_ n$-modules and we have transition maps $\mathop{N\! L}\nolimits _{B_{n + 1}/A_{n + 1}} \to \mathop{N\! L}\nolimits _{B_ n/A_ n}$. See Algebra, Section 10.134. Thus we can consider

as an inverse system of complexes of $B$-modules and a fortiori as an inverse system in $D(B)$. Furthermore $R\mathop{\mathrm{lim}}\nolimits \mathop{N\! L}\nolimits _{B_ n/A_ n}$ is a homotopy limit of this inverse system, see Derived Categories, Section 13.34.

Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge / J$. This defines presentations

where

The two term complex $J_ n/J_ n^2 \longrightarrow \bigoplus B_ n \text{d}x_ i$ represents $\mathop{N\! L}\nolimits _{B_ n/A_ n}$, see Algebra, Section 10.134. By Artin-Rees (Algebra, Lemma 10.51.2) in the Noetherian ring $A[x_1, \ldots , x_ r]^\wedge $ (Lemma 86.2.2) we find a $c \geq 0$ such that we have canonical surjections

for all $n \geq c$. A moment's thought shows that these maps are compatible with differentials and we obtain maps of complexes

compatible with the transition maps of the inverse systems $\{ \mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B B/I^ nB\} $ and $\{ \mathop{N\! L}\nolimits _{B_ n/A_ n}\} $. This proves part (1) of the lemma.

By part (1) and since pro-isomorphic systems have the same $R\mathop{\mathrm{lim}}\nolimits $ in order to prove (2) it suffices to show that $\mathop{N\! L}\nolimits _{B/A}^\wedge $ is equal to $R\mathop{\mathrm{lim}}\nolimits \mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B B/I^ nB$. However, $\mathop{N\! L}\nolimits _{B/A}^\wedge $ is a two term complex $M^\bullet $ of finite $B$-modules which are $I$-adically complete for example by Algebra, Lemma 10.97.1. Hence $M^\bullet = \mathop{\mathrm{lim}}\nolimits M^\bullet /I^ nM^\bullet = R\mathop{\mathrm{lim}}\nolimits M^\bullet /I^ n M^\bullet $, see More on Algebra, Lemma 15.86.1 and Remark 15.86.6. $\square$

Lemma 86.3.4. Let $(A_1, I_1) \to (A_2, I_2)$ be as in Remark 86.2.3 with $A_1$ and $A_2$ Noetherian. Let $B_1$ be in (86.2.0.2) for $(A_1, I_1)$. Let $B_2$ be the base change of $B_1$. Then there is a canonical map

which induces and isomorphism on $H^0$ and a surjection on $H^{-1}$.

**Proof.**
Choose a presentation $B_1 = A_1[x_1, \ldots , x_ r]^\wedge /J_1$. Since $A_2/I_2^ n[x_1, \ldots , x_ r] = A_1/I_1^{cn}[x_1, \ldots , x_ r] \otimes _{A_1/I_1^{cn}} A_2/I_2^ n$ we have

where we use $I_2$-adic completion on both sides (but of course $I_1$-adic completion for $A_1[x_1, \ldots , x_ r]^\wedge $). Set $J_2 = J_1 A_2[x_1, \ldots , x_ r]^\wedge $. Arguing similarly we get the presentation

for $B_2$ over $A_2$. As a consequence obtain a commutative diagram

The induced arrow $J_1/J_1^2 \otimes _{B_1} B_2 \to J_2/J_2^2$ is surjective because $J_2$ is generated by the image of $J_1$. This determines the arrow displayed in the lemma. We omit the proof that this arrow is well defined up to homotopy (i.e., indepedent of the choice of the presentations up to homotopy). The statement about the induced map on cohomology modules follows easily from the discussion (details omitted). $\square$

Lemma 86.3.5. 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

See proof for elucidation.

**Proof.**
Observe that taking the tensor product $\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C$ makes sense as $\mathop{N\! L}\nolimits _{B/A}^\wedge $ is well defined up to homotopy by Lemma 86.3.1. Also, $(B, IB)$ is pair where $B$ is a Noetherian ring (Lemma 86.2.2) and $C$ is in the corresponding category (86.2.0.2). Thus all the terms in the $6$-term sequence are (well) defined.

Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. 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 _{B/A}^\wedge $ with $\text{id}_ C$. The lemma follows by applying the snake lemma (Algebra, Lemma 10.4.1). $\square$

Lemma 86.3.6. With assumptions as in Lemma 86.3.5 assume that $B/I^ nB \to C/I^ nC$ is a local complete intersection homomorphism for all $n$. Then $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C) \to H^{-1}(\mathop{N\! L}\nolimits _{C/A}^\wedge )$ is injective.

**Proof.**
For each $n \geq 1$ we set $A_ n = A/I^ n$, $B_ n = B/I^ nB$, and $C_ n = C/I^ nC$. We have

The first equality follows from More on Algebra, Lemma 15.98.1 and the fact that $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C)$ is a finite $C$-module and hence $I$-adically complete for example by Algebra, Lemma 10.97.1. The second equality is trivial. The third holds by Lemma 86.3.3. The maps $H^{-1}(\mathop{N\! L}\nolimits _{B_ n/A_ n} \otimes _{B_ n} C_ n) \to H^{-1}(\mathop{N\! L}\nolimits _{C_ n/A_ n})$ are injective by More on Algebra, Lemma 15.33.6. The proof is finished because we also have $H^{-1}(\mathop{N\! L}\nolimits _{C/A}^\wedge ) = \mathop{\mathrm{lim}}\nolimits H^{-1}(\mathop{N\! L}\nolimits _{C_ n/A_ n})$ similarly to the above. $\square$

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