**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

\[ \ldots \to \mathop{N\! L}\nolimits _{B_3/A_3} \to \mathop{N\! L}\nolimits _{B_2/A_2} \to \mathop{N\! L}\nolimits _{B_1/A_1} \]

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

\[ B_ n = B/I^ nB = A_ n[x_1, \ldots , x_ r]/J_ n \]

where

\[ J_ n = JA_ n[x_1, \ldots , x_ r] = J/(J \cap I^ nA[x_1, \ldots , x_ r]^\wedge ) \]

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 88.2.2) we find a $c \geq 0$ such that we have canonical surjections

\[ J/I^ nJ \to J_ n \to J/I^{n - c}J \to J_{n - c},\quad n \geq c \]

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

\[ \mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B B/I^ nB \to \mathop{N\! L}\nolimits _{B_ n/A_ n} \to \mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B B/I^{n - c}B \to \mathop{N\! L}\nolimits _{B_{n - c}/A_{n - c}} \]

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 15.87.1 and Remark 15.87.6.
$\square$

## Comments (0)