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

$(J/J^2) \otimes _ B B^\wedge = (J/J^2)^\wedge = J^\wedge /(J^\wedge )^2$

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). $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).