Lemma 88.3.2. 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 (88.2.0.2) by Lemma 88.2.2. 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). $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)