Lemma 87.3.6. With assumptions as in Lemma 87.3.5 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 15.100.1 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 87.3.3. 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. $\square$

## 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Your email address will not be published.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)