Lemma 88.3.6. With assumptions as in Lemma 88.3.5 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
\begin{align*} H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C) & = \mathop{\mathrm{lim}}\nolimits H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C_ n) \\ & = \mathop{\mathrm{lim}}\nolimits H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B B_ n \otimes _{B_ n} C_ n) \\ & = \mathop{\mathrm{lim}}\nolimits H^{-1}(\mathop{N\! L}\nolimits _{B_ n/A_ n} \otimes _{B_ n} C_ n) \end{align*}
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 88.3.3. 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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