Lemma 15.84.9. Let $I$ be an ideal of a ring $R$. Let $\alpha : K \to K'$ be a morphism of $D(R)$. Assume
$H^ i(K) = H^ i(K') = 0$ for $i \not\in \{ -1, 0\} $
$H^0(\alpha )$ is an isomorphism and $H^{-1}(\alpha )$ is surjective.
If $K$ satisfies the equivalent conditions (1), (2), and (3) of Lemma 15.84.5, then $K'$ does too.
Proof. Set $M = \mathop{\mathrm{Ker}}(H^{-1}(\alpha ))$. Then $\alpha $ fits into a distinguished triangle
For any $R$-module $N$ this determines an exact sequence
Since $\mathop{\mathrm{Ext}}\nolimits ^0_ R(M[1], N) = \mathop{\mathrm{Ext}}\nolimits ^{-1}_ R(M, N) = 0$ we see that $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K', N)$ is a submodule of $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$. Hence if $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$ is annihilated by $I$ so is $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K', N)$. $\square$
Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)