Lemma 51.5.8. Let $A$ be a ring and let $T, T' \subset \mathop{\mathrm{Spec}}(A)$ subsets stable under specialization. For $K \in D^+(A)$ there is a spectral sequence
as in Derived Categories, Lemma 13.22.2.
Lemma 51.5.8. Let $A$ be a ring and let $T, T' \subset \mathop{\mathrm{Spec}}(A)$ subsets stable under specialization. For $K \in D^+(A)$ there is a spectral sequence
as in Derived Categories, Lemma 13.22.2.
Proof. Let $E$ be an object of $D_{T \cap T'}(A)$. Then we have
The first equality by the adjointness property of $R\Gamma _ T$ and the second by the adjointness property of $R\Gamma _{T'}$. On the other hand, if $J^\bullet $ is a bounded below complex of injectives representing $K$, then $H^0_{T'}(J^\bullet )$ is a complex of injective $A$-modules representing $R\Gamma _{T'}(K)$ and hence $H^0_ T(H^0_{T'}(J^\bullet ))$ is a complex representing $R\Gamma _ T(R\Gamma _{T'}(K))$. Thus $R\Gamma _ T(R\Gamma _{T'}(K))$ is an object of $D^+_{T \cap T'}(A)$. Combining these two facts we find that $R\Gamma _{T \cap T'} = R\Gamma _ T \circ R\Gamma _{T'}$. This produces the spectral sequence by the lemma referenced in the statement. $\square$
Your email address will not be published.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)