Lemma 46.4.10. With notation as in Lemma 46.4.2 we have R^ pQ(F) = 0 for all p > 0 and any adequate functor F.
Proof. Choose an exact sequence 0 \to F \to \underline{M^0} \to \underline{M^1}. Set M^2 = \mathop{\mathrm{Coker}}(M^0 \to M^1) so that 0 \to F \to \underline{M^0} \to \underline{M^1} \to \underline{M^2} \to 0 is a resolution. By Derived Categories, Lemma 13.21.3 we obtain a spectral sequence
Since Q(\underline{M^ q}) = \underline{M^ q} it suffices to prove R^ pQ(\underline{M}) = 0, p > 0 for any A-module M.
Choose an injective resolution \underline{M} \to I^\bullet in the category \mathcal{P}. Suppose that R^ iQ(\underline{M}) is nonzero. Then \mathop{\mathrm{Ker}}(Q(I^ i) \to Q(I^{i + 1})) is strictly bigger than the image of Q(I^{i - 1}) \to Q(I^ i). Hence by Lemma 46.3.6 there exists a linearly adequate functor L and a map \varphi : L \to Q(I^ i) mapping into the kernel of Q(I^ i) \to Q(I^{i + 1}) which does not factor through the image of Q(I^{i - 1}) \to Q(I^ i). Because Q is a left adjoint to the inclusion functor the map \varphi corresponds to a map \varphi ' : L \to I^ i with the same properties. Thus \varphi ' gives a nonzero element of \mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {P}(L, \underline{M}) contradicting Lemma 46.4.9. \square
Comments (0)