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Let $F,G, H: Mod \to Mod$ be three left exact functors such that $R^iF(-)\cong R^iG(-)$ for all $i\in\mathbb{N}$. We consider the exact sequence $\cdots\to R^iF(M)\to R^iG(M)\to R^iH(M)\to R^{i+1}F(M)\to R^{i+1}G(M)\to R^{i+1}H(M)\to\cdots$ where $M$ is an $R$-module ($R$ is a commutative Noetherian ring).

From the above exact sequence, can we have $R^iH(M)=0$?

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    Where exactly does your exact sequence come from?2012-04-10

1 Answers 1

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Derived functors don't matter in this case. You just have an exact sequence:

$ \dots \longrightarrow A \stackrel{f}{\longrightarrow} B \stackrel{g}{\longrightarrow} C \stackrel{h}{\longrightarrow} D \stackrel{i}{\longrightarrow} E \longrightarrow \dots $

in which $f$ and $i$ are isomorphisms.

But this means that morphisms $g$ and $h$ are zero:

$ B = \mathrm{im}\ f = \mathrm{ker}\ g \qquad \Longrightarrow \qquad g = 0 $

and

$ \mathrm{im}\ h = \mathrm{ker}\ i = 0 \qquad \Longrightarrow \qquad h = 0 $

So, your exact sequence at $C$ is just

$ 0 \longrightarrow C \longrightarrow 0 \ . $

Hence

$ C = 0 \ . $

So, indeed

$ R^iH(M) = 0 . $