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? – Keenan Kidwell Apr 10 '12 at 23:55

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

$$R^iH(M) = 0 .$$

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