Im having troubles in proving the following result: Let $C^{\bullet}$ be a complex of $R$-modules ($R$ noetherian ring) with non-zero modules in positive degree, and let $M$ be an $R$-module. Assume that $H^i(C^{\bullet})=0$ for $i
Cohomology of hom
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homological-algebra
1 Answers
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This may not be true, because $Hom(M, -)$ is not in general exact.
Consider the following example. Let $R = \mathbb{Z}$ be the ring of integers and consider the exact complex $C ^{\bullet}$ given by
$\ldots \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} _{2} \rightarrow 0 \rightarrow \ldots$,
where the first non-trivial map is multiplication by $2$. Let $M = \mathbb{Z} _{2}$. Since the cohomology of $C^{\bullet}$ is trivial, we also have $Hom (\mathbb{Z} _{2}, H^{i}(C^{\bullet})) = 0$ for all $i$. But $Hom(\mathbb{Z} _{2}, C^{\bullet})$ looks like this
$\ldots \rightarrow 0 \rightarrow 0 \rightarrow 0 \rightarrow \mathbb{Z} _{2} \rightarrow 0 \rightarrow \ldots$,
so it has non-trivial cohomology.
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0Thanks a lot! you are right, it doesn't happen in general, thanks again!! – 2012-12-28