tensor product of chain complexes
TensorProductOfChainComplexes
2006-09-07 04:30:20
2007-01-08 14:44:26
Definition
tensor product of chain complexes
chain complex
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Let $ C'=\gbra{C_n',\pap n}$ and $C''=\gbra{C_n'',\papp n}$ be two chain complexes of $R$-modules, where $R$ is a commutative ring with unity. Their \emph{tensor product} $C'\otimes_R C''=\gbra{(C'\otimes_R C'')_n,\pa n}$ is the chain complex defined by
$$ (C'\otimes_R C'')_n = \bigoplus_{i+j=n}(C_i'\otimes_R C_j''), $$
$$ \pa n(t'_i\otimes_R s''_j) = \pap i(t'_i)\otimes_R s''_j + (-1)^i\, t'_i\otimes_R \papp j(s''_j),\ \ \ \forall t'_i\in C_i',\ s''_j\in C_j'',\ (i+j=n),$$
where $C_i'\otimes_R C_j''$ denotes the \PMlinkname{tensor product}{TensorProduct} of $R$-modules $C_i'$ and $C_j''$.
Indeed, this defines a chain complex, because for each $t'_i\otimes_R s''_j\in C_i'\otimes_R C_j''\subseteq (C'\otimes_R C'')_{i+j}$ we have
$$\pa{i+j-1} \pa {i+j}(t'_i\otimes_R s''_j) = \pa{i+j-1}\cbra{ \pap i(t'_i)\otimes_R s''_j + (-1)^i\, t'_i\otimes_R \papp j(s''_j) }= $$
$$ = (-1)^{i-1}\, \pap i(t'_i)\otimes_R \papp j(s''_j)+(-1)^i \pap i(t'_i)\otimes_R \papp j(s''_j)=0, $$
thus $C'\otimes_R C''$ is a chain complex.