|
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 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 tensor product 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.
|
