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I just started learning the subject, so the question should be basic.

A complex $\mathbf{F}$ over a ring $R$ is a sequence of homomorphisms of $R$-modules $$\mathbf{F}: \cdots \to F_i\overset{d_i}{\to}F_{i-1}\to\cdots \to F_1\overset{d_1}{\to}F_0\to\cdots$$ such that $d_{i-1}d_i=0$ for $i\in\mathbf{Z}$. Then the book says we can consider $\mathbf{F}$ as an $R$-module and the differential as a homomorphism $d:\mathbf{F}\to\mathbf{F}$. But how do we think of the chain complex, $\mathbf{F}$, as an $R$-module?

D. Huang
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1 Answers1

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Let $$\mathbf{F}=\bigoplus_{i\in\mathbb{Z}}F_i.$$ $\mathbf{F}$ is a $\mathbb{Z}$-graded $R$-module and $d$ is a homomorphism of graded $R$-modules (that is $d$ preserves the grading induced by the sum).

Dan Rust
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  • One more question: why does $d$ has homological degree -1? – D. Huang Sep 22 '13 at 17:38
  • Do you have a definition for the homological degree of a module homomorphism? Otherwise I think your text may be referring to the fact that if $f\in F_i$, then $d(f)=f'$ for some $f'\in F_{i-1}$. (abusing notation slightly because really it should be $d(\ldots,0,0,f,0,0,\ldots)=(\ldots,0,f',0,0,0,\ldots)$.) – Dan Rust Sep 22 '13 at 18:03