Let $R$ be a commutative ring with unit $1_R$, $M$ be a $R-$module.
I have a small question about different definitions of projective resolutions of $M$ (and I'm confused with the degrees of the boundary maps and modules here):
1)A projective resolution of $M$ is a long exact sequence
$$..\to P_2\xrightarrow{d_1}P_1\xrightarrow{d_0}P_0\xrightarrow{\epsilon} M\to 0,$$where the $P_i$ are projective $R$-modules for all $i\in\mathbb{N} $.
2)A projective resolution of $M$ is a pair $(P,\epsilon)$ with a chain complex $P=(P_i,d_i)_{i\in\mathbb{N}}$ of projective $R$-modules $P_i$ and a $R$-linear map $\epsilon:P_0\to M$, such that $H_k(P_*)=0$ if $k>0$ and $H_0(P_*)\cong M$ (here is $H_k(P_*)$ is the $k-$th singular homology group of $P_*$).
My question is: are the two definitions equivalent?
$1\Rightarrow 2$: it is $\operatorname{im}(d_{i+1})=\ker(d_i)$ for all $i$ and $\operatorname{im}(d_{0})=\ker(\epsilon)$, therefore it is $d_i\circ d_{i+1}=0$ and $H_i(P_*)=\frac{\ker(d_i)}{\operatorname{im}(d_{i+1})}=0$ for $i>0$. But it should be $H_0(P_*)=\frac{\ker(d_0)}{\operatorname{im}(d_{1})}=0$
as well? It seems that something is wrong with the degree here $H_i(P_*)=\frac{\ker(d_i)}{\operatorname{im}(d_{i+1})}=0$. Why is $H_0(P_*)\cong \frac{P_0}{\operatorname{ker}(\epsilon)}$?
$2\Leftarrow 1$The same here.. It's clear that $$..\to P_2\xrightarrow{d_1}P_1\xrightarrow{d_0}P_0$$ is exact but I'm trouble with $\epsilon$.
Best
Best Answer
You are going well, but I think you're just mistaking degrees of boundary maps. A homological complex is usually of the form $$\cdots \to M_{n}\overset{f_n}{\to}M_{n-1}\to\cdots \to M_0\overset{f_0}\to 0$$
so it's straightforward to have $H_0(M^\bullet)=M_0/\mathrm{im}(f_1)$.