Let's $(K^{\bullet}, d^{\bullet})$ is the complex over field $A$ (i.e. all $K^{i}$ are vector spaces over this field) and $(L^{\bullet}, {\delta}^{\bullet})$ such that $$L^{i}=H^{i}(K)~\text{and all}~{\delta}^{i}=0.$$ Why this two complexes $K$ and $L$ are quasi-isomorphic? Why it's wrong for complex over ring?
Thanks a lot!
The result you mention is true for complexes of modules over a ring of global dimension at most one. It is not entirely trivial to prove, see section 2.5 in these lecture notes by Keller. Note that fields have global dimension zero.
To show that $K$ and $L$ are quasi-isomorphic, try splitting each vector space $K^n$ as $\operatorname{im} d^{n-1}\oplus H^n(K)\oplus (K^n/\ker d^n)$, such that $d^n$ acts in an appropriate way. Then show that the maps $H^n(K)\to K^n$ define a quasi-isomorphism.
To show that this fails over a ring, you need to find a chain complex that doesn't split in this way.
The general statement for why this may be false over a ring goes like this: for two $R$-modules $A,B$ consider $\mathrm{Ext}^2(A,B)$. This parameterizes extensions $0\to A\to X\to Y \to B\to 0$ under Baer sum. It is straightforward to check that for a non-zero class in $\mathrm{Ext}$ given by $0\to A\to X\to Y\to B\to 0$, $0\to A\to B \to 0$ with 0 differential has the same homology as $0\to X\to Y\to 0$ with the same differential is in the 4-term sequence, but it may happen that there are no maps between $A$ and $X$, and so there cannot be a quasi-isomorphism. Unfortunately, an illuminating example does not spring to mind right now.
