Computation of Ext as a cohomologies of certain complex

Question

Let $R$ be a ring and $K^\bullet$ be a complex of $R$-modules such that $K^\bullet$ has only one nontrivial cohomology $H^0(K^\bullet)=M$. Suppose that $R$-module $N$ is such that $\text{Ext}^i_R(K_j,N)=0$ for all $j$ and $i\geq1$. How to prove that for all $i\geq0$ $\text{Ext}^i_R(M,N)=H^i(\text{Hom}_R(K^\bullet,N))$?

I guess that I should apply spectral sequences but I have only a little experience with them.

Answer 1

It is a standard fact of homological algebra that derived functors can be computed from acyclic resolutions. In other words, if $F$ is a left exact functor and $K^\bullet \to M$ is a resolution by $F$-acyclic objects (ie. $R^iF(K^j) = 0$ for all $i > 0$), then $R^iF(M) = H^i(F(K^\bullet))$. You can find a proof of this (for the left derived case, actually) in this Q&A (or look up "derived functor acyclic resolution" in Google or your favorite textbook, eg. Weibel, 2.4.3 & 3.2.8).

Here the functor $F$ is $\hom_R(-, N)$; $K^\bullet \to M$ is a resolution of $M$, and the condition $\operatorname{Ext}^i_R(K^j, N) = 0$ for $i > 0$ precisely means that it is an acyclic resolution, thus $$\operatorname{Ext}^i_R(M,N) = H^i(\hom_R(K^\bullet, N)).$$

---

To make this answer a bit more self-contained, here is a sketch of proof. Choose projective resolution $P^{i, \bullet} \to K^i$ of the $K^i$. Because they are projective resolutions, you can extend the differentials $d : K^i \to K^{i+1}$ to chain maps $P^{i,\bullet} \to P^{i+1,\bullet}$ that square to zero (you might need to choose the $P^{i, \bullet}$ inductively for this). So you get a bicomplex $P^{\bullet, \bullet}$, and you can apply $\hom_R(-, N)$ to it to get a new bicomplex $Q^{\bullet,\bullet}$. I will compute its homology in two different ways (behind the scenes you have the two spectral sequences associated to the bicomplex appearing):

• Because $P^{i,\bullet} \to K^i$ is a projective resolution, the vertical cohomology of this bicomplex $Q$ is equal to $\operatorname{Ext}^i_R(K^j, N))$, but by hypothesis this is concentrated in degree $i=0$, and so the horizontal cohomology of the vertical cohomology is equal to $H^j(\hom_R(K^\bullet, N))$.
• If you compute the horizontal homology first, since $K^\bullet \to M$ is a resolution, a bit of diagram chasing tells you that it is concentrated in bidegree $(i,0)$ and you find something of the form $\hom(p^\bullet, N)$ for some projective resolution of $M$; by definition the vertical cohomology of this yields $\operatorname{Ext}^i_R(M,N)$.

In both cases the spectral sequences collapse at the $E_2$ page, and by general theorems the two ways of computing the cohomology of $Q$ yield the same results, so you finally find $$\operatorname{Ext}^i_R(M,N) = H^i(\hom_R(K^\bullet, N)).$$