Derived finite coefficients and the Bockstein sequence

Let $P$ be the complex $\mathbb{Z}\stackrel{n}{\to }\mathbb{Z}$ in degrees $-1$ and $0$. Its augmentation to $\mathbb{Z}\mathrm{/}n$ is a quasi-isomorphism because multiplication by $n$ on $\mathbb{Z}$ is injective and has cokernel $\mathbb{Z}\mathrm{/}n$. For any complex $A$, the total complex $A\otimes P$ is isomorphic to $\mathrm{Cone}(n:A\to A)$: in degree $m$ send the cone pair $(y,x)$ to $(y,(-1{)}^{m+1}x)$. The cone sits in a degreewise split short exact sequence $0\to A\to \mathrm{Cone}(n)\to A[1]\to 0$. The resulting cohomology exact sequence, obtained by lifting cocycles and taking their differentials, shows that this cone is acyclic whenever $A$ is acyclic. Thus $P$ is K-flat: tensoring it with an acyclic complex gives an acyclic complex. Consequently this free resolution computes the derived tensor product with $\mathbb{Z}\mathrm{/}n$, and the displayed isomorphism identifies it with $C\mathrm{/}n$.

The inclusion $C\to \mathrm{Cone}(n)$ sends $y$ to $(y,0)$, and projection to the second coordinate sends a cone cocycle $(y,x)$ to the class of $x$ in $H^{m+1}(C)$. These maps are well defined on cohomology: the cone cocycle equations are $dx=0$ and $dy=-nx$, and a cone boundary changes $x$ by a boundary. The kernel of the first map on $H^m(C)$ is $n{H}^m(C)$: if $(y,0)=d(v,u)$, then $du=0$ and $y=dv+nu$; the converse follows from $d(0,u)=(nu,0)$ for a cocycle $u$. A class in the kernel of the projection has a representative with $x=0$ after subtracting a cone boundary, so comes from $H^m(C)$. Finally, the image of the projection is precisely $H^{m+1}(C)[n]$: the cocycle equation gives $n[x]=0$, and conversely if $nx=-dy$ for a cocycle $x$, then $(y,x)$ is a cone cocycle. This proves the short exact sequence. If $H^m(C\mathrm{/}p)=0$, its surjection onto $H^{m+1}(C)[p]$ makes that subgroup zero, which is exactly injectivity of multiplication by $p$.