Example of a nongraded chain complex A chain complex is an abelian group $A$ with an endomorphism $d \in End(A)$, s.t. $d^2=0$ called the differential.  
I am trying to come up with an example of a nongraded chain complex with nonzero cohomology for notes on spectral sequences that I am writing.  Do you know of one?
From the fact that the Bockstein spectral sequence only has one grading, if I could realize the Bockstein spectral sequence as the spectral sequence of a filtered chain complex then I would have an example of a nongraded chain complex.
If this does not work for abelian groups I'd be happy with any example of a nongraded chain complex over an arbitrary ring with nonzero cohomology
 A: Ok this is really stupid.  Just take any ring with with $r \in R$ such that $ann(r) \supsetneq (r)$.  Then $R$ as an $R$ module is a chain complex with the multiplication by $r$ map as the differential.
A: The most basic example is just to take any nonzero abelian group $A$ and let $d=0$.  This may seem trivial, but if, for instance, you restrict your attention to vector spaces over a field, it is essentially the only example: every (ungraded) chain complex of vector spaces over a field is a direct sum of a chain complex with $d=0$ and an exact chain complex.
More generally, you can of course just take any graded chain complex and forget the grading (the example above comes from doing that to $0\to A\to 0$).  For a simple example that cannot be made into a graded chain complex, let $A=\mathbb{Z}/(p^3)$ and let $d$ be multiplication by $p^2$.
The following general perspective may be helpful.  A chain complex of $R$-modules is just a module over the ring $S=R[d]/(d^2)$.  (In particular, when $R$ is a field, every $S$-module is a direct sum of copies of $S/(d)$ and $R$, which gives the statement at the end of the first paragraph.)  The homology of such a module $A$ is then just $\operatorname{Ext}^n_S(S/(d),A)$ for any $n>0$.  Indeed, this is immediate from the free resolution $$\dots S\stackrel{d}\to S\stackrel{d}\to S\to S/(d)\to 0$$ of $S/(d)$.
