Suppose I have a dg algebra $(A,d)$ and a chain complex $M^\bullet$ of semi-free $(A,d)$ modules. I am hoping it is true that $ Tot^\coprod (M^\bullet)$ is again a semi-free $(A,d)$ module. Is this so? And if yes, what filtration should one choose?


The definition of semi-free I am using is: $X$ is semi-free if $X$ is the union of an increasing sequence of dg $A$-modules $X\_i$ , such that $ i \in \mathbb{N}$, $X\_0 = 0$ and $ X\_{i+1}/X\_i$ is a free dg $A$-module. 



Does semi-free behave well under totalization
If $M\_\bullet$ is not bounded above, then the totalization may not be semi-free.


For example, let $k$ be a field, and $A=k[x]/(x^2)$, considered as a DG algebra concentrated in degree zero, with trivial differential. Then the chain complex
$$\dots\to A\to A\to A\to A\to\dots,$$
where all the differentials are multiplication by $x$, is not semi-free considered as a DG $A$-module, as it has no free DG submodules.


