There exists a Quillen equivalence between $HRModSpectra$ (model category of ring spectra over Eilenberg-MacLane spectra $EM(R)$, where $R$ is a commutative ring, with stable model structure) and $Ch$ (model category of unbounded chain complexes of $R$-modules).


I was wondering what the Quillen functors are that give the above Quillen equivalence.


One can start with an unbounded chain complex $X$ and apply the Dold-Kan functor $\Gamma$ to chain complex $X\_{\geq 0}$ to get a simplicial abelian group $\Gamma(X\_{\geq 0})$, and then consider $\Gamma(X[-n]\_{\geq0})$ (shifting $X$ to the left by n places, truncating and then applying $\Gamma$). This way one gets an $\Omega$-spectrum $Y$ = {${Y\_{0}, Y\_{1},...}$}, with $Y\_{n} = \Gamma(X[-n]\_{\geq0})$. 


How does then one proceed to prove that $Y$ is a symmetric spectrum? For that one needs an action of symmetric group $S\_{n}$ on $Y\_{n}$. Now each $Y\_{i}$ is in fact as a simplicial set equivalent to $\prod K(\pi\_{k}(Y\_{n}), k)$ and $K(\pi\_{n}(Y\_{n}), n)$ has an $S\_{n}$ action, though I'm not sure if this action will satisfy the compatibility conditions that are required of a symmetric spectrum. 



Stable Dold-Kan correspondence and symmetric group actions
I'm posting a CW answer so that this doesn't remain open and unanswered, even though the OP said in the comments he figured it out. I just posted an [answer to an analogous question](https://mathoverflow.net/a/334622/11540) and in doing so found this question. The relevant paper is "[Stable model categories are categories of modules](http://homepages.math.uic.edu/~bshipley/classTopFinal.pdf)" by Schwede and Shipley. The construction the OP describes (Dold-Kan plus shift) is on page 39, where it is also shown how to make this into an $HR$-module (via the Alexander-Whitney map). Remark B.1.10 discusses the symmetric group actions. This functor $\mathcal{H}$ takes a chain complexes to a *naive* $HR$-module, not a symmetric spectrum. It is not an extension of the usual Eilenberg-MacLane functor, $H$. The authors claim that there is no way to make $\mathcal{H}R$ into a symmetric spectrum that is level equivalent to $HR$. This explains why the Quillen equivalence has to zig-zag through naive $HR$-modules. 


