Stable Dold-Kan correspondence and symmetric group actions

Question

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.

Answer 1

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 and in doing so found this question. The relevant paper is "Stable model categories are categories of modules" 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.