If $M$ is an $R$-module, how can I show that the Eilenberg-Maclane spectrum $HM$ is an $HR$-module spectrum

Question

Let $R$ be a commutative ring with identity, and let $M$ be an $R$-module. I want to show that the Eilenberg-Maclane spectrum $HM$ is an $HR$-module spectrum. Specifically, I want to know how to construct the action of $HR$ on $HM$: $$HR \wedge HM \to HM$$ Here's something related which I believe I understand (please let me know if this is incorrect). Let $S$ be a commutative ring with identity. If we have an action of $S$ on $M$, $$f:S \to End(M)$$ $$x \mapsto f_x$$ then, by applying $H$ to each $f_x$, we get the maps $$Hf_x : HM \to HM$$ Hence, we get an action of $S$ on $HM$. But doing the same thing with the action of $R$ on $M$ clearly doesn't show that $HM$ is an $HR$-module spectrum. I'd appreciate any help you all can give.

Answer 1

Here you can find the definition of $HA$ for symmetric spectra (Example 1.14), the construction of the smash product $\wedge$ on the category $\mathcal S^\Sigma$ of symmetric spectra (Section I.5) and that $H$ defines a lax symmetric monoidal functor $$(\mathbf{Ab},\otimes)\rightarrow(\mathcal S^\Sigma,\wedge)$$ (Example 5.28). An $A$-module structure on $M$ is a group homomorphism $A\otimes M\rightarrow M$, so by applying $H$ we get the $HA$-module structure on $HM$ $$HA\wedge HM\rightarrow H(A\otimes M)\rightarrow HM$$