Why is the Moore spectrum $S\mathbf{Z}_{(J)}$ a ring spectrum?

Question

Let $J$ be a set of primes and consider the Moore spectrum $S\mathbf{Z}_{(J)}$. In his paper 'The localization of spectra with respect to homology', Bousfield writes that $S\mathbf{Z}_{(J)}$ is a ring spectrum. Why is this the case? I first thought that it would be possible to apply the construction $S(-)$ to the multiplication map $\mathbf{Z}_{(J)} \otimes_{\mathbf{Z}} \mathbf{Z}_{(J)} \to \mathbf{Z}_{(J)}$, but then realized that $S(-)$ is not a functor.

Answer 1

Indeed Moore spectra need not be ring spectra, or at least $E_n$-ring spectra, see for example here and here. In your case, your spectra are $E_\infty$-ring spectra because they are the localizations of the sphere spectrum. Indeed you can notice that $$\mathbb S_{(T)}\simeq M(\mathbb Z_{(T)})$$ for $T$ a set of primes, and $M(A)$ the Moore spectrum of the abelian group $A$. By the properties of localization, they all carry an $E_\infty$ stucture induced from that of $\mathbb S$ (essentially because localization commutes with smash product).