Homotopy limit and Bousfield localization

Question

Suppose $E$ is a spectrum and $p$ is a prime. We can then $(H\mathbb{Z}/p)_*$-localize to obtain $L_{H\mathbb{Z}/p}E$. Is it true that the natural map $L_{H\mathbb{Z}/p}E\rightarrow\text{holim}_n(L_{H\mathbb{Z}/p}E)/p^n$ is an $(H\mathbb{Z}/p)_*$-equivalence?

Answer 1

Yes, $H\mathbb{F}_p$-localizations are $p$-complete. We have the cofiber sequence $$ \Sigma^{-1}\mathbb{S}/p^\infty \xrightarrow{j} \mathbb{S}\to \mathbb{S}[p^{-1}] \to \mathbb{S}/p^\infty $$ where $\mathbb{S}/p^\infty\approx \mathrm{colim}_n \mathbb{S}/p^n$, from which it is not hard to see that the $p$-completion map $X\to \mathrm{holim}_n X/p^n=X^\wedge_p$ is equivalent to $$ F(\mathbb{S},X) \xrightarrow{j^*} F(\Sigma^{-1}\mathbb{S}/p^\infty, X). $$ In particular, $X\to X^\wedge_p$ is an equivalence iff $F(\mathbb{S}[p^{-1}], X)=0$.

We have that $H\mathbb{F}_p\wedge \mathbb{S}[p^{-1}]=0$, so $F(\mathbb{S}[p^{-1}], L_{H\mathbb{F}_p}X)=0$.