Let $R$ be a ring spectrum (in the world of EKMM $S$-modules) and let $E$ be a smashing $R$-module. Denote by $R\_E$ the $E\_\*$-localization of $R$. By a theorem of Wolbert (Theorem 2 in [Classifying modules over $K$-theory spectra](http://www.sciencedirect.com/science/article/pii/S0022404996001120)), the derived category $\mathrm{Der}(R)[E^{-1}]$ of $E\_\*^R$-local $R$-modules is equivalent to the derived category $\mathrm{Der}(R\_E)$ of $R\_E$-modules. I wonder: is this equivalence induced by a Quillen equivalence
$$
L\_E \mathrm{Mod}(R)\simeq\_Q\mathrm{Mod}(R\_E)\; ?
$$



Does Wolbert's derived equivalence between $E_*^R$-local $R$-modules and $R_E$-modules come from a Quillen equivalence?
This is an elaboration on Lennart's comment.


This can be made to come from a Quillen equivalence. Here are the ingredients you'd usually need to show it. (Sorry, I don't have my copy of EKMM handy and so I can't provide theorem numbers.) The problem is that you haven't specified $R\_E$ as an actual object yet rather than just as a homotopy type.


There is a localization map $R \to R\_E$. This can be chosen as a map of $S$-algebras and with $R\_E$ cofibrant as a right $R$-module (which is possible up to equivalence, but not automatically satisfied). In this case, you can produce a Quillen equivalence. If you've got a representative for $R\_E$ that's not a cofibrant $R$-module, then you're typically going to have to replace it with an equivalent algebra $R'\_E$ which is, and this will give a zigzag of Quillen equivalences $L\_E Mod(R) \sim Mod(R'\_E) \sim Mod(R\_E)$. (Really, all we'll need is a "flatness" property.)


Now let's give some details of the proof.


Unless I'm mistaken, the ordinary model structures on $Mod(R)$ and $Mod(R\_E)$ are lifted from $Mod(S)$: a map is a weak equivalence or a fibration if and only if it is so in $S$-modules. For this reason, the forgetful map is automatically a right Quillen functor. Then $R\_E \wedge\_R (-)$ is a left Quillen functor, preserving cofibrations and weak equivalences.


Now let's talk about the localization. This is usually taken to be a left Bousfield localization on the model category level: $L\_E Mod(R)$ has the same underlying category, and the same underlying cofibrations, as $Mod(R)$, but new weak equivalences (the $E$-equivalences) and new fibrations.


Since it's the same underlying category, we still have the forgetful functor from $Mod(R\_E)$ to $L\_E Mod(R)$, with left adjoint $R\_E \wedge\_R (-)$. The left adjoint still preserves cofibrations.


More, it preserves weak equivalences. Since this is a smashing localization, a map $X \to Y$ is an $E$-equivalence if and only if it's an equivalence after taking derived smash product over $R$ with $R\_E$. Since $R\_E$ is cofibrant, smash products represent derived smash products $\wedge^{\mathbb L}\_R$. Therefore, $X \to Y$ is an $E$-equivalence if and only if $R\_E \wedge\_R X \to R\_E \wedge\_R Y$ is a weak equivalence.


Now we finally need to show that this is a Quillen equivalence. Suppose $X$ is a cofibrant $R$-module and $Y$ is a fibrant $R\_E$-module. We already showed that a map $X \to Y$ of $R$-modules is an $E$-equivalence if and only if the map $R\_E \wedge\_R X \to R\_E \wedge\_R Y$ is an equivalence. The unit map $R\_E \wedge\_R Y \to Y$ is always an equivalence because $Y$ is $E$-local and this is a smashing localization. Therefore, $X \to Y$ is an equivalence in $L\_E Mod(R)$ if and only if its adjoint $R\_E \wedge\_R X \to R\_E \wedge\_R Y \to Y$ is an equivalence in $Mod(R\_E)$, making this into a Quillen equivalence.


