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), 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 $$ L_E \mathrm{Mod}(R)\simeq_Q\mathrm{Mod}(R_E)\; ? $$

$\begingroup$ There's work of Schwede and Shipley on rigidification problems, and I'll bet this is covered in their work since everything in sight is a subcategory of the stable homotopy category. Perhaps their paper with a phrase like "stable equivalences" in the title, or Schwede's Annals paper. If I knew more I'd post as an answer; this is just a suggestion for where to look $\endgroup$ – David White Aug 11 '13 at 22:18

3$\begingroup$ I bet that the answer is yes and that you don't need Schwede's rigidity result. You have a Quillen pair between $Mod(R)$ and $Mod(R_E)$ giving by smashing $\wedge_R R_E$ and forgetting. With appropriate model structure, this should descent to a Quillen pair between $L_E Mod(R)$ and $Mod(R_E)$. What Wolbert is showing, is essentially that the derived functor of the left Quillen is an equivalence. $\endgroup$ – Lennart Meier Aug 12 '13 at 8:39
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.

2$\begingroup$ Just as an addition: A standard source for model categories of modules is math.unibonn.de/~schwede/AlgebrasModules.pdf , Section 4 and 5. And indeed, fibrations and weak equivalences are defined via the underlying map of spectra. The argument is rather modelindependent and can be done also in symmetric or orthogonal spectra. $\endgroup$ – Lennart Meier Aug 12 '13 at 13:57

$\begingroup$ Right, the reference Lennart gives axiomatizes what needs to be shown to construct model categories of algebras and modules. Tyler uses EKMM, which predates Lennart's reference but verifies the relevant axioms. One can also work with symmetric or orthogonal spectra, but the fact that all objects in EKMM are fibrant is convenient here. $\endgroup$ – Peter May Aug 12 '13 at 15:42

$\begingroup$ Many thanks to you three for these helpful replies! $\endgroup$ – Rasmus Aug 12 '13 at 19:09

$\begingroup$ I have a question about your second paragraph (greeting 0th paragraph not counted). How can you get an $R_E$ which is an Ralgebra and a cofibrant Rmodule (or something equivalent to $R_E$ satisfying these two conditions)? $\endgroup$ – Bruno Stonek Oct 23 '18 at 12:19

$\begingroup$ @BrunoStonek It looks like I need R to be cofibrant as an associative algebra in order to do this. A cofibrant replacement of R_E as an Ralgebra is formed by an iterated sequence of pushouts of algebras of B < A(x) > A(y), where x > y is a generating cofibration and A is the free algebra functor. Such a pushout of algebras can be constructed inductively as an iterated pushout along B^(k+1) smashed with the kfold pushout product of x>y. (This is, I believe, in SchwedeShipley.) If we assume inductively that B is a cofibrant Rmodule and spectrum then this is a cofibration... $\endgroup$ – Tyler Lawson Oct 24 '18 at 16:29