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Let $L_E$ denote Bousfield localisation with repsect to the cohomology theory $E$. I am trying to follow through some calculations in Hovey-Strickland's paper Morava $K$-theories and localisation

Claim 7.10(e) is that

$$L_{K(n)}X = \underset{\leftarrow}{\text{holim}}_I L_{E(n)}X \wedge S/I$$

where the homotopy limit is over a tower of generalised Moore spectra (the Moore spectra $S/I$ is defined in 4.12 and the tower in 4.22)

Define $$E_*^{\vee}:=\pi_{*} (L_{K(n)} (E \wedge X))$$

the $K(n)$-local version of Morava $E$-theory (where, I believe, Morava $E$-theory here is what I might call a completed Johnson-Wilson theory, but I don't believe it really matters).

The claim (8.04) is then that we can extract a Milnor exact sequence from this:

$$0 \to \varprojlim_I {}^1 (E/I)_{\ast+1}(X) \to E_*^\vee X \to \varprojlim_I (E/I)_*X \to 0 $$

I'm not sure how to show this. I would like to think that we can get a sequence

$$0 \to \varprojlim_I {}^1 \pi_{\ast+1}(L_{E(n)}(E \wedge X) \wedge S/I)\to E_{\ast}^\vee X \to \varprojlim_I \hspace{1mm} \pi_* (L_{E(n)}(E \wedge X) \wedge S/I) \to 0 $$

and then if you drop the $E(n)$-localisation it seems to work, but I'm not really sure about this.

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In this case, this follows because $E(n)$-localization is a smashing localization. Specifically, $$ L_{E(n)}(X) = L_{E(n)}(\mathbb{S}) \wedge X $$ for any $X$. In particular, this means that the smash product of any spectrum with an $E(n)$-local spectrum is already $E(n)$-local.

The fact that $E(n)$-localization is smashing was one of the Ravenel conjectures; specifically, conjecture 10.6 in his paper "Localization with respect to certain periodic homology theories". I don't have a copy of his orange book handy and so can't give you a reference for the theorem statement.

As a result, as the Morava $E$-theories are $E(n)$-local already, so are the smash products $E \wedge X$ in your equation.

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    $\begingroup$ Thanks Tyler! In its simplest form: $L_{E(n)}(E \wedge X) = L_{E(n)}(\mathbb{S}) \wedge E \wedge X = L_{E(n)} (E) \wedge X = E \wedge X$? $\endgroup$ – Drew Heard Mar 30 '12 at 2:11
  • $\begingroup$ Drew, exactly right. $\endgroup$ – Tyler Lawson Mar 30 '12 at 2:45

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