Does the spectrum of Morava E-theory depend only on height? I almost expect the answer to this question to be no, but I can't find it explicitly said anywhere.
Given a formal group law $f$ of height $n$ over a perfect field $k$ of characteristic $p$, we can construct the Morava E-theory $E(n)$. The resulting cohomology theory is
$$E(n)^m (X) =  MP^m(X)\otimes _{MP(*)} R$$
where $R = W(k)[[v_1,...,v_{n-1}]]$ is the Lubin-Tate ring of $f$ and $MP$ is complex cobordism. Also
$$ \pi_\bullet (E) = E(n)^\bullet (*) = R[\beta, \beta^{-1}]$$
so at least the homotopy groups of the spectrum are independent of our choice of formal group law. This suggests that maybe the spectrum is the same, and the difference is the multiplicative structure on it. I think my confusion also might come from the fact that people usually refer to it as "Morava E-theory" rather than "a Morava E-theory", somewhat implying uniqueness.
However, the main problem I see with this is that the map $MP(*) \to R$ from the Lazard ring classifying the universal deformation of $f$ (which makes $R$ a $MP(*)$-module) will be different for different formal group laws, so I would expect different values (as a graded abelian group) of the cohomology theory depending on the formal group chosen.
Can someone please link an appropriate reference?
 A: Here's an argument that Eric Peterson and I came up with showing that the homotopy type of Morava $E$-theory only depends on the choice of perfect char $p$ field $k$ and the height $n<\infty$.
$\textbf{Lemma}$: Let $R$ be a ring with two Landweber exact formal group laws $e,f:\text{Laz}\rightarrow R$ and let $E$ and $F$ be the spectra corresponding to these two formal group laws under the Landweber exact functor theorem. Then $E$ and $F$ have the same homotopy type if there is a ring extension $u: R\rightarrow S$ which is split as an $R$-module map, and such that the formal group laws $u\circ e$ and $u\circ f$ are isomorphic and Landweber exact.
$\textbf{Proof}$: First we show that $E\simeq F$ if the map $\eta:E_*\rightarrow F_*E$ (induced by $1\wedge id: \mathbb{S}\wedge E\rightarrow F\wedge E$) is split (as a map of $F_*$-modules). Note that $E_*$ is canonically an $F_*$ module by the \textit{equality} $E_*=R=F_*$. We claim that $F_*E$ is flat over $F_*$. That boils down to interpreting Landweber exactness as flatness of maps to $\mathcal{M}_{fg}$, noting that flatness is preserved under pullback, and that (Spec of) $E_*$, $F_*$ $F_*E$ and $\mathcal{M}_{fg}$ fit into a pullback square (with $\mathcal{M}_{fg}$ in the bottom right and Spec of $F_*E$ in the top left). It follows that $F\wedge E$ represents the cohomology theory $X\mapsto F^*X\otimes_{F_*}F_*E$, and then Brown representability (and Strickland's result about the absence of phantom maps between Landweber exact spectra) produces a map $\phi: \text{Mod}_{F_*}(F_*E,F_*)\rightarrow F^*E$. Now suppose $s$ is a splitting of $\eta$. Then there is a map $\phi(s):E\rightarrow F$ which on homotopy groups (set $X=\text{pt}$ above) is the composite $s\circ\eta$ which is the identity map (recall that $E_*$ and $F_*$ are canonically isomorphic to $R$), so $E\simeq F$.
Now we show that $\eta:E_*\rightarrow F_*E$ is split if there is a ring extension $u: R\rightarrow S$ which is split as an $R$-module map, and such that the formal group laws $u\circ e$ and $u\circ f$ are isomorphic and Landweber exact. Write $g:=u\circ e$ and $h:=u\circ f$. Assume they are isomorphic and Landweber exact. Let $G$ and $H$ be the spectra associated to $g$ and $h$. Then we have the following commutative diagram of $F_*(=R)$-modules
$\require{AMScd}$
\begin{CD}
S=G_* @>1\otimes 1\otimes id>> H_*G= S\otimes_{MU_*}MU_*MU\otimes_{MU_*}S\\
@A u A A @AA u\otimes id\otimes u A\\
R=E_* @>>1\otimes 1\otimes id> F_*E=R\otimes_{MU_*}MU_*MU\otimes_{MU_*}R
\end{CD}
Since the right vertical map is split (as a map of $F_*$-modules), splitting the bottom horizontal map (as a map of $F_*$-modules) is equivalent to splitting the diagonal map (as a map of $F_*$-modules). Moreover, the top horizontal map is split (as an $S=H_*$-algebra (and hence $F_*$-algebra) map in fact) by first choosing an isomorphism of $g$ and $h$, which induces an $H_*$-algebra isomorphism $H_*G\simeq G_*G$ and then using the multiplication map $G\wedge G\rightarrow G$ to split $G_*\rightarrow G_*G$. Therefore splitting the diagonal map is equivalent to splitting the left vertical map (as a map of $F_*$-modules).
$\textbf{End of proof}.$
Now we apply the lemma to show that at height $n<\infty$ the homotopy type of Morava $E$-theory depends only on the choice of a perfect characteristic $p$ field. In other words, let $k$ be a perfect characteristic $p$ field and let $E_1(n)$ and $E_2(n)$ be two Morava $E$-theory spectra corresponding to two arbitrary height $n< \infty$ formal group laws over $k$. Then we'll show that as spectra, $E_1(n)\simeq E_2(n)$.
It suffices to check the conditions of the lemma. Let $\overline{k}$ be the algebraic closure of $k$. Let $R$ and $S$ be the (2-periodic) Lubin-Tate deformation rings $W(k)[[u_1,...,u_{n-1}]][\beta^\pm]$ and $W(\overline{k})[[u_1,...,u_{n-1}]][\beta^\pm]$. Let $\tilde{e}$ and $\tilde{f}$ be formal group laws over $R$ universally deforming $e_1$ and $e_2$. Then $\tilde{e}_1$ and $\tilde{e}_2$ are Landweber exact and the corresponding spectra are the Morava $E$-theory spectra $E_1(n)$ and $E_2(n)$. Let $u:R\rightarrow S$ be the map that extends coefficients by $W(k)\rightarrow W(\overline{k})$, sends $u_i$ to $u_i$, and $\beta$ to $\beta$. Then the formal group laws $\tilde{g}:=u\circ \tilde{e}_1$ and $\tilde{h}:=u\circ \tilde{e}_2$ over $S$ are isomorphic and Landweber exact, since they universally deform the isomorphic formal group laws $g:u\circ e_1$ and $h:u\circ e_2$ over $\overline{k}$.  Finally, the map $u:R\rightarrow S$ is split as an $R$-module map, because the map $W(k)\rightarrow W(\overline{k})$ is split as a $W(k)$-module map (See (ill Sawin's comments at Splitting the Witt vectors of $\overline{\mathbb{F}_p}$))
