Let $E$ be a Landweber exact ring spectrum. That is, we have a map of homotopy ring spectra $MU\rightarrow E$ and an isomorphism of homology theories $E\_\*X\simeq MU\_\*X\otimes\_{MU\_\*}E\_\*$. Is the homotopy type of $E$ determined by the graded ring $E\_\*$?


My best guess is that the answer is "no" which would ideally be proved by finding a (graded) ring $R$ and two (graded), Landweber exact, formal group laws (graded ring maps) $e,f:MU\_\*\rightarrow R$, such that the spectra representing the two homology theories are not homotopy equivalent.



Are Landweber exact spectra determined by their coefficient ring?
Put $R\_\*=\mathbb{Z}\_{(p)}[x,y,y^{-1}]$ with $|x|=|y|=2$. Define $f,g\colon BP\_\*\to R\_\*$ by
\begin{align\*}
 f(v\_i) &= \begin{cases} 
 y^{p-1} & \text{ if } i = 1 \\
 0 & \text{ if } i > 1
 \end{cases} \\
 g(v\_i) &= \begin{cases} 
 x^{p-1} & \text{ if } i = 1 \\
 y^{p^2-1} & \text{ if } i = 2 \\
 0 & \text{ if } i > 2
 \end{cases} \\
\end{align\*}
This gives Landweber exact ring spectra $A\_f$ and $A\_g$ of heights one and two respectively, so for a finite spectrum $X$ of type $2$ we have $A\_f\wedge X = 0 \neq A\_g\wedge X$, so $A\_f$ and $A\_g$ are not equivalent as spectra.


