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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 homoto …

The answer is no. I guess I owe Fernando Muro 10 dollars. Let $\mathbb{S}\rightarrow\Sigma^{-2}\mathbb{CP}^\infty$ be the inclusion of the bottom cell, and let $f:F\rightarrow\mathbb{S}$ be the fiber. …

Recall that the Bousfield class of a spectrum $E$, written $\langle E\rangle$, is the class of spectra $X$ such that $X\wedge E$ is not contractible. For example the Bousfield class of any of the sphe …

A ring spectrum $E$ is complex oriented if it is equipped with a ring map $MU\rightarrow E$. It is complex orientable if such a ring map exists. An $E_\infty$-ring $E$ is $E_\infty$-complex oriented i …

Suppose $E$ is a complex-oriented ring spectrum whose formal group law is isomorphic to the additive one. As the title suggests, we might as well change the complex orientation so that the formal grou …

This is an answer to the question in the title, which is what I had meant to ask: is an $E$ as in the question body an $H\mathbb{Z}$-module? (the last sentence of the question body is stronger and lik …

To every complex-oriented ring spectrum $E$ there is associated a formal group law, which is a power series $F_E(x,y)\in E_*[[x,y]]$. Suppose $E$ and $F$ are two complex-oriented ring spectra and supp …

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 $n<\infty$. $\textb …