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 group law is literally the additive one. Is $E$ an $H\mathbb{Z}$-algebra?



Is every complex oriented ring spectrum with additive FGL an Eilenberg-Maclane spectrum?
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 likely has a negative answer)


The answer is yes. Here is an outline of the proof. The vast majority of the following was outlined to me by Robert Burklund and Eric Peterson, but as usual all the errors are mine, etc, etc.


The first step is to reduce to the connective case by noting that $E$ being an $H\mathbb{Z}$-module is equivalent to asking that the unit map $\mathbb{S}\rightarrow E$ factor through $H\mathbb{Z}$, as a map of spectra, and the unit map of $E$ always factors through its connective cover.


The next step is to show that the coation of the mod $p$ dual Steenrod algebra $(H\mathbb{F}\_p)\_\*H\mathbb{F}\_p$ on $(H\mathbb{F}\_p)\_\*E$ (and also $(H\mathbb{F}\_p)\_\*(E/p)$ which is needed later) on is cofree for all $p$. There are probably a lot of ways to do that, my favorite being to take the "formal geometry" perspective by considering the action-of-a-group-scheme-on-another-scheme that the coaction map in question corepresents, and constructing an equivariant map to a scheme on which the group scheme clearly acts freely. This is where you use the fact that you can choose a complex orientation of $E$ in which the formal group law of $E$ is equal to the additive formal group law.


The next step is to make some arguments using the Adams spectral sequence: the previous step implies that the $H\mathbb{F}\_p$-ASS for $E/p$ is concentrated on the zero line. Since $E$ is connective, $E/p$ is $H\mathbb{F}\_p$-nilpotent which implies that the Hurewicz map $E/p\rightarrow H\mathbb{F}\_p\wedge E/p$ is an injection on homotopy groups. Choosing a basis of $\pi\_\*(E/p)$ and extending it to a basis of $\pi\_\*(H\mathbb{F}\_p\wedge E/p)$ defines an equivalence from a wedge of shifts of $H\mathbb{F}\_p$ $X\_p$ to $H\mathbb{F}\_p\wedge E/p$. Let $Y\_p$ be the summand of $X\_p$ corresponding to basis elements of $\pi\_\*(E/p)$. Then the map $E/p\rightarrow H\mathbb{F}\_p\wedge E/p\simeq X\_p\rightarrow Y\_p$ is a weak equivalence, and $Y\_p$ is an $H\mathbb{F}\_p$-module.


The final step is to invoke a lemma that says that if $E/p$ is an $H\mathbb{F}\_p$-module for all $p$, then $E$ is an $H\mathbb{Z}$-module, which involves an argument about $k$-invariants being visibile mod $p$ for some $p$.


