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?
Algebraic Topology – Is Every Complex Oriented Ring Spectrum with Additive FGL an Eilenberg-Maclane Spectrum?
at.algebraic-topologychromatic-homotopyformal-groupshomotopy-theory
Related Solutions
The other major class of cases where one can construct $E$ is when $R$ is a localised regular quotient of $L$, or in other words $R=L[S^{-1}]/I$, where the ideal $I$ can be generated by a regular sequence. There is a long history of general results of this type. I think that the sharpest versions are in my papers "Products on $MU$-modules" and "Realising formal groups" (although the methods used are not so different from earlier work). Note that the latter paper works with the periodic spectrum $MP=\bigvee_{n\in\mathbb{Z}}\Sigma^{2n}MU$ rather than $MU$ itself; this is more natural for many purposes. Note that in this context we focus on $\pi_0$ and note that $\pi_0(MP)=L$.
There is one more useful construction which is less often discussed in the literature. Choose generators $a_1,a_2,\dotsc$ for $L=\pi_0(MP)$. Let $A$ be the monoid (under multiplication) generated by these elements, and put $T=\Sigma^\infty_+A$, or equivalently $T=\bigvee_{a\in A}S^0$. Now $\pi_*(T)=\pi_*(S)\otimes L$, and there is an evident map $f\colon T\to MP$ of naive ring spectra, which is an isomorphism on $\pi_0$. If we want to make an $MP$-algebra $E$, we could hope to start by making a $T$-algebra $E'$, and then put $E=E'\wedge_TMP$. In particular, suppose that $I$ is an ideal in $L$ that is generated by some subset of monomials in the generators $a_i$. Then it is easy to construct $T/I$ as $\bigvee_{b\in B}S^0$ for a suitable subset $B\subseteq A$, and we can hope to construct $MP/I$ as $T/I\wedge_TMP$. In particular, we can choose lifts in $\pi_0(MP)$ of the chromatic generators $v_k$, and use these as some of the integral generators $a_i$; then we can form $MP/(v_1^2,v_1v_2,v_2^2)$, which is similar but not identical to the thing that @LennartMeier mentioned in his comment.
However, to make this work, we need more highly structured versions of $T$, $MP$ and $f$ (because there is no general construction of smash products of modules over unstructured ring spectra). We can construct strictly commutative versions of $T$ and $MP$, and also of $T/I$ when $I$ is generated by monomials, but unfortunately $f\colon T\to MP$ cannot be a map of strictly commutative rings, because $\pi_0(MP)$ has interesting power operations and $\pi_0(T)$ does not. There are some pitfalls with model category structures that can cause trouble here, and I have not checked all the details, but I think one can do the following, for example in the category of EKMM spectra.
- In the EKMM context, the $0$-sphere spectrum $S$ is not cofibrant. We write $C$ for the cofibrant replacement, and $C^m$ for the $m$-fold smash power.
- Let $A_i$ denote the submonoid of $A$ generated by $a_i$ and put $T_i=\Sigma^\infty_+A_i=\bigvee_mS$. The spectrum $T=\Sigma^\infty_+A$ is then the smash product of all the $T_i$ (by which I mean, the colimit over $n$ of $\bigwedge_{i=1}^nT_i$).
- Let $T_i^c$ denote the cofibrant replacement of $T_i$, in the category of strictly commutative ring spectra. This has no simple description. Let $T^c$ denote the smash product of the $T_i^c$; I think this is the cofibrant replacement of $T$ in the commutative category.
- Put $T_i^a=\bigvee_mC^m$. This is the free associative ring generated by $C$, and is the cofibrant replacement of $T_i$ in the associative category. There is a natural weak equivalence $T_i^a\to T_i^c$ of associative rings.
- Let $T^a$ be the smash product of all the objects $T^a_i$. This is an associative ring, but is not cofibrant as such. I think that does not matter.
- Now take a strictly commutative model of $MP$. Using the freeness property of $T_i^a$ we get a map $f^a_i\colon T^a_i\to MP$ carrying the generator to $a_i$. As $MP$ is commutative, we can smash these together and pass to a colimit to get a map $T^a\to MP$ of strictly associative ring spectra. This gives us an associative $T^a$-algebra. To be on the safe side, we should probably take the cofibrant replacement of this in the category of $T^a$-algebras, to get an object $M^a$. We can then take $M^c=T^c\wedge_{T^a}M^a$. This is an associative $T^c$-algebra, and I think we have done enough cofibrant replacement to ensure that the underlying homotopy type is still $MP$.
- Now let $I$ be an ideal in $L$ that is generated by monomials, and let $B$ denote the set of monomials that do not lie in $I$. There is a fairly direct way to make $\Sigma^\infty_+B$ into a strictly commutative ring spectrum, and we let $T^c/I$ denote the cofibrant replacement. This has a natural map from $T^c$. Now we can define $MP/I=T^c/I\wedge_TM^c$.
The following is a communal answer from the algebraic topology Discord [1], primarily put forward by Irakli Patchkoria (correcting previous half-answers by Tyler Lawson and me). Kiran suggested it be recorded here to ease future reference.
The idea is to produce two topological realizations $M$, $N$ of a single $MU_*$–module by finding two distinct resolutions whose effect on homotopy is the same. The two associated square-zero extensions then give a counterexample. We'll reduce complexity first by considering $ku$–modules rather than $MU$–modules, and second by aiming for a $ku$–module whose homotopy cleaves into small even and odd parts, forcing its $ku_*$–module structure to trivialize.
$\DeclareMathOperator{\Sq}{Sq} \newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\HFtwo}{H\F_2} \newcommand{\Susp}{\Sigma} \newcommand{\co}{\colon\thinspace}$ The nice homotopy groups of complex $K$–theory, $ku_* = \Z[u]$, can be used to show that its bottom $k$–invariant $\kappa_{ku}$ is $ku$–linear: in the diagram $$ \begin{array}{ccccc} & & \Susp^4 ku \\ & & u \downarrow \\ \Susp^{-1} H\Z & \to & \Susp^2 ku & \to & \Susp^2 H\Z \\ & & u \downarrow \\ & & ku & \to & H\Z, \end{array} $$ the vertical maps are multiplication by homotopy elements, hence are $ku$–linear; in turn the horizontal co/fibers are also $ku$–linear; and, finally, the $k$–invariant appears as the middle composite, hence is also $ku$–linear. Similarly, we can show the $ku$–linearity of the bottom $k$–invariant of $ku/2$ and of the Bockstein map $\beta\co \HFtwo \to \Susp H\Z$ (relying on the $ku$–linearity of $2\co H\Z \to H\Z$).
Stringing some of these together gives a $ku$–linear composite $$\HFtwo \xrightarrow{\kappa_{ku/2}} \Susp^3 \HFtwo \xrightarrow{\beta} \Susp^4 H\Z \to \Susp^4 \HFtwo.$$ The bottom $k$–invariant $\kappa_{ku/2}$ of $ku/2$ is given as the Milnor primitive $Q_2 = \Sq^3 + \Sq^2 \Sq^1$, the composite of the latter two maps is given as $\Sq^1$, and hence the whole composite is the nontrivial Steenrod operation $$\Sq^1 Q_2 = \Sq^1(\Sq^3 + \Sq^2 \Sq^1) = \Sq^3 \Sq^1.$$ Meanwhile, the homotopy groups of the cofiber $M$ of this composite are $\Susp \F_2 \oplus \Susp^4 \F_2$, which splits as a $ku_*$–module — hence this $ku_*$–module could alternatively be modeled by $N = \Susp H\F_2 \oplus \Susp^4 \HFtwo$ (i.e., the cofiber of the zero map). To finish, set $E = ku \oplus M$ and $F = ku \oplus N$.
Best Answer
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 $p$.