Is $MU/I_\infty$ an $E_\infty$ ring?

Question

Fix a prime $p$, and suppose that $p>2$ for simplicity, although many things should also work for $p=2$. Let $F$ be the usual formal group law defined over $MU_*$, and let $I_\infty$ be the ideal in $MU_*$ generated by the coefficients of $[p]_F(x)$. Thus $MU_*/I_\infty$ is the universal example of a ring equipped with an FGL for which the $p$-series is zero, and it is known that any such FGL is isomorphic to the additive FGL.

My question: is $MU/I_\infty$ an $E_\infty$ ring spectrum, or an $E_\infty$ algebra over $MU$ or over $\mathbb{F}_p\otimes MU$?

Here are two possible versions that are more detailed.

First, $I_\infty$ can be generated by a regular sequence, so a standard result shows that there is an essentially unique object $MU/I_\infty\in\operatorname{CAlg}(\operatorname{Ho}(\operatorname{Mod}_{MU}))$ with $\pi_*(MU/I_{\infty})=MU_*/I_\infty$. Work of Angeltveit shows that this admits an $A_\infty$ structure, so it can be promoted to an object in $\operatorname{Ho}(\operatorname{Alg}_{MU})$ for which the product is commutative up to homotopy. The question is whether we can promote it further to $\operatorname{Ho}(\operatorname{CAlg}_{MU})$.

Alternatively, we can take the ring $R=\mathbb{F}_p\otimes MU$ (which means the smash product of $MU$ with the mod $p$ Eilenberg-MacLane spectrum) as the ground ring instead of $MU$. This is an $E_\infty$ algebra over $MU\otimes MU$ whose underlying $(MU\otimes MU)$-module can be identified with $BP\otimes MU/I_\infty$. (This follows easily from the fact that $I_\infty$ is an invariant ideal with $BP/I_\infty=\mathbb{F}_p$.) One can check that there is a canonical regular sequence in $H_*(MU;\mathbb{F}_p)=R_*$ generating an ideal $J$ such that the composite $MU_*\to R_*\to R_*/J$ is surjective with kernel $I_\infty$. The same results as mentioned in the previous paragraph give an object $R/J$ in $\operatorname{Ho}(\operatorname{Alg}_R)$ which can also be thought of as a model for $MU/I_\infty$. This perspective makes it clear that the underlying spectrum of any model for $MU/I_\infty$ is a module over $\mathbb{F}_p$, and thus a sum of suspended copies of $\mathbb{F}_p$. An even better solution to my question would be to promote $R/J$ to $\operatorname{Ho}(\operatorname{CAlg}_R)$.

If any of this works then we will get various kinds of power operations related to $MU/I_\infty$, so we should check whether this is algebraically plausible. There is a well-known relationship between power operations and isogenies of formal groups. The spectrum $MU$ is associated with the stack of pairs consisting of a formal group $G$ equipped with a coordinate $x$ on $G$. (This glosses over the difference between connective $MU$ and the periodic version, but I will ignore that issue here.) Any pair $(G,x)$ gives rise to a formal group law $F$, and $[p]_F(t)=0$ iff $p.1_G=0$, so this condition depends only on $G$ and not $x$. Given an isogeny $q\colon G_0\to G_1$ and a coordinate $x_0$ on $G_0$ then there is a norm construction giving a coordinate $x_1=N_q(x_0)$ on $G_1$, which is morally given by $x_1(b)=\prod_{q(a)=b}x_0(a)$. One can check that if $x_0$ is additive then so is $x_1$. Also, as $q$ is an epimorphism, if $p.1_{G_0}=0$ then $p.1_{G_1}=0$. All this is consistent with the existence of the power operations that would arise from constructing $MU/I_\infty$ as an object in $\operatorname{Ho}(\operatorname{CAlg}_R)$, and might even give a construction of such power operations with a bit of extra work. This might even give an $H_\infty$ structure, i.e. a compatible system of maps $(MU/I_\infty)^{\otimes k}_{h\Sigma k}\to MU/I_\infty$ in the homotopy category of spectra. However, methods like this are probably too lax to give an $E_\infty$ structure.

Answer 1

I don't know what happens at odd primes. Here is a proof that if you do this at p=2, you cannot make $MU / I_\infty$ into an $E_\infty$ $MU$-algebra.

First, as in the comments: $p$-locally, the Quillen idempotent gives rise to a map $BP \to MU$ of $E_2$-rings due to work of Chadwick and Mandell. Using the augmentation $BP \to H$ to the mod-$p$ Eilenberg-Mac Lane spectrum, we can form the relative smash product $$ H \otimes_{BP} MU $$ and because the original maps were maps of $E_2$-rings, this has the structure of an $E_1$-ring. Moreover, on homotopy groups it is formed by starting with $MU$ and taking the quotient by the regular sequence $(p,v_1,\dots)$ by the Kunneth formula, and so it must be a model for $MU / I_\infty$.

Applying mod-$p$ homology, the Kunneth spectral sequence $$ Tor^{H_* BP} (H_* H, H_* MU) \Rightarrow H_*(H \otimes_{BP} MU) $$ degenerates (in fact, both sides are free over $H_* BP$) to an isomorphism $$ A_* \otimes_{H_* BP} H_* MU \cong H_*(MU / I_\infty) $$ where $A_*$ is the dual Steenrod algebra.

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Now let's specialize to $p=2$. In this case, there are classical computations: $$ \begin{align*} A_* &\cong \Bbb F_2[\xi_1, \xi_2, \xi_3, \dots]\\ H_* BP &\cong \Bbb F_2[\xi_1^2, \xi_2^2, \xi_3^2, \dots]\\ H_* MU &\cong \Bbb F_2[b_1,b_2,b_3,\dots] \end{align*} $$ Here $|\xi_j| = 2^j - 1$ and $|b_i| = 2i$. Moreover, the map $H_* BP \to H_* MU$ is well-behaved: $H_* MU$ is a polynomial algebra over $H_* BP$ on the generators $b_i$ for $i$ not of the form $2^j - 1$ (ie: $\xi_j^2 \equiv b_{2^j-1}$ mod decomposables). Roughly, this tells us that the map $H_* MU \to H_* (MU / I^\infty)$ adjoins square roots to certain generators in the polynomial ring.

Now we take the indecomposables $Q(-)$ in these graded rings. The map $Q H_*(MU) \to Q H_*(MU / I^\infty)$ is a map $$ \bigoplus_i \Bbb F_2 \cdot b_i \to \left(\bigoplus_{i \neq 2^j-1} \Bbb F_2 \cdot b_i\right) \oplus \left(\bigoplus_j \Bbb F_2 \cdot \xi_j\right). $$ The kernel $K$ is generated by the classes $b_{2^j - 1}$.

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I claim that this is incompatible with the map $MU \to MU/I_\infty$ being a map of $E_\infty$-rings. If it was, then the map $H_* MU \to H_*(MU/I^\infty)$ would be compatible with the Dyer-Lashof operations. By the Cartan formula, Dyer-Lashof operations preserve decomposables, and so the map $Q H_* MU \to Q H_*(MU/I_\infty)$ on indecomposables would be compatible with them; in particular, the kernel $K$ would be closed under Dyer--Lashof operations.

However, Kochman proved (Theorem 6) that $$ Q^{2r}(b_n) \equiv \binom{r-1}{n} b_{n+r} $$ mod decomposables. This implies that $Q^8 b_1 \equiv b_5$ mod decomposables; but $b_1$ is in the kernel $K$ and $b_5$ is not. (I guess this excludes the possibility that the map $MU \to MU/I_\infty$ is a map of $E_7$-algebras.)