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Let $\mathbb S[z]$ be the free $E_\infty$-ring spectrum generated by the commutative monoid $\mathbb N$. That is, $\mathbb S[z] = \Sigma^\infty_+ \mathbb N$.

In Bhatt-Morrrow-Scholze II (https://arxiv.org/abs/1802.03261), they define a map out of this spectrum by choosing an element of the target. In particular, they use the uniformizer $\varpi$ of $\mathcal O_K$ to define a map $\mathbb S[z] \to \mathcal O_K$ by $z \mapsto \varpi$. This occurs in the paragraph after Theorem 11.2. By all indications, this seems to be an $E_\infty$-ring map.

This is not the only place in the literature where such a property of $\mathbb S[z]$ is used. In this paper of Krause--Nikolaus (https://arxiv.org/abs/1907.03477), they define the same map at the beginning of section 3.

However, we know the universal property of $\mathbb S[z]$: it is the free $E_1$-algebra on a single generator. This does not often coincide with the free $E_\infty$-algebra on a single generator. See here or the second answer here.

My questions are the following:

  1. Are these maps $\mathbb S[z] \to \mathcal O_K$, $z \mapsto \varpi$ actually $E_\infty$-ring maps? Or just $E_1$?
  2. How do you obtain $E_\infty$-ring maps out of $\mathbb S[z]$? In particular, for which elements of $\Omega^\infty A \simeq \operatorname{Map}_{E_1}(\mathbb S[z], A)$ can we lift an $E_1$-map to an $E_\infty$-map?
  3. When the target is an Eilenberg--Maclane spectrum, can we obtain an $E_\infty$-map $\mathbb S[z] \to HB$ for any element of $B = \pi_0 HB$?

Ideally, the answer to this second question would come in the form of a natural transformation $\Omega^\infty \to \operatorname{Map}_{E_\infty}(\mathbb S[t], -)$. But this is probably too optimistic.

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    $\begingroup$ I think the answer to 3 is yes: E_infty maps out of S[z] are the same as E_infty maps from N to \Omega^{infty}X. Any map of (strictly) commutative monoids is also E_infty, so if X is Eilenberg-MacLane on a commutative ring then you’re good. That also answers 1. The answer to 2 is ‘that’s very hard’. See, eg, mathoverflow.net/questions/135641/… $\endgroup$ Commented Oct 3, 2019 at 0:17
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    $\begingroup$ For 3: If $R$ is any $(-1)$-connected $E_\infty$-ring, then $\mathrm{Map}_{E_\infty}(R,HB)=\mathrm{Hom}(\pi_0 R,B)$. $\endgroup$ Commented Oct 3, 2019 at 1:25

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$\newcommand{\E}{\mathbf{E}}$Dylan answered question 3 (and hence question 1) in the comments, but here's another equivalent way to see it: $\E_\infty$-maps $S^0[z]\to R$ with $R$ a discrete ring (i.e., viewed as an Eilenberg-Maclane spectrum) are the same data as $\E_\infty$-maps $\tau_{\leq 0} S^0[z] = H\mathbf{Z}[t] \to R$ of $\E_\infty$-rings, i.e., maps $\mathbf{Z}[t]\to R$ of rings. Question 2 is hard: in general, $\E_n$-maps $\Sigma^\infty_+ \Omega^n X\to A$ (where $X$ is a topological space and $A$ is an $\E_n$-ring) are the same data as maps $X\to B^n\mathrm{GL}_1(A)$ of spaces. If $n=\infty$, that's saying that $\E_\infty$-maps $\Sigma^\infty_+ \mathbf{Z}\to A$ are the same things as maps $H\mathbf{Z}\to gl_1(A)$ of spectra. (In Lurie's elliptic cohomology papers, the space of such maps is denoted $\mathbf{G}_m(A)$.) Lurie's answer in the question linked to in the comments is concerned with computing $\mathbf{G}_m(A)$ in the special case of $A$ being a Morava E-theory.

In fact, that result can be made explicit at height $1$; see http://math.uchicago.edu/~amathew/notes_thursday.pdf. Here's a summary of what goes on. If $R$ is a $K(1)$-local $\E_\infty$-$K_p$-algebra, it's a little easier to study $\E_\infty$-maps $S^0[z]\to R$. Such maps factor through the $K(1)$-local spectrum $L_{K(1)} S^0[z]$, which admits a concise presentation as a $K(1)$-local $\E_\infty$-ring. Recall (e.g. from https://web.math.rochester.edu/people/faculty/doug/otherpapers/knlocal.pdf) that the $p$-adic K-theory of the free $K(1)$-local $\E_\infty$-ring $L_{K(1)} S^0\{z\}$ on a generator in degree zero is simply the polynomial ring $\mathbf{Z}_p[z, \theta(z), \theta^2(z), \cdots]$, where the $K(1)$-local power operation $\theta$ takes $\theta^{n-1}(z)$ to $\theta^n(z)$. The element $\theta(z)\in \pi_0 (K_p \wedge L_{K(1)} S^0\{z\}) = \pi_0 L_{K(1)} K_p\{z\}$ defines a map $S^0\to L_{K(1)} K_p\{z\}$ which extends (by the universal property of free $\E_\infty$-rings) to a map $L_{K(1)} K_p\{z\} \xrightarrow{\theta} L_{K(1)} K_p\{z\}$. Since this map must commute with power operations, it sends $\theta^{n-1}(z)$ to $\theta^n(z)$ on homotopy.

The $\E_\infty$-cone $L_{K(1)} K_p\{z\}/\!\!/\theta(z)$ on this element is defined via the pushout of the diagram $$K_p \leftarrow L_{K(1)} K_p\{z\} \xrightarrow{\theta} L_{K(1)} K_p\{z\}.$$ A simple calculation with the long exact sequence on homotopy shows that there is in fact an equivalence $L_{K(1)} K_p\{z\}/\!\!/\theta(z) \simeq L_{K(1)} K_p[z]$ of $\E_\infty$-rings. This is the only compact presentation of any strict spectral polynomial ring that I know of.

As a side remark, here's a curious observation. There is an element $\psi^p(z)\in L_{K(1)} K_p\{z\}$, so we can consider the $\E_\infty$-cone $L_{K(1)} K_p\{z\}/\!\!/\psi^p(z)$. Using the relation $\psi^p(z) = z^p + p\theta(z)$, you can show that this $\E_\infty$-ring has divided powers on $z$.

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  • $\begingroup$ Thanks! Does the translation between $E_\infty$-maps $\mathbb S[z] \to HR$ and ring maps $Z[z] \to R$ come about because of an adjunction between $\tau_{\leq 0}$ and $H$? I want to say something like $\tau_{\leq 0}$ is left inverse to $H$ as a functor from connective spectra to abelian groups, but this doesn't quite typecheck. $\endgroup$ Commented Oct 3, 2019 at 16:56
  • $\begingroup$ @DavidMehrle Yes: in general, $\mathrm{Map}(X, Y) \simeq \mathrm{Map}(\tau_{\leq n} X, Y)$ if $Y\simeq \tau_{\leq n} Y$. $\endgroup$
    – skd
    Commented Oct 3, 2019 at 23:36

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