A bit of context: for any ordinary associative algebra $A$ and element $x \in A$, the subalgebra spanned by the powers of $x$ is commutative. In the universal example, this says that the free associative algebra on one element is commutative. Or if we prefer: any associative algebra is the union of its commutative subalgebras.


I expect any analogue of these statements to be false for ring spectra, and moreover I no longer have a clear idea of what a "subalgebra" ought to be. But I haven't been able to produce an example of what goes wrong.


So here's a tentative definition. Let $A$ be an $A\_\infty$ ring spectrum. We'll say that $x \in \pi\_\*(A)$ is *of commutative origin* in case there exists an $E\_\infty$ ring spectrum $B$, equipped with an $A\_\infty$ map $f: B \to A$ such that $x \in \text{image}(\pi\_\*(f)).$


My question is: is it easy to produce an example of an associative ring spectrum $A$ and an element $x \in \pi\_\*(A)$ which is **not** of commutative origin in this sense?


The question arose in thinking about whether one could reduce nilpotence questions in $A$, which are controlled by $MU$ according to the theorems of Devinatz-Hopkins-Smith, to nilpotence questions in a lift to $B$, where one has the simpler(?) May nilpotence machinery which only requires integral homology. 


(May's nilpotence conjecture was affirmed by Mathew-Naumann-Noel in <https://arxiv.org/abs/1403.2023>)



commutative "subalgebras" of associative ring spectra
Let $A$ be an $\mathbf{E}\_1$-ring, and let $x\in \pi\_n A$. There are two distinct cases to consider. First, if $n = 0$, then the answer to your question is that $x$ is in the image of an $\mathbf{E}\_1$-map from an $\mathbf{E}\_\infty$-ring. Indeed, then $x:S^0\to A$ extends to a map $\Sigma^\infty\_+ \mathbf{Z}\_{\geq 0}\to A$, essentially because $B\mathbf{Z}\_{\geq 0} = S^1$ (and so is evidently in the image of this map). Now observe that $\Sigma^\infty\_+ \mathbf{Z}\_{\geq 0}$ admits the structure of an $\mathbf{E}\_\infty$-ring (where the commutative monoid $\mathbf{Z}\_{\geq 0}$ is regarded as an $\mathbf{E}\_\infty$-space).


Next, if $n>0$, then $x$ is not necessarily in the image of an $\mathbf{E}\_1$-map from an $\mathbf{E}\_\infty$-ring. Consider the universal case, when $A = \Omega S^{n+1}\_+$ is the free $\mathbf{E}\_1$-ring on one generator in degree $n$, and $x\in \pi\_n(A)$ is given by the suspension $E:S^n\to \Omega S^{n+1}$. The question, then, is whether one can find an $\mathbf{E}\_\infty$-ring $B$ and an $\mathbf{E}\_1$-map $f:B\to \Omega S^{n+1}\_+$ such that $E$ lifts to $B$. If there was a lift $\widetilde{E}:S^n\to B$ of $E$, then there must be an $\mathbf{E}\_k$-map $\widetilde{E}:\Omega^k S^{n+k}\_+\to B$ for every $k\geq 0$. Moreover, the $\mathbf{E}\_1$-composite
$$\Omega S^{n+1}\_+\to \Omega^k S^{n+k}\_+\xrightarrow{\widetilde{E}} B\xrightarrow{f} \Omega S^{n+1}\_+$$
would be the identity on $\Omega S^{n+1}\_+$. In particular, $\Omega S^{n+1}\_+$ would be an $\mathbf{E}\_1$-summand of $\Omega^k S^{n+k}\_+$. The James splitting gives an equivalence $\Omega S^{n+1}\_+ = \bigvee\_{k\geq 0} S^{nk}$, and so $S^n$ would be a summand of $\Omega^k S^{n+k}\_+$. I think this fails, for instance, once $k\geq 2$ (recall that $n>0$). (By the way: the proof of May nilpotence relies on the nilpotence theorem. I can talk more about this over email if you're interested.)


The essence of this answer is that $S^1$ is the only sphere which admits the structure of an infinite loop space.


One other thing to mention that might be of interest is the following. Although it's very hard (and often impossible) to lift to $\mathbf{E}\_\infty$-rings, it's sometimes possible to prove liftings to $\mathbf{E}\_k$-rings for finite $k$. For instance, let us work $p$-locally, where $p$ is an odd prime (slight variants also work at $p=2$). A hard theorem of Cohen, Moore, and Neisendorfer says that the map $p:S^{2n-1}\to S^{2n-1}$ factors through the double suspension $E^2:S^{2n-1}\to \Omega^2 S^{2n+1}$. In particular, there's a map $\Omega^2 S^{2n+1}\to S^{2n-1}$ which is degree $p$ on the bottom cell. Looping, we get that $p\cdot E\in \pi\_{2n-2} \Omega S^{2n-1}$ factors through an $\mathbf{E}\_1$-map $\Omega^3 S^{2n+1} \to \Omega S^{2n-1}$. Iterating, you find that $p^k\cdot E\in \pi\_{2n-2} \Omega S^{2n-1}$ factors through an $\mathbf{E}\_1$-map $\Omega^{2k+1} S^{2n+2k-1} \to \Omega S^{2n-1}$. Stabilizing, you get (for each $k$) an example of an element in the homotopy of an $\mathbf{E}\_1$-ring which lifts to an $\mathbf{E}\_k$-ring.


