$\newcommand{\Sph}{\mathbb S} \newcommand{\Z}{\mathbb Z} \newcommand{\F}{\mathbb F}$ In this question I will abuse notation by writing $A$ for the (generalized) Eilenberg-MacLane spectrum associated to the abelian group (chain complex) $A$.
Let $R$ be a usual (associative, unital) ring. My question is about the two natural $\Z$-algebra structures on $\Z\otimes_\Sph R$ : the first one "by inclusion to the left", and the second one by $\Z\to R$ followed "by the inclusion to the right".
A priori, one shouldn't expect these to be equivalent $E_1-\Z$-algebra structures, and in fact in general they aren't - I think I have an argument that proves it for $R=\mathbb F_2$ (I'll sketch the argument below)
They are however equivalent in certain cases, e.g. when $R$ is a generalized monoid algebra (over $\Z$), and so my (perhaps too broad) question is :
What are some interesting conditions on $R$ for them to be equivalent ? inequivalent ?
(to make the statement precise, I'm mainly interested in the $E_1-\Z$-algebra structure; but I'm also interested in the $E_n$-structure when $R$ is commutative, $n\in\mathbb N_{\geq 2}\cup\{\infty\}$)
Perhaps a first step would be to try to generalize the fact for $\mathbb F_2$ to odd primes :
Is it true that they're inequivalent for $R=\mathbb F_p$ for all odd primes $p$ ?
I suspect this should be true : indeed suppose they were equivalent as $E_1-\Z$-algebras; then by tensoring with $\F_p$ over $\Z$ we get $\F_p \otimes_\Sph \F_p \simeq \Z\otimes_\Sph (\F_p\otimes_\Z \F_p) \simeq \Z\otimes_\Sph \F_p[\epsilon]$ where $\epsilon^2= 0$ and $|\epsilon|=1$.
Then we get, by the Thom isomorphism and using a description of $\F_p$ and $\Z$ as Thom spectra as described e.g. in A simple universal property of Thom ring spectra (Antolin-Camarena and Barthel) : $$\F_p\otimes_\Sph\Sigma^\infty_+\Omega^2S^3 \simeq \F_p[\epsilon]\otimes_\Sph \Sigma^\infty_+\Omega^2S^3\langle 3\rangle$$ as $E_1$-algebras (where $S^3\langle 3\rangle$ is the $3$-connected cover of $S^3$)
So essentially this would be saying that "removing" the degree $1$ class on the left and "adding it freely, with a zero square" wouldn't change the algebra structure, which seems rather unlikely.
For $p=2$, we know (by a description of $\pi_*(\F_2\otimes_\Sph \F_2)$) that the degree $1$ class on the left is not nilpotent, so it indeed fails badly, which gives the result I mentioned earlier.
For odd $p$, such a simple argument does not work (as the degree $1$ class on the left obviously squares to $0$), but I still suspect that the equivalence does not hold. In fact the term on the left has $\pi_*$ the dual mod $p$ Steenrod algebra so I think a careful analysis of that one could lead to a proof, but I don't know enough about it.
I'm also interested in other examples - I've been told that there might be examples of inequivalence when $R$ is a ring of algebraic integers, but unfortunately I don't know how to attack that question so :
Are there nice examples of rings of algebraic integers where the inequivalence holds ? Is there some characterization of those ?
