Consider the endomorphism ring spectrum $R = \mathrm{End}\_S(H\mathbb{F}\_p)$ of the mod $p$ Eilenberg-MacLane spectrum $H\mathbb{F}\_p$. The homotopy groups of $R$ are the Steenrod algebra $A^\*$ with reversed grading:
$$\pi\_n R = [\Sigma^n H\mathbb{F}\_p, H\mathbb{F}\_p] = A^{-n}.$$
This spectrum $R$ is an associative $S$-algebra (or $A\_{\infty}$ ring spectrum). Moreover, $R$ is an $H\mathbb{F}\_p$-module spectrum, say, using the $H\mathbb{F}\_p$-module structure on the target $H\mathbb{F}\_p$. In particular, $R$ is an $H\mathbb{Z}$-module spectrum. However, $R$ is known *not* to be an $H\mathbb{F}\_p$-algebra spectrum.



> 
> **Question**. Is $R = \mathrm{End}\_S(H\mathbb{F}\_p)$ an $H\mathbb{Z}$-algebra spectrum?
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My hunch is that the answer is no, but I couldn't find that statement in the literature. Perhaps it can be shown using an invariant of structured ring spectra, some flavor of $THH$. Or perhaps a dg-algebra over $\mathbb{Z}$ doesn't have enough room to encode the homotopical structure of the Steenrod algebra [1].


**Remark**. For the sake of definiteness, feel free to pick a model of spectra such as $S$-modules or symmetric spectra. The question is meant to be about the underlying symmetric monoidal $\infty$-category. In light of [2], working with your favorite model of spectra should be fine.


[1] *Shipley, Brooke*, [**$H\Bbb Z$-algebra spectra are differential graded algebras**](http://dx.doi.org/10.1353/ajm.2007.0014), Am. J. Math. 129, No. 2, 351-379 (2007). [ZBL1120.55007](https://zbmath.org/?q=an:1120.55007).


[2] *Mandell, M.A.; May, J.P.; Schwede, S.; Shipley, B.*, [**Model categories of diagram spectra**](http://dx.doi.org/10.1112/S0024611501012692), Proc. Lond. Math. Soc., III. Ser. 82, No.2, 441-512 (2001). [ZBL1017.55004](https://zbmath.org/?q=an:1017.55004).



Endomorphism ring spectrum of the Eilenberg-MacLane spectrum
No, $A$ is not an $H\Bbb Z$-algebra.


Suppose $R$ is an $H\Bbb Z$-algebra. Then the category of left $R$-modules is $H\Bbb Z$-linear: for any $R$-modules $M$ and $N$, the function spectrum $F\_R(M,N)$ naturally has the structure of an $H\Bbb Z$-module. One reason for this is that $R$ is now an algebra object in the symmetric monoidal closed category of $H\Bbb Z$-modules, and the internal $R$-module function objects can be given weakly equivalent constructions *there* instead of in the category of $\Bbb S$-modules. In particular, the unit $\Bbb S \to F\_R(M,M)$ of the endomorphism ring factors through $\Bbb S \to H\Bbb Z \to F\_R(M,M)$ because the endomorphism ring is now an algebra in $H\Bbb Z$-modules.


However, if we take $R$ to be the Steenrod algebra spectrum and $M = H\Bbb F\_p$ with the action it has by definition, there is an equivalence of the $R$-linear endomorphisms of $M$ with the $p$-completed sphere:
$$
F\_A(H\Bbb F\_p, H\Bbb F\_p) \simeq \Bbb S^\wedge\_p
$$
(This, for example, is what gives rise to the Adams spectral sequence.) In particular, the unit $\Bbb S \to \Bbb S^\wedge\_p$ doesn't factor through $H\Bbb Z$ for any prime.


I think that many people find this pretty surprising when they first encounter it; I certainly did.


