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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?

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, Am. J. Math. 129, No. 2, 351-379 (2007). ZBL1120.55007.

[2] Mandell, M.A.; May, J.P.; Schwede, S.; Shipley, B., Model categories of diagram spectra, Proc. Lond. Math. Soc., III. Ser. 82, No.2, 441-512 (2001). ZBL1017.55004.

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    $\begingroup$ I am a bit confused: since $H\mathbb{F}_p$ is a $H\mathbb{F}_p$-module spectrum, then there is a map of $E_1$-algebras $H\mathbb{F}_p\to \mathrm{End}(H\mathbb{F}_p)$. $\endgroup$ – Denis Nardin May 8 '17 at 17:07
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    $\begingroup$ Precisely. In fact for any $E_1$-ring spectrum $R$, an $R$-module structure on $M$ is the same thing as an $E_1$-ring map $R\to \mathrm{End}(M)$ (if I recall correctly this is how Lurie defines the $E_1$-structure on $\mathrm{End}(M)$). $\endgroup$ – Denis Nardin May 8 '17 at 17:20
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    $\begingroup$ I am afraid I do not understand. I am claiming $\mathrm{End}(H\mathbb{F}_p)$ is, in a canonical way, an $H\mathbb{F}_p$-algebra (and more generally that $\mathrm{End}_S(R)$ has a canonical $R$-algebra structure), so I don't understand how can $\mathrm{End}(H\mathbb{F}_p)$ be "known" not to be an algebra over $H\mathbb{F}_p$. What am I missing? $\endgroup$ – Denis Nardin May 8 '17 at 17:28
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    $\begingroup$ When you have a ring map $k\to R$, do you say that $R$ is an $k$-algebra? I think that most of us do not, unless $k$ is commutative and its image is in the center of $R$. In other words, by a $k$-algebra (for a commutative ring $k$) we mean a $k$-module equipped with a $k$-bilinear multiplication. At least I am guessing that Martin would say this, and that that's what Denis needs to know to clear up the confusion. $\endgroup$ – Tom Goodwillie May 8 '17 at 17:36
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    $\begingroup$ For $p=2$, one version of the statement can be found in [H.J. Baues, The algebra of secondary cohomology operations; Theorem 4.6.5], though phrased in a different language. $\endgroup$ – Martin Frankland May 8 '17 at 18:20
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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.

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    $\begingroup$ Here's a follow-up question. The displayed equivalence seems to rely on the following fact. Denote the Steenrod algebra spectrum $A = \mathrm{End}_S(H\mathbb{F}_p)$. Taking mod $p$ cohomology is the functor sending a spectrum $X$ to the left $A$-module $F_S(X,H\mathbb{F}_p) = H\mathbb{F}_p^X$. If $X$ and $Y$ are finite spectra, then the map of spectra $F_S(X,Y) \to F_{A}(H\mathbb{F}_p^Y, H\mathbb{F}_p^X)$ is a $p$-completion. I see that this is more or less the convergence of the Adams spectral sequence, but do you know a reference that states it this way? $\endgroup$ – Martin Frankland May 8 '17 at 22:53
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    $\begingroup$ @Martin: Good question. The homological version of this is more popular because of continuity issues. It appears like this in Baker-Lazarev's "On the Adams spectral sequence for R-modules" and Carlsson's "Derived completions in stable homotopy theory". Dwyer-Greenlees-Iyengar also mention this perspective in section 5 of their duality paper. $\endgroup$ – Tyler Lawson May 8 '17 at 23:04
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    $\begingroup$ Tyler's answer also shows that $H\mathbb F_p$ is not an $MU$--algebra, which is a bit less intuitive, as the Steenrod algebra arises when one considers the endomorphism ring of the additive formal group. $\endgroup$ – Nicholas Kuhn May 10 '17 at 10:42
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    $\begingroup$ I am confused, I thought $\mathrm{H}\mathbb{F}_p$ was an $MU$-algebra. I even know how to construct a map of $E_{\infty}$-ring spectra from $MU$ to $\mathrm{H}\mathbb{F}_p$. Maybe you mean the "spectral" Steenrod algebra isn't an $MU$-algebra? $\endgroup$ – Sean Tilson May 10 '17 at 12:30
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    $\begingroup$ Right. That's what I meant to mean. $\endgroup$ – Nicholas Kuhn May 10 '17 at 19:10

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