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Given an $\mathbb E_\infty$-ring (highly structured commutative ring spectrum if you want) $R$, we have the free $R$-algebra (on one generation) $R\{t\}\simeq \bigoplus_{n\ge 0} R_{\mathrm h\Sigma_n}$ vs. the polynomial $R$-algebra $R[t]\simeq \bigoplus_{n\ge 0} R.$ It is well-known that $R\{t\}\simeq R[t]$ when $R$ is rational, i.e. a $H\mathbb Q$-algebra. It is also very easy to verify that this is not the case in many other cases, for instance when $R= H\mathbb Z$ and $H\mathbb F_p$.

Q: Does $R\{t\}\simeq R[t]$ for a (connective?) $\mathbb E_\infty$-ring imply that $R$ is a $H\mathbb Q$-algebra?

In case not, I would be very happy with some example of non-rational ring spectra $R$ for which polynomial and free $R$-algebras agree. Also, in that case, what condition on an $\mathbb E_\infty$-ring $R$ does the condition that $R\{t\}\simeq R[t]$ imply?

In the special case of a discrete ring $R$ = Eilenberg-MacLane $\mathbb E_\infty$-ring $HR,$ this question reduces to one about group cohomology of symmetric groups. More precisely:

Q': Does the condition that $\mathrm H^i(\Sigma_n, R)\simeq 0$ for all $n$ and all $i\ge 1$ for a commutative ring $R$ imply that $R$ is a $\mathbb Q$-algebra?

This sounds like a possibly-very-easy piece of classical algebra, but to my shame, I do not know the answer even in this particular case.

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    $\begingroup$ Have you looked at the paper of Birgit Richter on the AHSS for TAQ? In the case of discrete rings she identifies $H_*(Hk;\mathbb{Z})$ as a key object which measures the difference between a free resolution in commutative $k$-algebras and $Hk$-algebras. I think this might be relevant to your question. In some sense you are asking about this spectral sequence that Richter develops. $\endgroup$ – Sean Tilson Apr 23 '17 at 18:40
  • $\begingroup$ Thanks, I was not aware of the paper and will look into it. $\endgroup$ – A Rock and a Hard Place Apr 23 '17 at 22:27
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    $\begingroup$ At least if $R$ is connective, the canonical map $R\{t\} \to R[t]$ is invertible iff $\pi_0(R)$ is a $\mathbf{Q}$-algebra. Since the map is compatible with extensions of scalars, it suffices to assume $R$ is discrete. Consider any residue field $R \to k$: if $k$ is of characteristic $p > 0$ then $k\{t\} \to k[t]$ cannot be invertible (since $\mathbf{F}_p\{t\} \to k\{t\}$ is faithfully flat, and you know $\mathbf{F}_p\{t\} \to \mathbf{F}_p[t]$ is not invertible), which means that $k$ must be of characteristic zero. Thus $R$ is purely of characteristic zero, i.e. a $\mathbf{Q}$-algebra. $\endgroup$ – AAK Apr 25 '17 at 15:09
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    $\begingroup$ Actually, the non-connective case then follows by taking connective covers (note that the map $R\{t\} \to R[t]$ is also compatible with formation of connective covers). $\endgroup$ – AAK Apr 25 '17 at 15:14
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    $\begingroup$ Dear Adeel, after writing and deleting some stupid comments revealing my own sheer stupidity, I have come to realize that you really have answered my question just about as completely as I could possibly have hoped for. I may have formulated the question in terms of equivalences of $\mathbb E-\infty$-rings, but surely if the canonical map isn't an equivalence, that is already a sufficient amount of pathology. If you want to re-post your comments in the form of an answer, I will gladly accept it. $\endgroup$ – A Rock and a Hard Place Apr 27 '17 at 3:32
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Any morphism of $R$-algebras $\varphi : R\{t\} \to R[t]$ is determined up to homotopy by an element of $\pi_0(R[t]) \approx \pi_0(R)[t]$. If $\varphi$ is an equivalence, then this element must be the generator $t$, so we may as well assume $\varphi$ is the canonical map $\varepsilon_R : R\{t\} \to R[t]$.

Claim: Let $R$ be an $E_\infty$-ring spectrum. Then the map $\varepsilon_R$ is invertible if and only if $R$ is a $\mathbf{Q}$-algebra.

Useful observations:

1) The map $\varepsilon_R$ is compatible with extensions of scalars. Therefore if $\varepsilon_R$ is invertible, then $\varepsilon_{R'}$ is invertible for any $R$-algebra $R'$.

2) $R$ is a $\mathbf{Q}$-algebra if and only if $\pi_0(R)$ is a $\mathbf{Q}$-algebra (see comments).

Sufficiency: Suppose that $R$ is a $\mathbf{Q}$-algebra. The map $\varepsilon_R$ is obtained by extension of scalars along the connective cover $R_{\ge 0} \to R$, so we may assume $R$ is connective. The $\infty$-category of connective $E_\infty$-algebras over $\mathbf{Q}$ is equivalent to that of simplicial commutative $\mathbf{Q}$-algebras. Under this equivalence, the map $\varepsilon_R$ corresponds to the identity morphism $R[t] \to R[t]$ (where by abuse of notation $R$ also denotes the corresponding simplicial commutative $\mathbf{Q}$-algebra).

Necessity: Suppose $\varepsilon_R$ is an equivalence. Since the map $\varepsilon_R$ is compatible with formation of connective covers, we may replace $R$ by its connective cover. We may also extend scalars along $R \to \pi_0(R)$ to assume $R$ is discrete.

Consider any residue field $R \to k$. If $k$ is of characteristic $p>0$, then $\varepsilon_k$ cannot be invertible. Indeed, it is well-known that $\varepsilon_{\mathbf{F}_p}$ is not invertible, and $\varepsilon_k$ is the extension of scalars along the faithfully flat map $\mathbf{F}_p\{t\} \to k\{t\}$. Thus every residue field $k$ must be of characteristic zero, so $R$ is a $\mathbf{Q}$-algebra.

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  • $\begingroup$ Wonderful, thank you very much! Does $\pi_0R$ being a $\mathbb Q$-algebra suffice for $R$ (maybe connective?) to be a $\mathbb Q$-algebra, as an $\mathbb E_\infty$-ring? Is that something that has to do with the motto: "the rational sphere is the rational Eilenberg-MacLane"? $\endgroup$ – A Rock and a Hard Place Apr 28 '17 at 0:29
  • $\begingroup$ I do think that should be the case: if $\pi_0R$ is a $\mathbb Q$-algebra, then all $\pi_nR$ are rational vector spaces, so $R$ is $\mathbb Q$-local. Since rationalization is a smashing equivalence and $L_\mathbb QS\simeq \mathbb Q,$ then $R\simeq L_\mathbb QR\simeq \mathbb Q\otimes R$ which is clearly an $\mathbb E_\infty$-$\mathbb Q$-algebra. $\endgroup$ – A Rock and a Hard Place Apr 28 '17 at 5:01
  • $\begingroup$ Ah, good point. That simplifies the proof a bit as well, thanks! $\endgroup$ – AAK Apr 28 '17 at 11:25

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