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.