Properties of coefficients of ring spectra

Question

This is an awkwardly backwards question, but bear with me here: Suppose I have a graded ring $R$ with unit, which has an invertible element $u$ in degree $2$. The multiplicative formal group law $f(x,y) = x + y + u\,x\,y$ yields a homomorphism $MU_* \to R$. Suppose that I know the following things

• The functor $X \mapsto MU^*(X) \otimes_{MU_*} R$ is a multiplicative generalized cohomology theory (in particular, it is exact).
• The above functor is represented by an $E_{\infty}$-ring spectrum.

What properties can I deduce from this about the ring $R$?

For example: Is it true that $R$ is torsion-free (in each degree)? I vaguely seem to remember that $K$-theory with mod p-coefficients does not have an $E_{\infty}$-ring structure, so this seems at least plausible. Does $R_0$ have to be a subring of $\mathbb{Q}$ if I throw into the mix that $R_0$ is countable?

Answer 1

You have given yourself an invertible element $u\in\pi_2(R)$ and a coordinate $x\in R^2(\mathbb{C}P^\infty)$ with $\psi(x)=x\otimes 1 + 1\otimes x + ux\otimes x$. This means that the class $m=1+ux\in R^0(\mathbb{C}P^\infty)$ satisfies $\psi(m)=m\otimes m$. In other words, $m$ can be regarded as a map of ring spectra from the ring spectrum $T=\Sigma^\infty_+\mathbb{C}P^\infty$ to $R$. Snaith defined an element $v\in \pi_2(T)$ and proved that $T[v^{-1}]$ is equivalent to $K$. You are also implicitly assuming that $x$ restricts to the standard generator of $\widetilde{R}^2(\mathbb{C}P^1)\simeq R^0(\text{point})$. I think that this means that $m_*(v)=u$, which is invertible, so $m$ induces a ring map $K=T[v^{-1}]\to R$. Under plausible additional assumptions, this map $K \to R$ will in fact be $E_\infty$. Thus, $R$ will always be a $K$-algebra, and probably an $E_\infty$ $K$-algebra. The non-$E_\infty$ version could also be extracted from the standard Landweber exactness technology, at least up to phantoms.

If we have an $E_\infty$ map $K\to R$ and $R$ is $p$-complete then we can check that $R^0(B\Sigma_p)/\text{tr}(1)$ is isomorphic to $R^0$, and we can use this to define a power operation $\psi^p$ on $R^0(X)$ for all spaces $X$, which is a ring map satisfying $\psi^p(t)=t^p\pmod{p}$. The existence of $\psi^p$ excludes many possibilities for $R^0$ (or the $p$-completion of $R^0$, if $R^0$ is not already $p$-complete). For example, the ring $A=\mathbb{Z}[e^{2\pi i/p}]$ does not admit a map $\psi^p$ of the required type.