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.



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



Properties of coefficients of ring spectra
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.


