TITLE: Is a wedge of spheres an $E_\infty$ ring spectrum? 
QUESTION [6 upvotes]: The wedge sum $\bigvee_{k \in 2 \mathbb{Z}} S^{k}$ is an $A_\infty$-ring spectrum: the connective cover is the free $A_\infty$-ring on the sphere $S^2$, if I'm not mistaken, and then one inverts the element in $\pi_2$. It is also homotopy commutative. Can it be made into an $E_\infty$-ring spectrum? 
More generally, given an $E_\infty$-ring spectrum $R$, when can $\bigvee_{k \in 2 \mathbb{Z}} \Sigma^k E$  be made into an $E_\infty$-ring?

REPLY [14 votes]: Ah, tracked it down.  Here is an argument.  I should mention that Peter once pointed me towards an original source due to McClure, or page 238, Prop. 6.1, of SLN 1176 (the $H_\infty$ book).
Suppose you had such a ring object $R$.  We examine its mod-2 homology.  This has several features:

It is the ring $\mathbb Z/2[t^{\pm 1}]$.
It has trivial action of the Steenrod operations $P_r$ (this is dual to the cohomology action), because it's a wedge of spheres.
From the $H_\infty$ book, it has Dyer-Lashof operations $Q^s$.  These satisfy $Q^{|x|} x = x^2$, and the Nishida relations
$$
P_r Q^s = \sum \binom{s - r}{r - 2i} Q^{s-r+i}P_i.
$$

In particular, these together would say 
$$
0 = P_2 Q^4 t = \binom{2}{2} Q^{2}P_0 t + \binom{2}{0} Q^{4}P_2 t = Q^2 t = t^2.
$$
If $E$ is a commutative $MU$-algebra, then you can use the map from $MU$ to its periodic version $MUP$ to produce $R \wedge_{MU}MUP$, which is 2-periodified.  Unfortunately, we know very few genuine $MU$-algebras.  Barry Walker proved that complex K-theory is one of them, based on Matthew Ando's work on studying $H_\infty$ structures on Lubin-Tate cohomology theories.  My understanding is that the problem is still open for almost all of the Lubin-Tate cohomology theories.