It's well-known that complex cobordism $MU^\ast$ is universal among complex-oriented associative, graded-commutative cohomology theories $E$. This means that if $E$ is a multiplicative cohomology cohomolgoy theory whose coefficients from an associative (unital) graded-commutative ring, then $E$ receives a map of multiplicative cohomology theories from $MU$, uniquely if we ask for compatibility with the complex orientation.
I'm curious what happens if we relax the commutativity assumption on the multiplication on $E$. I'm willing to strengthen the other hypotheses:
Question: Let $E$ be an $E_1$ ring spectrum which is complex oriented (i.e. the unit map $S^0 \to E$ extends to $\Sigma^{\infty-2} \mathbb C \mathbb P^\infty \to E$). Then does $E$ receive a map of some kind from $MU$? Can we at least deduce that $E \otimes MU \neq 0$ if $E \neq 0$?
(At least if we just want $E \otimes MU \neq 0$, the question about $E_1$ rings is equvialent to the same question about weak rings (= half-unital, not-necessarily-associative ring spectra) since we can upgrade the latter to the former via the stable version of the James construction.)
Best Answer
The initial example of such an $E$ is the Thom spectrum $M\xi$ associated to the $E_1$-map $\Omega \Sigma BU(1) \to BU$, studied by Baker and Richter in "Quasisymmetric functions from a topological point of view" and "Some properties of the Thom spectrum over loop suspension of complex projective space". The $p$-localization $M\xi_{(p)}$ is a wedge sum of suspensions of $BP$, and receives an essential map from $MU$, cf. Proposition 7.1 in the former paper. There is no map of ring spectra from $MU$ to $M\xi$, cf. Remark 4.2 in the latter paper.