Pfthb

Let $MU$ be the complex bordism spectrum and let $HmathbbZ$ be the Eilenberg-Maclane spectrum.

Is it know what the structure of the complex cobordism cohomology $MU^*(HmathbbZ)$ is?

EDIT: What if instead $HmathbbZ$, one consider $HmathbbZ/(p)$ for a prime $p$?

One can prove that $mathrmMap(HmathbfF_p,MU)$ is contractible. We know that $HmathbfF_p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories"), but $MU$ is harmonic (Theorem 4.2 of that paper). Since dissonant spectra (resp. harmonic spectra) are by definition $E$-acyclic (resp. local) for the spectrum $E = bigvee_p E_p$, where $E_p = bigvee_0leq n<infty K(n)$, the claim follows. (I just realized that this is Corollary 4.10 of Ravenel's paper.)

It is, however, not the case that $mathrmMap(HmathbfZ,MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $HmathbfZ$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $mathrmMap(HmathbfZ,MU)$ and $mathrmMap(L_E HmathbfZ,MU)$. We therefore need to understand $L_E HmathbfZ$. By the discussion at this question, we can conclude that $L_E_p HmathbfZ simeq HmathbfQ_p$. It therefore suffices to understand $MU^ast(HmathbfQ)$. But $HmathbfQ$ is the colimit of multiplication by $2,3,5,7,cdots$ on the sphere, so $MU^ast(HmathbfQ)$ admits a description in terms of $lim^0$ and $lim^1$ of multiplication by $2,3,5,7,cdots$ on $pi_ast MU$. In particular, the $lim^1$ term is $mathrmExt^1_mathbfZ(mathbfQ,Z) cong widehatmathbfZ/mathbfZ$.