Let $MU$ be the complex bordism spectrum and let $H\mathbb{Z}$ be the Eilenberg-Maclane spectrum.



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> Is it know what the structure of the complex cobordism cohomology $MU^{\*}(H\mathbb{Z})$ is? 
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EDIT: What if instead $H\mathbb{Z}$, one consider $H\mathbb{Z}/(p)$ for a prime $p$?



The structure of complex cobordism cohomology of the Eilenberg-Maclane spectrum
One can prove that $\mathrm{Map}(H\mathbf{F}\_p,MU)$ is contractible. We know that $H\mathbf{F}\_p$ 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 $E\_p = \bigvee\_{0\leq n<\infty} K(n)$, the claim follows. (I just realized that this is Corollary 4.10 of Ravenel's paper.)


It is, however, not the case that $\mathrm{Map}(H\mathbf{Z},MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $H\mathbf{Z}$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $\mathrm{Map}(H\mathbf{Z},MU)$ and $\mathrm{Map}(L\_E H\mathbf{Z},MU)$. We therefore need to understand $L\_E H\mathbf{Z}$. By the discussion at [this question](https://mathoverflow.net/questions/220952/localization-at-the-johnson-wilson-spectrum-and-rationalization), we can conclude that $L\_{E\_p} H\mathbf{Z} \simeq H\mathbf{Q}\_p$. It therefore suffices to understand $MU^\ast(H\mathbf{Q})$. But $H\mathbf{Q}$ is the colimit of multiplication by $2,3,5,7,\cdots$ on the sphere, so $MU^\ast(H\mathbf{Q})$ 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 $\mathrm{Ext}^1\_\mathbf{Z}(\mathbf{Q,Z}) \cong \widehat{\mathbf{Z}}/\mathbf{Z}$.


