# The structure of complex cobordism cohomology of the Eilenberg-Maclane spectrum

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

Is it know what the structure of the complex cobordism cohomology $$MU^{*}(H\mathbb{Z})$$ is?

EDIT: What if instead $$H\mathbb{Z}$$, one consider $$H\mathbb{Z}/(p)$$ for a prime $$p$$?

• Hrmm.. we know $MU_*H\mathbb{Z}=H\mathbb{Z}_*MU=\mathbb{Z}[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated. Nov 14 '18 at 19:23
• If I remember correctly, it is $0$, but I don't remember a reference off the head. Nov 14 '18 at 19:24

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, 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 $$\mathrm{Ext}^1_\mathbf{Z}(\mathbf{Q,Z}) \cong \widehat{\mathbf{Z}}/\mathbf{Z}$$.
• I don't think that the last bit of this is right. $MU^0(H\mathbb{Q}_p)$ is not a ring. We can write $H\mathbb{Q}=S\mathbb{Q}$ as the telescope of multiplication by $2,3,4,5,\dotsc$ on $H$ or on $S$. From the first description together with $F(H/n,MU)=0$ we get $F(H,MU)=F(H\mathbb{Q},MU)=F(S\mathbb{Q},MU)$. The second description relates $[S\mathbb{Q},MU]_*$ to $\lim^0$ and $\lim^1$ of multiplication by $2,3,4,\dotsc$ on $\pi_*(MU)$. Here $\lim^0=0$ but $\lim^1$ involves $\text{Ext}(\mathbb{Q},\mathbb{Z})=\widehat{\mathbb{Z}}/\mathbb{Z}$. Nov 14 '18 at 20:59
• Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $H\mathbb{F}_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $\mathcal{A}^*$; this is already enough to show that $MU^*(H\mathbb{F}_p) = 0$. Then the sequence $\mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$ tells you that $Y^*\mathrm{H}\mathbb{Z}$ will always be a humongous sum of $\mathrm{Ext}(\mathbb{Q}, ?)$'s if $Y^*(H\mathbb{F}_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this. Nov 14 '18 at 23:21