The $K$-theory homology of the Eilenberg-MacLane spectrum

Question

Let $KU$ be the complex $K$-theory spectrum and $H\mathbb{Z}$ be the Eilenberg-MacLane spectrum.

For $n\in \mathbb{Z}$, it is known what the homology groups $KU_{n}(H\mathbb{Z})$ are?

Answer 1

Since for any two spectra $E,F$ we have $$ E_n(F)=[\mathbb{S}^n,E\wedge F]\cong [\mathbb{S}^n,F\wedge E] = F_n(E), $$ you may as well ask about $H_n(KU;\mathbb{Z})$, the integral homology of the complex $K$-theory spectrum.

This calculation is carried out e.g. in Chapter 16 of the book

Switzer, Robert M., Algebraic topology – homology and homotopy., Classics in Mathematics. Berlin: Springer. xii, 526 p. (2002). ZBL1003.55002.

In particular, Theorem 16.25 gives the Pontrjagin algebra structure as $$ H_*(KU;\mathbb{Z})\cong \mathbb{Q}[u,u^{-1}],$$ the ring of finite Laurent series over $\mathbb{Q}$ in an element $u\in H_2(KU;\mathbb{Z})$.

Answer 2

We have $KU_*(H\mathbb Z) = \pi_*(KU\wedge H\mathbb Z)$; this ring is concentrated in even dimensions and carries an isomorphism between the additive and multiplicative formal group law, hence is rational. Thus the map $$ KU\wedge H\mathbb Z\to (KU\wedge H\mathbb Z)\wedge H\mathbb Q\cong KU\wedge H\mathbb Q $$ is an isomorphism. The Chern character $KU\to \sum_{i\in\mathbb Z} \Sigma^{2i}H\mathbb Q$ is a rational equivalence, so that $$ KU_{2n}(H\mathbb Z) \cong\mathbb Q, KU_{2n+1}(H\mathbb Z) \cong 0\ . $$

Alternatively, the Snaith theorem $KU\cong \Sigma^{\infty}\mathbb{CP}^{\infty}[\beta^{-1}]$, where $\beta\in \pi_2(\mathbb{CP}^{\infty})$ is the Bott element, shows that $$ H_*(KU)\cong H_*(\mathbb{CP}^{\infty})[\beta^{-1}]\ . $$ It's easy to calculate that $H_*(\mathbb{CP}^{\infty})$ is a divided power algebra on the generator $\beta\in H_2(\mathbb{CP}^{\infty})$. Since $\beta^n$ is divisible by $n!$, this ring is the Laurent series over $\mathbb Q$ generated by $\beta$.