Really blown away by the following calculation of the homology of complex K theory. It's a really unexpected demonstration of how even basic chromatic technology can just obliterate certain calculational problems..

1. We want to compute the homotopy groups of \(H\mathbb{Z} \otimes \mathrm{KU}\)

2. \(H\mathbb{Z} \otimes \mathrm{KU}\) carries two complex orientations; one from \(H\mathbb{Z}\) and one from \(\mathrm{KU}\).

3. The associated formal group laws are therefore isomorphic. Since these are the additive and multiplicative formal group laws, we must have that the associated graded ring is rational.

4. This tells us that \(H\mathbb{Z} \otimes \mathrm{KU}\) is rational, so we may tensor with \(H\mathbb{Q}\) and proceed. This immediately kills that \(H\mathbb{Z}\), since that's rationally equivalent to the sphere.

5. Finally, Chern character gives us a rational equivalence \(\mathrm{KU} \to \prod_i \Sigma^{2i}H\mathbb{Q}\).

All told, we conclude that the homology of \(\mathrm{KU}\) is \(\mathbb{Q}\) in even dimensions and vanishes in odd dimensions with absolutely no messing about with Bott periodicity or calculations with decomposables (as is done in the classical proof).