The homology algebra $H\_\*( K(\mathbb{Z},2n); \mathbb{Z})$ contains a 
divided polynomial algebra on a generator $x$ of dimension $2n$.  

I suppose I could read through the Cartan seminar for a proof, but I'm hoping
someone knows of a nice simple argument for this fact.



Good reference for homology of $K(\mathbb{Z}, 2n)$?
We need to do three things: (1) show that the homology contains a polynomial algebra, (2) show that powers of the generator are sufficiently divisible, and (3) show that torsion doesn't interfere.


Let me repeat Hatcher's path through (1), since we need the particular generator to show that it is divisible. According to Dold-Thom (Hatcher 4.K; cf Dold-Kan1), a model for $K(Z,m)$ is the free commutative monoid on pointed2 $S^m$. According to James (Hatcher 4.J), the homology of the free associative monoid $JS^m$ on $S^m$ is a polynomial algebra on one generator in degree $m$. Thus there is a map of monoids $JS^m\to K(Z,m)$ so its map on homology is compatible with the ring structure. It is easy to see (eg, by the Serre spectral sequence) that the rational homology of $K(Z,m)$ is the signed-commutative algebra free on a generator in degree $m$, so this map is a rational isomorphism if $m=2n$.


Now for step (2), where we divide the James class by $k!$. If we restrict to words of length $k$ in these monoids, the affect of imposing commutativity is to quotient by the $k$th symmetric group. This piece of the James construction is a pseudo-manifold representing the class in degree $nk$ which is the $k$th power of the generator. The action of the symmetric group is generically free, so this cycle has multiplicity $k!$ when it lands in $K(Z,2n)$. Thus it is divisible. In other words, choose a fundamental domain for the action of the symmetric group on the $k$th filtration of the James construction. This yields is an $kn$-chain which when symmetrized is the whole of the $k$th filtration, and thus a cycle. Projecting it to the quotient is an alternative way of symmetrizing, so it is a cycle for $K(Z,2n)$.


I will leave step (3) as an exercise. What this shows so far is that the integral homology hits the divided power algebra in the rational homology, but the canonical choice of the divided element should make it plausible that these divisions are the right choices, ie, $x^{(k)}x^{(l)}=\binom {k+l}k x^{(k+l)}$.


1If one accepts the fragment of Dold-Kan of the agreement of the two things one can do to a simplicial abelian group (1) homology of (either) associated chain complex and (2) the homotopy groups of the geometric realization; then a version of Dold-Thom follows trivially. Indeed, the very definition of simplicial homology is the free abelian group (ie, simplicial group) on the simplicial set.


2 by "the free (commutative) monoid on a pointed set," I mean that we impose on the free object the relation that the basepoint is the identity for the monoid.


