If true this should probably be a consequence of one of Ravanel/Wilson's hopf ring papers. (Suppose $p>2$), Is it true that the cup product $BP_{\ast}(B \mathbb{Q}_p / \mathbb{Z}_p)^{\boxtimes k} \to BP_{\ast}(B^k \mathbb{Q}_p /\mathbb{Z}_p)$ exhibits $BP_{\ast}(B^k \mathbb{Q}_p /\mathbb{Z}_p)$ as the $k$-th exterior power (for the box tensor product corepresenting bilinear maps of hopf algebras) of the hopf algebra $BP_{\ast}(B \mathbb{Q}_p/\mathbb{Z}_p)$ ?

Persumably once the "exterior power" and the cup product maps are made sense of this could be reduced to the standard computation in morava k theory by a combination of algebraic and landweber flatness arguments?

Has anyone thought about this before?

@SaalHardali the most that’s known about this is in R-W-Yagita’s Brown Peterson cohomology from Morava K-theory (and a tiny bit at the very end of the R-W … Conner-Floyd conjecture paper). it’s not easily readable: there’s a big difference between homology and cohomology, and they also p-complete as it suits them

my memory is that it does not work at all how you’d like, but i don’t recall the argument exactly. the place to start, iirc, is to note that in order for K(Z, 3) to be K(1)-acyclic and not K(2)-acyclic, there must be big v1-divisible towers in its E(2)-homology, and from that everything begins to fall apart.

something else to note: even invoking technology like the box product, alternating powers of hopf algebras need not exist, even for hopf algebras as nice as formal groups of finite height and dimension > 1. that counterexample isn’t directly relevant here, but the general warning may well be