Skip to main content
2 of 2
Kunneth -> Künneth, and other tidying
LSpice
  • 12.9k
  • 4
  • 45
  • 69

If $A_\ast$ has a Künneth theorem, then is $A$ a module over Morava $K$-theory?

$\newcommand\Spt{\mathit{Spt}}\newcommand\GrAb{\mathit{GrAb}}$Let $A$ be a ring spectrum. Suppose that $A$ has a Künneth theorem — i.e. the homology theory $A_\ast : \Spt \to \GrAb$ is a strong monoidal functor [1].

Question: Does it follow that $A$ is a module over Morava $K$-theory $K(h)$ for some prime $p$ and some $0 \leq h \leq \infty$?

An affirmative answer would be a variation on the theorem that every ring spectrum which is a field is a module over some $K(h)$. See e.g. Lurie's notes Uniqueness of Morava $K$-theory.

[1] I have learned here that it's not so straightforward to say exactly what it means to "have a Künneth theorem" — right now I'm just assuming that there is some way to make $A_\ast$ into a strong monoidal functor, but it seems in general there need not a canonical way to make $A_\ast$ into a lax monoidal functor, unless $A$ is homotopy commutative. I think for a start, I'd be happy with an answer which assumes that $A$ is homotopy commutative, and assumes that the strong monoidal structure comes from the lax monoidal structure which exists canonically in this case.

Tim Campion
  • 64k
  • 13
  • 143
  • 384