Quick Background: The $p$-series of $F$ (where $F$ is a formal group law over a graded ring $R$) will be of the form $[p](x) = px + v_1x^{p^1} + ... + v_nx^{p^n} + ...$ ; $(F, R)$ is Landweber-exact if the sequence $(p, v_1, ..., v_n)$ is $R$-regular for all $p$ and $n$.

Recall the the Landweber-Ravenel-Stong Construction: $MU^*(X) \otimes_{L} R \simeq E^*(X)$, where $MU^* \simeq L$ and $R \simeq E^*(pt)$.

The Landweber-exact functor theorem states that if $(F, R)$ is Landweber-exact, then $E^*(X)$ will satisfy the generalized Eilenberg-Steenrod axioms.

An object $M$ in an abelian tensor category $C$ is 'flat' if for all $X \in Obj(C)$ , the functor $X \to X \otimes M$ preserves exact sequences.

What confuses me is the statement that requiring $L \to R$ to be Landweber-exact is a weaker condition than flat.

If we are requiring the functor $- \otimes_L R: MU^*(X) \to E^*(X)$ preserves exact sequences, then wouldn't it automatically be flat?

What is an example of a formal group law that is Landweber-exact but not flat?

  • 2
    $\begingroup$ Landweber exact means tensoring with it preserves exactness for comodules over the Hopf algebroid $(MU_*,MU_*MU)$(Which is automatically satisfied when it is $MU_*(X)$ for some spectrum $X$), which is a subclass of modules over the Lazard's ring. So Landweber-exact is weaker than flat. But I could not come up with any quick example in my mind. $\endgroup$ Apr 10, 2015 at 5:39
  • 6
    $\begingroup$ $\Bbb Q$ is Landweber flat, but certainly not flat over $L$. As Mingcong and Drew point out, what Landweber actually proved was that any $MU_* MU$-comodule is built out of a very restricted family of $L$-modules, and so you only need to test flatness against those basic building blocks rather than a general $L$-module. $\endgroup$ Apr 10, 2015 at 15:07
  • 4
    $\begingroup$ If $R$ is flat over $MU_*$, then $R$ has to be $MU_*$-torsion free and thus contains a copy $MU_*\cdot 1$ of $MU_*$. This cannot be true for any ring $R$ of finite dimension. So virtually all examples you can think of are not flat over $MU_*$. E.g. rational homology, K-theory, elliptic homology, Johnson--Wilson theories $E(n)$... $\endgroup$ Apr 10, 2015 at 16:31

1 Answer 1


Consider the functor that sends $X$ to $MU_*(X) \otimes_{MU_*} R$. If $R$ is a flat $MU_*$-module then this defines a homology theory. Note that the condition that $R$ is flat over $MU_*$ is equivalent to requiring that $\operatorname{Tor}_1^{MU_*}(-,R) = 0$. But we can get away with something weaker; namely we only require that $\operatorname{Tor}_1^{MU_*}(MU_*X,R) = 0$ for $X$ a finite complex. It is this requirement that leads (after some work) to Landweber's criterion.

Since you are interested in elliptic cohomology you can take the original Landweber-Ravenel-Stong construction (see Section 4), as an example of a cohomology theory arising from a ring that is not flat over $MU_*$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.