Contents

Idea

The Lazard ring is a commutative ring which is

and by Quillen's theorem also

Definition

The Lazard ring may be presented by generators $a_{i j}$ with $i,j \in \mathbb{N}$

$L = \mathbb{Z}[a_{i j}] / (relations\;1,2,3\;below)$

and relations as follows

1. $a_{i j} = a_{j i}$

2. $a_{10} = a_{01} = 1$; $\forall i \neq 1: a_{i 0} = 0$

3. the obvious associativity relations imposed by $\ell(x, \ell(y, z)) = \ell(\ell(x, y), z)$

where we write

$\ell(x,y) = \sum_{i,j} a_{i j} x^i y^j \in L[[x,y]].$

In other words, $\ell(x, y)$ becomes the universal 1-dimensional formal group law as a formal power series in two variables with coefficients in the Lazard ring, Theorem below.

Properties

As classifying ring for formal group laws

Theorem

For any ring $S$ with formal group law $g(x,y) \in S[ [x,y] ]$ there is a unique ring homomorphism $L \to S$ that sends $\ell$ to $g$.

review includes (Hopkins 99, theorem 2.3, theorem 2.5)

Remark

Passing to formal dual ring spectra, this says that $Spec(L)$ is something like the moduli space for formal groups. By Quillen's theorem on MU, the lift of $L$ to higher algebra is the E-infinity ring MU and the E-infinity ring spectrum $Spec(MU)$ is something like the derived moduli stack for formal group laws.

Lazard’s theorem

Lazard's theorem states:

Theorem

The Lazard ring is isomorphic to a graded polynomial ring

$L \simeq \mathbb{Z}[t_1, t_2, \cdots]$

with the variable $t_i$ in degree $2 i$.

review includes (Hopkins 99, theorem 2.5, Lurie 10, lect 2, theorem 4)

As the complex cobordism cohomology ring

By Quillen's theorem on MU the Lazard ring is the cohomology ring of complex cobordism cohomology theory.

Theorem

Let $M P$ denote the peridodic complex cobordism cohomology theory. Its cohomology ring $M P(*)$ over the point together with its formal group law is naturally isomorphic to the universal Lazard ring with its formal group law $(L,\ell)$.

Remark

This can be used to make a cohomology theory out of a formal group law $(R,f(x,y))$. Namely, one can use the classifying map $M P({*}) \to R$ to build the tensor product

$E^n(X) := M P^n(X) \otimes_{M P({*})} R,$

for any $n\in\mathbb{Z}$. This construction could however break the left exactness condition. However, $E$ built this way will be left exact if the ring morphism $M P({*}) \to R$ is a flat morphism. This is the Landweber exactness condition (or maybe slightly stronger). See at Landweber exact functor theorem.