Lazard’s ring and Height

1 Lazard’s ring and consequences

Here a formal group law over a ring \(R\) means a one-dimensional commutative formal group law over \(R\). Lazard’s ring, denoted \(L\), is the ring with the universal formal group law, i.e. it represents the functor \(FGL\) sending a ring to the set of all formal group laws. We are interested in it because in the connection between formal group laws and complex oriented theories, \(L\) is \(MU_*\), and the universal formal group law is exactly the one on \(MU\). Lazard’s theorem states that \(L\) is a polynomial ring \(\ZZ [t_1,t_2,\dots ]\) on infinitely many generators.

One way to produce formal group laws is to change coordinates: i.e. one can take a formal group law \(f(x,y)\), and an invertible power series \(g(t) = t+b_1t^2+b_2t^3+\dots \), \(g(f(g^{-1}(x),g^{-1}(y)))\) is a formal group law. The universal way of doing this produces a formal group law over the ring \(C=\ZZ [b_1,b_2,\dots ]\), i.e. a map \(L \to C\). In fact, \(C\) with its formal group law is \(H_*(MU)\), and the map from \(L\) is the Hurewicz map!

An important observation is that \(L\) and \(C\) come with a natural grading, so that \(L \to C\) is a graded map. Namely, \(L\) also represents formal group laws on \(\ZZ \)-graded rings. A formal group law on a \(\ZZ \)-graded ring is one where \(f(x,y) = \sum _{i,j}c_{ij}x^iy^j\) has \(c_{ij}\) in degree \((2(i+j-1))\). This is so that if \(x,y\) each have degree \(-2\), then \(f(x,y)\) also has degree \(-2\). The factor of \(2\) is to agree with topological gradings.

Let \(I,J\) be the ideals of elements of positive degree on \(L,C\) respectively. Then the main idea leading to understanding Lazard’s ring is to linearize the problem:

Lemma 1.1. The map \(L \to C\) induces a map \(I/I^2 \to J/J^2\) that is an isomorphism in degree \(2n\) when \(n+1 \neq p^k\) for any prime \(p\), \(k > 0\), and is the inclusion of a subgroup of index \(p\) otherwise.

Let \(t_i\) be homogeneous in \(I\) and project to the generator of \(I/I^2\) in degree \(2i\).

Corollary 1.2. \(L\) is \(\ZZ [t_1,t_2,\dots ]\). Moreover \(L \otimes \QQ \) maps isomorphically to \(C \otimes \QQ \).

Proof. \(L\) is generated by the \(t_i\) since \(L\) is \(\ZZ \) in degree \(0\), \(I/I^2\) is generated by \(t_i\), and the \(t_i\) are in positive degree. To see that there are no relations, note that the since the map \(I/I^2 \to J/J^2\) is rationally an isomorphism, which again since the generators are in different degrees implies that the map \(L \otimes \QQ \to \CC \otimes \QQ \) is an isomorphism. Now the fact that there are no relations among the \(t_i\) follows from graded dimension counting of \(C\otimes \QQ \) and \(\ZZ [t_1,t_2,\dots ]\). □

We should expect that the map rationally \(L \otimes \QQ \cong C\otimes \QQ \) since the Hurewicz map is a rationally an isomorphism for spectra. However, it has the following consequence:

Corollary 1.3. Let \(R\) be a \(\QQ \)-algebra. Then for any two formal group laws over \(R\), there is a unique strict isomorphism between them.

Later, the isomorphism will be made explicit.