Quick Preliminaries: 


A *commutative formal group law* is a formal power series $F(x,y)=\sum\_{ij}c\_{ij}x^iy^j$ that satisfies:


1. Commutativity: $F(x,y) = F(y,x)$
2. Identity: $F(x,0)=x=F(0,x)$
3. Associativity: $F(F(x,y),z) = F(x,F(y,z))$


The restrictions imposed by the formal group law axioms on the coefficient ring of the formal power series generate an ideal $I$. 


The Lazard ring is the result of modding out our coefficient ring $c\_{ij} \in Z[c\_{ij}]$ by this ideal. $L=\mathbb{Z}[c\_{ij}]/I$.


Explicitly, the relations imposed on our coefficients are:


1. $F(x,y)=F(y,x)$ 


=> $c\_{ij}=c\_{ji}$
2. $F(x,0)=x=F(0,x)$ 


$F(x,0)=\sum\_{ij}c\_{ij}x^i0^j=x$ => $c\_{10}=1$


$F(0,x)=\sum\_{ij}c\_{ij}0^ix^j=x$ => $c\_{01}=1$


=> $c\_{10}=c\_{01}=1$
3. $F(F(x,y),z) = F(x,F(y,z))$ 


=> ????


**Here's my question:** What do the relations imposed by the associativity condition explicitly look like? 


Thusfar, in every source I've looked at (including Lazard's original papers), I've been unable to find this! It's driving me up the wall! I derived the following. However, I'm **unsure if it is correct, and would be deeply appreciative if someone could point out a simpler way express it**. 


My sad attempt:


Looking for the coefficient of $x^{\alpha}y^{\beta}z^{\gamma}$ in $F(x,F(y,z))$: 


We start by expanding the coefficients of $F(x, -)$ (with indices $i$ and $j$), then we must have $i = \alpha$ to get $x^\alpha$ (this is why $i$ does not appear).


Next we expand the coefficients of $F(y,z)$ (with indices $k$ and $\ell$) -- this whole thing gets raised to the power $j$. I think we need $j$ copies of it that multiply out correctly (which corresponds to the exponents adding up correctly):


$b\_{\alpha\beta\gamma} = \sum\_{j=0}^{\infty}\sum\_{(k\_{1}+...+k\_{j}=\beta)}$ $\sum\_{(\ell\_{1}+...+\ell\_{j}=\gamma)}c\_{\alpha j}c\_{k\_{1}\ell\_{1}}...c\_{k\_{j}\ell\_{j}}$


Repeating the procedure with the appropriate indices, we get the coefficients associated to $F(F(x,y), z)$:


$b\_{\alpha\beta\gamma} = \sum\_{i=0}^{\infty}\sum\_{(k\_{1}+...+k\_{i}=\alpha)}$ $\sum\_{(\ell\_{1}+...+\ell\_{i}=\beta)}c\_{i\gamma}c\_{k\_{1}\ell\_{1}}...c\_{k\_{i}\ell\_{i}}$




---


A related question: I get that we must make sure that each of our 3 constraints is homogeneous (this ensures that you still have a grading when you mod out by the ideal generated by them). However, I don't understand **why we impose the following grading:** 
$$\text{deg}c\_{ij} = i + j - 1$$


My only hint is looking at the coefficients of $F(x,F(y,z)$ (i.e. 
$c\_{\alpha j}c\_{k\_{1}\ell\_{1}}...c\_{k\_{j}\ell\_{j}}$).
If we add the degrees of each term in the product, we get:


$(\alpha + j - 1) + (k\_1 + l\_1 - 1) + ... + (k\_j + l\_j - 1)$
$\begin{align\*}
&= \alpha + (k\_1 + ... + k\_j) + (l\_1 + ... + l\_j) + j - j - 1
&= \alpha + \beta + \gamma - 1
\end{align\*}$


I'm guessing the two expressions for $b\_{ijk}$ that we have are homogeneous of the same degree, so that setting them equal to each other doesn’t ruin the homogeneity of the ideal.



What is the explicit ideal (wrt the Lazard ring) generated by the associativity of formal group laws?
Your computation of the relations looks correct to me.


The grading is that way because it comes from the symmetry where you multiply the variable by a constant. What I mean is that you take a formal group law:


$$z = F(x,y)$$


and you multiply every variable by $\lambda$:


$$\lambda z = F(\lambda x, \lambda y) $$


$$ z = \frac{ F( \lambda x, \lambda y)}{\lambda}$$


and you get another formal group law.


So $c\_{i,j}$ is sent to $\lambda^{i+j-1}c\_{i,j}$. This $\mathbb G\_m$-action gives you a grading.


