Let $L$ be the Lazard ring, i.e., the underlying ring of the universal one-dimensional formal group law. Let $M$ be the ring $\mathbb{Z}[c\_4, c\_6, 1/6]$ of Weierstrass curves over $\mathbb{Z}[1/6]$. There is a natural surjective ring map $L[1/6] \rightarrow M$ classifying the formal group law of a Weierstrass curve.


I recall hearing, years ago, that *the kernel of the map $L[1/6] \rightarrow M$ is generated by a regular sequence*. At the time I think I saw why this was true, but now I don't see the argument, although I still find the claim entirely plausible, with a small modification, explained below. The only written reference I have found is in a nice MathOverflow post of T. Lawson, the first bullet-point here: [Can we construct a Baas-Sullivan presentation of TMF?](https://mathoverflow.net/questions/208256/can-we-construct-a-baas-sullivan-presentation-of-tmf) (To translate between this post and that post: the ring $L$ is naturally isomorphic to the homotopy groups of the complex bordism spectrum $MU$, while $M$ is isomorphic to the homotopy groups of $6$-inverted $tmf$.)


My question: *is the kernel of the map $L[1/6] \rightarrow M$ generated by a regular sequence?* If so, is there a citeable reference for this fact already in the literature somewhere?


One note: "regular sequence" must be taken a bit impressionistically, here, since regular sequences are supposed to be of finite length, but $L\cong \mathbb{Z}[x\_1, x\_2, \dots]$, a polynomial algebra on countably infinitely many generators, so the kernel of $L[1/6] \rightarrow M$ won't be generated by any finite-length sequence! Instead, for the purposes of this question, let's say (contra standard usage in commutative algebra) that regular sequences are allowed to be infinite, so that a regular sequence in a commutative ring $R$ is a sequence $(r\_1, r\_2, \dots)$ such that, for each $n\geq 1$, there is no nonzero $r\_n$-torsion in $R/(r\_1, \dots ,r\_{n-1})$.


Thanks!



With 6 inverted, is the ring of Weierstrass curves a quotient of the Lazard ring by a regular sequence?
I'll refer to my notes on formal groups at <https://strickland1.org/courses/formalgroups/fg.pdf>. There are results about the formal group law of an elliptic curve in Section 19. That is written in terms of the general homogeneous Weierstrass form
$$ y^2 z + a\_1 x y z + a\_3 y z^2 - x^3 - a\_2 x^2 z - a\_4 x z^2 - a\_6 z^3 = 0, $$
but you can put $a\_1=a\_2=a\_3=0$ to get the special form that works better when $6$ is inverted, namely
$$ f(x,y,z) = y^2 z - x^3 - a\_4 x z^2 - a\_6 z^3 = 0. $$
It is not hard to see that there is a unique series $\xi(x)$ of the form $\sum\_{k\geq 3}\xi\_kx^k$ with $\xi\_3=1$ and $f(1,x,\xi(x))=0$, and any computer algebra package will calculate terms of this series with reasonable efficiency. By putting $a\_1=a\_3=0$ in Proposition 19.2 we get $[-1]\_F(x)=-x$. It is also shown in Section 19 that $x\_0+\_Fx\_1+\_Fx\_2$ is a unit multiple of the series $\chi(x\_0,x\_1,x\_2)=\sum\_{i,j,k\geq 0}\xi\_{i+j+k+2}x\_0^ix\_1^jx\_2^k$, so the series $m(x\_0,x\_1)=[-1]\_F(x\_0+\_Fx\_1)=-(x\_0+\_Fx\_1)$ is characterised by the fact that $\chi(x\_0,x\_1,m(x\_0,x\_1))=0$. This leads to a reasonably efficient calculation of $x+\_Fy$ as
$$ x+\_Fy = x+y-2a\_4c\_5(x,y)-3a\_6c\_7(x,y)\pmod{(x,y)^8}, $$
where $c\_p(x,y)=((x+y)^p-x^p-y^p)/p$ for prime $p$. By the standard analysis of the structure of the Lazard ring, we see that the generators in degrees $8$ and $12$ map to $\pm 2a\_4$ and $\pm 3a\_6$ mod decomposables.


Now invert $6$. From the above it follows that we can choose elements $b\_4\in L[\tfrac{1}{6}]\_8$ and $b\_6\in L[\tfrac{1}{6}]\_{12}$ that map to $a\_4$ and $a\_6$ respectively. For $k\neq 4,6$ we can then choose a generator in $L[\tfrac{1}{6}]\_{2k}$ and subtract a polynomial in $b\_4,b\_6$ if necessary to get a generator $b\_{2k}$ that maps to zero in $\mathbb{Z}[\tfrac{1}{6}][a\_4,a\_6]$. We now have $L[\tfrac{1}{6}]=\mathbb{Z}[\tfrac{1}{6}][b\_1,b\_2,b\_3,\dotsc]$, and the sequence of generators other than $b\_4$ and $b\_6$ is a regular sequence that generates the required ideal.


It follows that there is an essentially unique commutative ring object in the homotopy category of $MU$-modules whose homotopy ring is $\mathbb{Z}[\tfrac{1}{6}][a\_4,a\_6]$, considered as an $MU\_\*$-algebra using the above FGL. The sharpest version of this is Theorem 2.6 of my paper [Products on $MU$-modules](https://arxiv.org/pdf/math/0011122.pdf)


Note that most of the work above goes into proving that the map $L[\tfrac{1}{6}]\to\mathbb{Z}[\tfrac{1}{6}][a\_4,a\_6]$ is surjective, which you mention in your question as a known fact. If you are willing to assume that, then only a couple of the steps described above are needed.


