With 6 inverted, is the ring of Weierstrass curves a quotient of the Lazard ring by a regular sequence?

Question

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? (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!

Answer 1

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

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.