# Generating the coordinate ring of the Lubin-Tate formal group

Let $$K$$ be a $$p$$-adic local field with uniformizer $$\pi \in \mathcal{O}_{K}$$ and residue field $$k = \mathcal{O}_{K}/\pi$$. Let $$G$$ be a Lubin-Tate formal $$\mathcal{O}_{K}$$-module and $$G_{0}$$ its reduction to $$k$$.

Let $$G_{0}[\pi]$$ denote the $$\pi$$-torsion subgroup of $$G_{0}$$, this is a finite group scheme over $$k$$. The action of $$\mathcal{O}_{K}$$ on $$G_{0}$$ yields an action of the multiplicative monoid $$\Gamma = (k, \cdot)$$ on $$G_{0}[\pi]$$ and hence on its coordinate ring $$\mathcal{O}_{G_{0}[\pi]}$$.

Question: Does there always exist an element $$g \in \mathcal{O}_{G_{0}[\pi]}$$ which generates the coordinate ring as a module over the monoid ring $$k[\Gamma]$$? How to find such an element?

Example: Suppose that $$K = \mathbb{Q}_{p}$$, $$\pi = p$$ and $$G = \mathbb{G}^{\wedge}_{m}$$ the formal multiplicative group, so that

$$\mathcal{O}_{G[p]} \simeq \mathbb{F}_{p}[x]/((1+x)^{p}-1)$$.

Then an element $$[\gamma] \in \Gamma = (\mathbb{Z}/p, \cdot)$$ acts on $$g = x+1$$ via

$$[\gamma]g = [\gamma](x)+[\gamma](1) = (x+1)^{\gamma}-1+1 = (x+1)^{\gamma}$$

and hence the translates of $$g$$ along $$0, \cdots, p-1 \in \mathbb{F}_{p}$$ are of the form

$$(x+1)^{0}, (x+1)^{1}, \cdots, (x+1)^{p-1}$$

This is a $$\mathbb{F}_{p}$$-basis of $$\mathcal{O}_{G_{0}[\pi]}$$, so $$g = x+1$$ gives an element which generates the whole coordinate ring as needed.

Some thoughts: 1) In the case of $$\mathbb{G}_{m}^{\wedge}$$, where $$\mathbb{G}_{m}[p] = \mu_{p}$$, another way to find an element $$g$$ is to notice that Cartier duality gives an isomorphism

$$\mathcal{O}_{\mu_{p}} \simeq \mathcal{O}_{\mathbb{Z}/p}^{*} \simeq Fun(\mathbb{Z}/p, \mathbb{F}_{p})^{*} \simeq \mathbb{F}_{p}[\mathbb{Z}/p]$$,

(where $$*$$ denotes the dual vector space) so that the coordinate ring is just the regular representation of the multiplicative monoid $$\Gamma = \mathbb{Z}/p$$ and hence must be generated by a single element.

1. Also in the case of $$\mathbb{G}_{m}^{\wedge}$$, the invariant differential is given by $$\omega(x) dx = \frac{1}{1+x} dx$$. So perhaps it is possible to relate a generating element $$g$$ to invariant differentials?

2. We have verified that such an element also exists when $$G_{0}$$ is the Honda formal group law of height 2 over $$\mathbb{F}_{4} \simeq W(\mathbb{F}_{4})/p$$, and can be taken to be

$$g = 1 + x + x^{2} + x^{3} \in \mathcal{O}_{G_{0}[p]} \simeq \mathbb{F}_{4}[x]/(x^{4})$$

I'll assume that $$\pi=p$$; I guess that the general case is the same but I have not checked. We now have $$k=\mathbb{F}_{p^n}$$ and $$\mathcal{O}_K=W\mathbb{F}_{p^n}$$. The group of roots of unity in $$W\mathbb{F}_{p^n}$$ is cyclic of order $$p^n-1$$, with generator $$\omega$$ say, and maps by an isomorphism to $$\mathbb{F}_{p^n}^\times$$. Put $$f(x)=px+x^{p^n}$$. Lubin-Tate theory then says that there is a unique formal group law $$F$$ with $$f(F(x,y))=F(f(x),f(y))$$ and that this has $$[p](x)=f(x)$$ and $$[\omega^i](x)=\omega^ix$$. The ring $$R=\mathcal{O}_{G_0[\pi]}$$ is just $$W\mathbb{F}_{p^n}[[x]]/f(x)$$, which has basis $$B=\{x^i:0\leq i over $$W\mathbb{F}_{p^n}$$. Put $$a=1+x+\dotsb+x^{p^n-1}\in R$$. I claim that this is a generator of the required type. More explicitly, let $$b$$ be the image of $$a$$ under the substitution $$x\mapsto [0](x)=0$$ i.e. $$b=1$$, and let $$c_i$$ be the image of $$a$$ under the substitution $$x\mapsto[\omega^i](x)=\omega^ix$$. The claim is then that the list $$b,c_0,c_1,\dotsc,c_{p^n-2}$$ is a basis for $$R$$ over $$W\mathbb{F}_{p^n}$$. If we write this list in terms of the monomial basis, we get a matrix $$M$$ of shape $$p^n\times p^n$$, which we must prove to be invertible. The top row (corresponding to $$b$$) is just $$(1,0,0,\dotsc,0)$$, and the bottom right block of size $$(p^n-1)\times(p^n-1)$$ is the Vandermonde matrix with entries $$\omega^{ij}$$. The Vandermonde determinant formula shows that this is invertible.