Generating the coordinate ring of the Lubin-Tate formal group

Question

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.

2. 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?

3. 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

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

Answer 1

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 $B=\{x^i:0\leq i<p^n\}$ 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.