6
$\begingroup$

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

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

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks, this is very helpful! $\endgroup$ Nov 29, 2022 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.