Nonarchimedean dynamical systems and formal groups

TL;DR

This paper proves theorems confirming that certain commuting $p$-adic power series are related to formal groups, either as endomorphisms or semi-conjugate, advancing understanding of nonarchimedean dynamical systems.

Contribution

It establishes conditions under which commuting $p$-adic power series correspond to formal group endomorphisms or semi-conjugates, confirming Lubin's observation.

Findings

Existence of a formal group associated with commuting power series

Conditions under which power series are endomorphisms or semi-conjugate

Advancement in understanding nonarchimedean dynamical systems

Abstract

We prove two theorems that confirm an observation of Lubin concerning families of $p$-adic power series that commute under composition: under certain conditions, there is a formal group such that the power series in the family are either endomorphisms of this group, or semi-conjugate to endomorphisms of this group.

Equations

$${\mathrm{L}}^{\prime }(X)=\underset{n\to +\infty }{\lim }\frac{(f^{\circ n}{)}^{\prime }(X)}{f^{\prime }(0{)}^n}=\underset{n\to +\infty }{\lim }\frac{f^{\prime }(f^{\circ n-1}(X))}{f^{\prime }(0)}\cdots \frac{f^{\prime }(f(X))}{f^{\prime }(0)}\cdot \frac{f^{\prime }(X)}{f^{\prime }(0)},$$

$$a_j(X)=f^{\circ n}(X)\sum _{i⩾0}{s}_{j+i}(X)f^{\circ n}(X{)}^i=s_j(X)f^{\circ n}(X)+s_{j+1}(X)f^{\circ n}(X{)}^2+\cdots .$$

$${\mathrm{val}}_p((j+i)!)=\frac{j+i-s_p(j+i)}{p-1}⩽\frac{i}{p-1}+\frac{j}{p-1}.$$

$$f^{\circ n}(X)\in \pi X\cdot {\mathcal{O}}_K[[X]]+X^{q^n}\cdot {\mathcal{O}}_K[[X^{q^n}]],$$

$$\Vert f^{\circ n}(X){\Vert }_{{\rho }_n}⩽p^{-q^n/(e(q^n-1))}<p^{-1/e}⩽p^{-1/(p-1)}$$

$$s_j(X)+s_{j+1}(X)f^{\circ n}(X)+s_{j+2}(X)f^{\circ n}(X{)}^2+\cdots \in {\mathcal{O}}_K[[X]].$$

$$f^{\circ n}(X)=f(f^{\circ n-1}(X))\in \pi f^{\circ n-1}(X)\cdot {\mathcal{O}}_K[[X]]+f^{\circ n-1}(X{)}^q\cdot {\mathcal{O}}_K[[X]].$$

$$f\circ h(X)=f(X^m)=[c]\cdot X^m\cdot g_0(X^m{)}^m\cdot (1+w_0(X^m){)}^m=f_0(X{)}^m=h\circ f_0(X).$$

$$f^{\circ n}(X^m)=X^m\cdot \prod _{\alpha }(X^m-\alpha {)}^m\cdot v_n(X^m),$$

Full text

Introduction

Let $K$ be a finite extension of $\mathbf{Q}_{p}$, and let $\mathcal{O}_{K}$ be its ring of integers and $\mathfrak{m}_{K}$ the maximal ideal of $\mathcal{O}_{K}$. In [Lub94], Lubin studied nonarchimedean dynamical systems, namely families of elements of $X\cdot\mathcal{O}_{K}[\![X]\!]$ that commute under composition, and remarked (page 341 of ibid.) that “experimental evidence seems to suggest that for an invertible series to commute with a noninvertible series, there must be a formal group somehow in the background”. Various results in that direction have been obtained (by Hsia, Laubie, Li, Movahhedi, Salinier, Sarkis, Specter, …; see for instance [Li96], [Li97a], [Li97b], [LMS02], [Sar05], [Sar10], [SS13], [HL16], [Ber17], [Spe18]), using either $p$-adic analysis, the theory of the field of norms or, more recently, $p$-adic Hodge theory. The purpose of this article is to prove two theorems that confirm the above observation in many new cases, using only $p$-adic analysis.

If $g(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$, we say that $g$ is invertible if $g^{\prime}(0)\in\mathcal{O}_{K}^{\times}$ and noninvertible if $g^{\prime}(0)\in\mathfrak{m}_{K}$. We say that $g$ is stable if $g^{\prime}(0)$ is neither [math] nor a root of unity. For example, if $S$ is a formal group of finite height over $\mathcal{O}_{K}$ and if $c\in\mathbf{Z}$ with $p\nmid c$ and $c\neq\pm 1$, then $f(X)=[p](X)$ and $u(X)=[c](X)$ are two stable power series, with $f$ noninvertible and $u$ invertible, having the following properties: the roots of $f$ and all of its iterates are simple, $f\not\equiv 0\bmod{\mathfrak{m}_{K}}$ and $f\circ u=u\circ f$. Our first result is a partial converse of this. If $f(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$, let $\mathrm{U}_{f}$ denote the set of invertible power series $u(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ such that $f\circ u=u\circ f$, and let $\mathrm{U}_{f}^{\prime}(0)=\{u^{\prime}(0),\ u\in\mathrm{U}_{f}\}$. This is a subgroup of $\mathcal{O}_{K}^{\times}$.

Theorem A**.**

Let $K$ be a finite extension of $\mathbf{Q}_{p}$ such that $e(K/\mathbf{Q}_{p})\leqslant p-1$, and let $f(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ be a noninvertible stable series. Suppose that

1. (1)

the roots of $f$ and all of its iterates are simple, and $f\not\equiv 0\bmod{\mathfrak{m}_{K}}$; 2. (2)

there is a subfield $F$ of $K$ such that $f^{\prime}(0)\in\mathfrak{m}_{F}$ and such that $\mathrm{U}_{f}^{\prime}(0)\cap\mathcal{O}_{F}^{\times}$ is an open subgroup of $\mathcal{O}_{F}^{\times}$.

Then there is a formal group $S$ over $\mathcal{O}_{K}$ such that $f\in\mathrm{End}(S)$ and $\mathrm{U}_{f}\subset\mathrm{End}(S)$.

Condition (1) can be checked using the following criterion (proposition 1).

Criterion A**.**

If $f(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ is a noninvertible stable series with $f\not\equiv 0\bmod{\mathfrak{m}_{K}}$, and if $f$ commutes with a stable invertible series $u(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$, then the roots of $f$ and all of its iterates are simple if and only if $f^{\prime}(X)/f^{\prime}(0)\in 1+X\cdot\mathcal{O}_{K}[\![X]\!]$.

If $K=\mathbf{Q}_{p}$, condition (2) of Theorem A amounts to requiring the existence of a stable invertible series that commutes with $f$.

Corollary A**.**

If $f(X)\in X\cdot\mathbf{Z}_{p}[\![X]\!]$ is a noninvertible stable series such that the roots of $f$ and all of its iterates are simple and $f\not\equiv 0\bmod{p}$, and if $f$ commutes with a stable invertible series $u(X)\in X\cdot\mathbf{Z}_{p}[\![X]\!]$, then there is a formal group $S$ over $\mathbf{Z}_{p}$ such that $f\in\mathrm{End}(S)$ and $\mathrm{U}_{f}\subset\mathrm{End}(S)$.

There are examples of commuting power series where $f$ does not have simple roots, for instance $f(X)=9X+6X^{2}+X^{3}$ and $u(X)=4X+X^{2}$ with $K=\mathbf{Q}_{3}$ (more examples can be constructed following the discussion on page 344 of [Lub94]). It seems reasonable to expect that if $f$ and $u$ are two stable noninvertible and invertible power series that commute, with $f\not\equiv 0\bmod{\mathfrak{m}_{K}}$, then there exists a formal group $S$, two endomorphisms $f_{S}$ and $u_{S}$ of $S$, and a nonzero power series $h$ such that $f\circ h=h\circ f_{S}$ and $u\circ h=h\circ u_{S}$. We then say that $f$ and $f_{S}$ are semi-conjugate, and $h$ is an isogeny from $f_{S}$ to $f$ (see for instance [Li97a]).

The simplest case where this occurs is when $m$ is an integer $\geqslant 2$, and the nonzero roots of $f$ and all of its iterates are of multiplicity $m$ (for an example of a more complicated case, see remark 3). In this simplest case, we have the following.

Theorem B**.**

Let $K$ be a finite extension of $\mathbf{Q}_{p}$, let $f(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ be a noninvertible stable series and take $m\geqslant 2$. Let $h(X)=X^{m}$. Suppose that

1. (1)

the nonzero roots of $f$ and all of its iterates are of multiplicity $m$ 2. (2)

$f\not\equiv 0\bmod{\mathfrak{m}_{K}}$.

Then there exists a finite unramified extension $L$ of $K$ and a noninvertible stable series $f_{0}(X)\in X\cdot\mathcal{O}_{L}[\![X]\!]$ with $f_{0}\not\equiv 0\bmod{\mathfrak{m}_{K}}$, such that $f\circ h=h\circ f_{0}$, and the roots of $f_{0}$ and all of its iterates are simple.

If in addition $u$ is an element of $\mathrm{U}_{f}$ with $u^{\prime}(0)\equiv 1\bmod{\mathfrak{m}_{K}}$, then there exists $u_{0}\in\mathrm{U}_{f_{0}}$ such that $u\circ h=h\circ u_{0}$. Finally, if there is a subfield $F$ of $K$ such that $f^{\prime}(0)\in\mathfrak{m}_{F}$ and such that $\mathrm{U}_{f}^{\prime}(0)\cap\mathcal{O}_{F}^{\times}$ is an open subgroup of $\mathcal{O}_{F}^{\times}$, then $(f_{0}^{\circ m})^{\prime}(0)\in\mathfrak{m}_{F}$ and $\mathrm{U}_{f_{0}}^{\prime}(0)\cap\mathcal{O}_{F}^{\times}$ is an open subgroup of $\mathcal{O}_{F}^{\times}$.

Condition (1) can be checked using the following criterion (proposition 3).

Criterion B**.**

If $f(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ is a noninvertible stable series with $f\not\equiv 0\bmod{\mathfrak{m}_{K}}$, and if $f$ commutes with a stable invertible series $u(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$, then the nonzero roots of $f$ and all of its iterates are of multiplicity $m$ if and only if the nonzero roots of $f$ are of multiplicity $m$, and the set of roots of $f^{\prime}$ is included in the set of roots of $f$.

We have the following simple corollary of Theorem B when $K=\mathbf{Q}_{p}$.

Corollary B**.**

If $m\geqslant 2$ and $f(X)\in X\cdot\mathbf{Z}_{p}[\![X]\!]$ is a noninvertible stable series such that the nonzero roots of $f$ and all of its iterates are of multiplicity $m$ and $f\not\equiv 0\bmod{p}$, and if $f$ commutes with a stable invertible series $u(X)\in X\cdot\mathbf{Z}_{p}[\![X]\!]$, then there is an unramified extension $L$ of $\mathbf{Q}_{p}$, a formal group $S$ over $\mathcal{O}_{L}$ and $f_{S}\in\mathrm{End}(S)$ such that $f\circ X^{m}=X^{m}\circ f_{S}$.

Theorem A implies conjecture 5.3 of [HL16] for those $K$ such that $e(K/\mathbf{Q}_{p})\leqslant p-1$. It also provides a new simple proof (that does not use $p$-adic Hodge theory) of the main theorem of [Spe18]. Note also that Theorem A holds without the restriction “$e(K/\mathbf{Q}_{p})\leqslant p-1$” if $f^{\prime}(0)$ is a uniformizer of $\mathcal{O}_{K}$ (see [Spe17]). This implies “Lubin’s conjecture” formulated at the very end of [Sar10] (this conjecture is proved in [Ber17] using $p$-adic Hodge theory, when $K$ is a finite Galois extension of $\mathbf{Q}_{p}$) as well as “Lubin’s conjecture” on page 131 of [Sar05] over $\mathbf{Q}_{p}$ if $f\not\equiv 0\bmod{p}$.

The results of [HL16], [Ber17] and [Spe18] are proved under strong additional assumptions on $\operatorname{wideg}(f)$ (namely that $\operatorname{wideg}(f)=p$ in [Spe18], or that $\operatorname{wideg}(f)=p^{h}$, where $h$ is the residual degree of $K$, in [HL16] and [Ber17]). Theorem A is the first general result in this direction that makes no assumption on $\operatorname{wideg}(f)$, besides assuming that it is finite. It also does not assume that $f^{\prime}(0)$ is a uniformizer of $\mathcal{O}_{K}$.

Theorem A and its corollary are proved in section §2 and theorem B and its corollary are proved in section §3.

1. Nonarchimedean dynamical systems

Whenever we talk about the roots of a power series, we mean its roots in the $p$-adic open unit disk $\mathfrak{m}_{\mathbf{C}_{p}}$. Recall that the Weierstrass degree $\operatorname{wideg}(g(X))$ of a series $g(X)=\sum_{i\geqslant 1}g_{i}X^{i}\in X\cdot\mathcal{O}_{K}[\![X]\!]$ is the smallest integer $i\leqslant+\infty$ such that $g_{i}\in\mathcal{O}_{K}^{\times}$. We have $\operatorname{wideg}(g)=+\infty$ if and only if $g\equiv 0\bmod{\mathfrak{m}_{K}}$.

If $r<1$, let $\mathcal{H}(r)$ denote the set of power series in $K[\![X]\!]$ that converge on the closed disk $\{z\in\mathfrak{m}_{\mathbf{C}_{p}}$ such that $|z|_{p}\leqslant r\}$. If $h\in\mathcal{H}(r)$, let $\|h\|_{r}=\sup_{|z|_{p}\leqslant r}|h(z)|_{p}$. The space $\mathcal{H}(r)$ is complete for the norm $\|{\cdot}\|_{r}$. Let $\mathcal{H}=\projlim_{r<1}\mathcal{H}(r)$ be the ring of holomorphic functions on the open unit disk.

Throughout this article, $f(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ is a stable noninvertible series such that $\operatorname{wideg}(f)<+\infty$, and $\mathrm{U}_{f}$ denotes the set of invertible power series $u(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ such that $f\circ u=u\circ f$.

\lemmname** \the\smf@thm.**

A series $g(X)\in X\cdot K[\![X]\!]$ that commutes with $f$ is determined by $g^{\prime}(0)$.

Proof.

This is proposition 1.1 of [Lub94]. ∎

\propname** \the\smf@thm.**

If $\mathrm{U}_{f}$ contains a stable invertible series, then there exists a power series $g(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ and an integer $d\geqslant 1$ such that $f(X)\equiv g(X^{p^{d}})\bmod{\mathfrak{m}_{K}}$.

We have $\operatorname{wideg}(f)=p^{d}$ for some $d\geqslant 1$.

Proof.

This is the main result of [Lub94]. See (the proof of) theorem 6.3 and corollary 6.2.1 of ibid. ∎

\propname** \the\smf@thm.**

There is a (unique) power series $\mathrm{L}(X)\in X+X^{2}\cdot K[\![X]\!]$ such that $\mathrm{L}\circ f=f^{\prime}(0)\cdot\mathrm{L}$ and $\mathrm{L}\circ u=u^{\prime}(0)\cdot\mathrm{L}$ if $u\in\mathrm{U}_{f}$. The series $\mathrm{L}(X)$ converges on the open unit disk, and $\mathrm{L}(X)=\lim_{n\to+\infty}f^{\circ n}(X)/f^{\prime}(0)^{n}$ in the Fréchet space $\mathcal{H}$.

Proof.

See propositions 1.2, 1.3 and 2.2 of [Lub94]. ∎

\lemmname** \the\smf@thm.**

If $f(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ is a noninvertible stable series and if $f$ commutes with a stable invertible series $u$, then every root of $f^{\prime}$ is a root of $f^{\circ n}$ for some $n\gg 0$.

Proof.

This is corollary 3.2.1 of [Lub94]. ∎

\propname** \the\smf@thm.**

If $f(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ is a noninvertible stable series with $f\not\equiv 0\bmod{\mathfrak{m}_{K}}$, and if $f$ commutes with a stable invertible series $u$, then the roots of $f$ and all of its iterates are simple if and only if $f^{\prime}(X)/f^{\prime}(0)\in 1+X\cdot\mathcal{O}_{K}[\![X]\!]$.

Proof.

We have $(f^{\circ n})^{\prime}(X)=f^{\prime}(f^{\circ n-1}(X))\cdots f^{\prime}(f(X))\cdot f^{\prime}(X)$. If $f^{\prime}(X)/f^{\prime}(0)\in 1+X\cdot\mathcal{O}_{K}[\![X]\!]$, then the derivative of $f^{\circ n}(X)$ belongs to $f^{\prime}(0)^{n}\cdot(1+X\cdot\mathcal{O}_{K}[\![X]\!])$ and hence has no roots. The roots of $f^{\circ n}(X)$ are therefore simple.

By lemma 1, any root of $f^{\prime}(X)$ is also a root of $f^{\circ n}$ for some $n\gg 0$. If the roots of $f^{\circ n}(X)$ are simple for all $n\geqslant 1$, then $f^{\prime}(X)$ cannot have any root, and hence $f^{\prime}(X)/f^{\prime}(0)\in 1+X\mathcal{O}_{K}[\![X]\!]$. ∎

2. Formal groups

We now prove theorem A. Let $S(X,Y)=\mathrm{L}^{\circ-1}(\mathrm{L}(X)+\mathrm{L}(Y))\in K[\![X,Y]\!]$. By proposition 1, $S$ is a formal group law over $K$ such that $f$ and all $u\in\mathrm{U}_{f}$ are endomorphisms of $S$. In order to prove theorem A, we show that $S(X,Y)\in\mathcal{O}_{K}[\![X,Y]\!]$. Write $S(X,Y)=\sum_{j\geqslant 0}s_{j}(X)Y^{j}$.

\lemmname** \the\smf@thm.**

If $\mathrm{L}^{\prime}(X)\in\mathcal{O}_{K}[\![X]\!]$, then $s_{j}(X)\in j!^{-1}\cdot\mathcal{O}_{K}[\![X]\!]$ for all $j\geqslant 0$.

Proof.

This is lemma 3.2 of [Li96]. ∎

\lemmname** \the\smf@thm.**

If the roots of $f^{\circ n}(X)$ are simple for all $n\geqslant 1$, then $\mathrm{L}^{\prime}(X)\in\mathcal{O}_{K}[\![X]\!]$.

Proof.

This is sketched in the proof of theorem 3.6 of [Li96]. We give a complete argument for the convenience of the reader.

We have $(f^{\circ n})^{\prime}(X)=f^{\prime}(f^{\circ n-1}(X))\cdots f^{\prime}(f(X))\cdot f^{\prime}(X)$, and by proposition 1, $f^{\prime}(X)/f^{\prime}(0)\in 1+X\mathcal{O}_{K}[\![X]\!]$. We have $\mathrm{L}(X)=\lim_{n\to+\infty}f^{\circ n}(X)/f^{\prime}(0)^{n}$ by proposition 1, so that

[TABLE]

and hence $\mathrm{L}^{\prime}(X)\in 1+X\mathcal{O}_{K}[\![X]\!]$. ∎

\theoname** \the\smf@thm.**

If $e(K/\mathbf{Q}_{p})\leqslant p-1$, then $s_{j}(X)\in\mathcal{O}_{K}[\![X]\!]$ for all $j\geqslant 0$.

Proof.

For all $n\geqslant 1$, the power series $u_{n}(X)=S(X,f^{\circ n}(X))$ belongs to $X\cdot K[\![X]\!]$ and satisfies $u_{n}\circ f=f\circ u_{n}$. Since $\mathrm{U}_{f}^{\prime}(0)\cap\mathcal{O}_{F}^{\times}$ is an open subgroup of $\mathcal{O}_{F}^{\times}$, there exists $n_{0}$ such that if $n\geqslant n_{0}$, then $u_{n}^{\prime}(0)=1+f^{\prime}(0)^{n}\in\mathrm{U}_{f}^{\prime}(0)$. We then have $u_{n}\in\mathrm{U}_{f}$ by lemma 1.

In order to prove the theorem, we therefore prove that if $S(X,f^{\circ n}(X))\in\mathcal{O}_{K}[\![X]\!]$ for all $n\geqslant n_{0}$, then $s_{i}(X)\in\mathcal{O}_{K}[\![X]\!]$ for all $i\geqslant 0$. If $j\geqslant 1$, let

[TABLE]

We prove by induction on $j$ that $s_{0}(X),\ldots,s_{j-1}(X)$ as well as $a_{j}(X)$ belong to $\mathcal{O}_{K}[\![X]\!]$. This holds for $j=1$; suppose that it holds for $j$.

We claim that if $h\in\mathcal{H}(r)$ and $\|h\|_{r}<p^{-1/(p-1)}$, then $\sum_{i\geqslant 0}s_{j+i}(X)h(X)^{i}$ converges in $\mathcal{H}(r)$. Indeed, if $s_{p}(j+i)$ denotes the sum of the digits of $j+i$ in base $p$, then

[TABLE]

Let $\pi$ be a uniformizer of $\mathcal{O}_{K}$ and let $e=e(K/\mathbf{Q}_{p})$ so that $|\pi|_{p}=p^{-1/e}$. By proposition 1, we have

[TABLE]

where $q=p^{d}=\operatorname{wideg}(f)$, so that $\|f^{\circ n}(X)\|_{r}\leqslant\max(rp^{-1/e},r^{q^{n}})$. If $\rho_{n}=p^{-1/(e(q^{n}-1))}$, then

[TABLE]

and the series $\sum_{i\geqslant 0}s_{j+i}(X)f^{\circ n}(X)^{i}$ therefore converges in $\mathcal{H}(\rho_{n})$.

We have $f^{\circ n}(X)\in\pi X\cdot\mathcal{O}_{K}[\![X]\!]+X^{q^{n}}\cdot\mathcal{O}_{K}[\![X^{q^{n}}]\!]$, as well as $\operatorname{wideg}(f^{\circ n})=q^{n}$. By the theory of Newton polygons, all the zeroes $z$ of $f^{\circ n}(X)$ satisfy $\mathrm{val}_{p}(z)\geqslant 1/(e(q^{n}-1))$, and hence $|z|_{p}\leqslant\rho_{n}$. The equation $a_{j}(X)=f^{\circ n}(X)\sum_{i\geqslant 0}s_{j+i}(X)f^{\circ n}(X)^{i}$ holds in $\mathcal{H}(\rho_{n})$, and this implies that $a_{j}(z)=0$ for all $z$ such that $f^{\circ n}(z)=0$. Since all the zeroes of $f^{\circ n}(X)$ are simple and $f^{\circ n}(X)\not\equiv 0\bmod{\pi}$, the Weierstrass preparation theorem implies that $f^{\circ n}(X)$ divides $a_{j}(X)$ in $\mathcal{O}_{K}[\![X]\!]$, and hence that

[TABLE]

Choose some $0<\rho<1$ and take $n\geqslant n_{0}$ such that $\rho_{n}\geqslant\rho$. We have

[TABLE]

Therefore $\|f^{\circ n}(X)\|_{\rho}\to 0$ as $n\to+\infty$, and $\|s_{j+1}(X)f^{\circ n}(X)+s_{j+2}(X)f^{\circ n}(X)^{2}+\cdots\|_{\rho}\to 0$ as $n\to+\infty$. The series $s_{j}(X)$ is therefore in the closure of $\mathcal{O}_{K}[\![X]\!]$ inside $\mathcal{H}(\rho)$ for $\|{\cdot}\|_{\rho}$, which is $\mathcal{O}_{K}[\![X]\!]$.

This proves that $s_{j}(X)$ as well as $s_{j+1}(X)f^{\circ n}(X)+s_{j+2}(X)f^{\circ n}(X)^{2}+\cdots$ belong to $\mathcal{O}_{K}[\![X]\!]$. This finishes the induction and hence the proof of the theorem. ∎

Theorem A now follows: $S$ is a formal group over $\mathcal{O}_{K}$ such that $f\in\mathrm{End}(S)$. Any power series $u(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ that commutes with $f$ also belongs to $\mathrm{End}(S)$, since $u(X)=[u^{\prime}(0)](X)$ by lemma 1. In particular, $\mathrm{U}_{f}\subset\mathrm{End}(S)$.

To prove corollary A, note that we can replace $u$ by $u^{\circ p-1}$ and therefore assume that $u^{\prime}(0)\in 1+p\mathbf{Z}_{p}$. In this case, $u^{\circ m}$ is defined for all $m\in\mathbf{Z}_{p}$ by proposition 4.1 of [Lub94] and $\mathrm{U}_{f}^{\prime}(0)$ is therefore an open subgroup of $\mathbf{Z}_{p}^{\times}$.

3. Semi-conjugation

We now prove theorem B. Assume therefore that the nonzero roots of $f$ and all of its iterates are of multiplicity $m$. Let $h(X)=X^{m}$.

Since $q=\operatorname{wideg}(f)$ is finite, we can write $f(X)=X\cdot g(X)\cdot v(X)$ where $g(X)\in\mathcal{O}_{K}[X]$ is a distinguished polynomial and $v(X)\in\mathcal{O}_{K}[\![X]\!]$ is a unit. If the roots of $g(X)$ are of multiplicity $m$, then $g(X)=g_{0}(X)^{m}$ for some $g_{0}(X)\in\mathcal{O}_{K}[X]$. Write $v(X)=[c]\cdot(1+w(X))$ where $c\in k_{K}$ (and $[c]$ is its Teichmüller lift) and $w(X)\in(\mathfrak{m}_{K},X)$. Since $m\cdot\deg(g)=q-1$, $m$ is prime to $p$ and there exists a unique $w_{0}(X)\in(\mathfrak{m}_{K},X)$ such that $1+w(X)=(1+w_{0}(X))^{m}$. If $f_{0}(X)=[c^{1/m}]\cdot X\cdot g_{0}(X^{m})\cdot(1+w_{0}(X^{m}))$, then

[TABLE]

It is clear that $f_{0}\not\equiv 0\bmod{\mathfrak{m}_{K}}$. If we write $f^{\circ n}(X)=X\cdot\prod_{\alpha}(X-\alpha)^{m}\cdot v_{n}(X)$ with $v_{n}$ a unit of $\mathcal{O}_{K}[\![X]\!]$, and where $\alpha$ runs through the nonzero roots of $f^{\circ n}$, then

[TABLE]

so that all the roots of $f^{\circ n}(X^{m})$ have multiplicity $m$. Since $f^{\circ n}(X^{m})=f_{0}^{\circ n}(X)^{m}$, the roots of $f_{0}$ and all of its iterates are simple. This finishes the proof of the first part of the theorem, with $L=K([c^{1/m}])$.

If $u\in\mathrm{U}_{f}$ and $u^{\prime}(0)\in 1+\mathfrak{m}_{K}$, then there is a unique $u_{0}(X)\in 1+(\mathfrak{m}_{K},X)$ such that $u_{0}(X)^{m}=u(X^{m})$. We have $u_{0}^{\prime}(0)=u^{\prime}(0)^{1/m}$ and $(f_{0}\circ u_{0})^{m}=(u_{0}\circ f_{0})^{m}$ as well as $(f_{0}\circ u_{0})^{\prime}(0)=(u_{0}\circ f_{0})^{\prime}(0)$, so that $u_{0}\in\mathrm{U}_{f_{0}}$. This proves the existence of $u_{0}$. Since $f(X^{m})=f_{0}(X)^{m}$, we have $f^{\prime}(0)=f_{0}^{\prime}(0)^{m}$. We then have $(f_{0}^{\circ m})^{\prime}(0)=f_{0}^{\prime}(0)^{m}=f^{\prime}(0)\in\mathfrak{m}_{F}$. This finishes the proof of the last claim of theorem B.

Corollary B follows from theorem B in the same way that corollary A followed from theorem A.

\exemname* \the\smf@thm.*

If $p=3$ and $f(X)=9X+6X^{2}+X^{3}$ and $u(X)=4X+X^{2}$, so that $f\circ u=u\circ f$, then $f(X)=X(X+3)^{2}$ and $f^{\prime}(X)=3(X+3)(X+1)$. The nonzero roots of $f$ and all of its iterates are therefore of multiplicity $2$. We have $f(X^{2})=(X(X^{2}+3))^{2}$ so that $f_{0}(X)=3X+X^{3}$, and the corresponding formal group is $\mathbf{G}_{m}$ (this is a special case of the construction given on page 344 of [Lub94]).

\propname** \the\smf@thm.**

If $f(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$ is a noninvertible stable series with $f\not\equiv 0\bmod{\mathfrak{m}_{K}}$, and if $f$ commutes with a stable invertible series $u(X)\in X\cdot\mathcal{O}_{K}[\![X]\!]$, then the nonzero roots of $f$ and all of its iterates are of multiplicity $m$ if and only if the nonzero roots of $f$ are of multiplicity $m$ and the set of roots of $f^{\prime}$ is included in the set of roots of $f$.

Proof.

If the nonzero roots of $f$ and all of its iterates are of multiplicity $m$, then the nonzero roots of $f$ are of multiplicity $m$. Hence if $\alpha$ is a root of $f^{\circ n}(X)$ with $f(\alpha)\neq 0$, the equation $f(X)=f(\alpha)$ has simple roots. Since $\alpha$ is one of these roots, we have $f^{\prime}(\alpha)\neq 0$. By lemma 1, any root of $f^{\prime}(X)$ is also a root of $f^{\circ n}$ for some $n\geqslant 1$. This implies that the set of roots of $f^{\prime}$ is included in the set of roots of $f$.

Conversely, suppose that the nonzero roots of $f$ are of multiplicity $m$, and that $f^{\prime}(\beta)\neq 0$ for any $\beta$ that is not a root of $f$. If $\alpha$ is a nonzero root of $f^{\circ n}$ for some $n\geqslant 1$, then this implies that the equation $f(X)=\alpha$ has simple roots, so that the nonzero roots of $f$ and all of its iterates are of multiplicity $m$. ∎

\remaname* \the\smf@thm.*

If $p=2$ and $f(X)=4X+X^{2}$ and $u(X)=9X+6X^{2}+X^{3}$, then $f\circ u=u\circ f$. The roots [math] and $-4$ of $f$ are simple, but $f^{\circ 2}(X)=X(X+4)(X+2)^{2}$ has a double root. In this case, $f$ is still semi-conjugate to an endomorphism of $\mathbf{G}_{m}$, but via the more complicated map $h(X)=X^{2}/(1+X)$ (see the discussion after corollary 3.2.1 of [Lub94], and example 2 of [Li96]).