Automorphisms of complete discrete valuation ring

Question

Let ${\Bbb F}_2[[T]]$ be a c.d.v.r over ${\Bbb F}_2$. We consider the automorphism $\sigma$ of ${\Bbb F}_2[[T]]$ such that $\sigma \colon T \mapsto T + c_2T^2 + c_3T^3 + \cdots$ with $c_i \in {\Bbb F}_2$.

Question: For any given $\sigma$, does such an element $S \in {\Bbb F}_2[[T]]$ always exist as $\sigma(S) = S + S^2 + S^3$ ?

Answer 1

You seem to be working in the complete discrete valuation ring $\mathbb F_2[[T]]$ of all power series over $\mathbb F_2$.

If $\sigma$ is an automorphism of the form you state, then we can think of $\sigma(T)$ as a power series $\gamma=\gamma(T)$, and the action of $\sigma$ on $\eta=\eta(T))=\sum_ia_iT^i$ is to send this to $\sum_ia_i\gamma^i$, in other words the composition $\eta\circ\gamma$ of the two power series. So we’re really working in the group of $\mathbb F_2$-series, under composition, and its action on the ring $R=\mathbb F_2[[T]]$, by right composition.

The group is often called the Nottingham group, but I wish it had a more descriptive name; I’ll call it $G$ here. Be that as it may, you’re asking for an $S\in R$ such that $S\circ\gamma=S+S^2+S^3=(T+T^2+T^3)\circ S$. It’s because $T+T^2+T^3$ is a polynomial that you can plug $S$ into it. In case $S$ happens to be in $G$, we can do more, but that’s for later.

First let’s deal with the case that $S$ has a nonzero constant term. Then since our field has only two elements, that constant is $1$, and we can write $S=1+S_1$, where $S_1\in G$. Our desire is to find $S_1$ such that $$ (1+S_1)\circ\gamma=(T+T^2+T^3)\circ(1+S_1)=1+S_1^3\,, $$ and thus $S_1\circ\gamma=S_1^3$. Now $S_1$ is in $TR$, and if it’s nonzero, say its initial degree is $m$. Then the initial degree of $S_1\circ\gamma$ will be $m$, while the initial degree of $S_1^3$ is $3m$, impossible, so $S_1$ is zero. So your only possibility for $S$ here is $S=1$.

On the other hand, if $S\in TR$, i.e. if $S$ has no constant term, we can deal with it as an element of $G$, and we’re asking the much more interesting question of whether it’s possible for $S\circ\gamma$ to equal $(T+T^2+T^3)\circ S$. In other words, given our $\gamma=\sigma(T)$, is $\gamma$ conjugate in $G$ to $T+T^2+T^3$? This is asking whether there’s $S$ with $T+T^2+T^3=S\circ\gamma\circ S^{-1}$. There are such $\sigma$’s, but they are rare. For instance, you can easily see that $c_2=1$, and all other $c_i=0$ gives a series not conjugate to $T+T^2+T^3$.