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Let $\mathbf{F}_q$ be a finite field, and let $A=\mathbf{F}_q [[ x^{1/q^\infty} ]]$ be the completion of $\mathbf{F}_q[x^{1/q^\infty}]$ with respect to the $x$-adic topology. Then the $q$th power Frobenius map $\pi\colon A\to A$ is an automorphism. If $F$ is the local field $\mathbf{F}_q((\pi))$, there is an action $F^\times\to \text{Aut} A$ by continuous automorphisms; an element of $F$ sends $x$ to the corresponding linear fractional power series. Let $\mathcal{O}_F=\mathbf{F}_q[[\pi]]$.

My question is: Can anyone produce an explicit nonconstant element of $A$ which is $\mathcal{O}_F^\times$-invariant?

I assure you that such elements exist! Here's why: Let $f(X)=\pi X+X^q\in F[X]$. Let $x_1,x_2,\dots$ be a compatible family of roots of iterates of $f$, with $x_1\neq 0$. Let $F_n=F(x_n)$. Then by Lubin-Tate theory, $\text{Gal}(F_n/F)\cong (\mathcal{O}_F/\pi^n)^\times$. If $F_\infty$ is the union of the $F_n$, then $F_\infty/F$ is a maximal totally ramified abelian extension of $F$, with Galois group $\mathcal{O}_F^\times$. Let $K$ be the $\pi$-adic completion of $F_\infty$.

I claim that $\mathcal{O}_K$ is isomorphic to $A$. Indeed, it isn't hard to see that the sequence $x_1^q,x_2^{q^2},\dots$ converges in $\mathcal{O}_K$ to an element $x$, all of whose $q$th power roots lie in $\mathcal{O}_K$. Observe that $x^{1/q^n}-x_n$ is divisible by $\pi$, hence by $x^{1/q}$, and therefore (since $\mathcal{O}_K$ is topologically generated by the $x_n$) we have $\mathcal{O}_K=\mathbf{F}_q[x^{1/q^\infty}]+x^{1/q}\mathcal{O}_K$. This shows that $\mathbf{F}_q[x^{1/q^\infty}]$ is dense in $\mathcal{O}_K$, which proves the claim.

The Galois action of $\mathcal{O}_F^\times$ on $F_\infty$ extends to the completion $K$. It's a fun exercise to see that the isomorphism $\mathcal{O}_K\cong A$ respects the action of $\mathcal{O}_F^\times$. Meanwhile, we have the element $\pi\in \mathcal{O}_K\cong \mathbf{F}_q[[x^{1/q^\infty}]]$. This element is invariant under the Galois group $\mathcal{O}_F^\times$. What is its power series development in $x$?

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  • $\begingroup$ I don't understand the group action you are describing, could you elaborate on exactly what it does? $\endgroup$
    – Will Sawin
    Commented Apr 25, 2012 at 5:54
  • $\begingroup$ @Will: The power series $\sum a_n \pi^n$ preserves constants, and sends $x$ to $\sum a_n x^{q^n}$. $\endgroup$
    – S. Carnahan ♦
    Commented Apr 25, 2012 at 7:18
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    $\begingroup$ Let $R_1=\mathbb{F}_q[[X]],R_2=R_1[[Y]]$. Set $h_1=-X^{q-1}\in R_1$, so that $\pi=h_1(x_1)$. Let $g_2(Y)=h_1(X^q+YX)\in R_2$. Then $\lim_{n\rightarrow\infty}g_2^{(n)}(0)$ converges in $R_1$ (where $g_n^{(n)}$ is $n$-fold composition), and denote this limit by $h_2$. We have $h_2(x_2)=\pi$. We can continue similarly: $g_{i+1}(Y)=h_i(X^q+YX)$, $h_{i+1}=\lim_{n\rightarrow\infty}g_{i+1}^{(n)}(0)$, $h_{i+1}(x_{i+1})=\pi$. Magma computations show that $H=\lim_{n\rightarrow\infty} h_n(X^{1/q^{n}})$ converges in $\mathbb{F}_q[[X^{1/q^\infty}]]$. So finally $H(x)=\pi$. $\endgroup$
    – Dror Speiser
    Commented Apr 25, 2012 at 15:17
  • $\begingroup$ For example, for $q=3$ we have $\pi=-x^{2/3}-x^{10/9}+x^{14/9}+x^{46/27}-x^2-x^{58/27}+...$ $\endgroup$
    – Dror Speiser
    Commented Apr 25, 2012 at 15:33
  • $\begingroup$ Dror: This is top-rate work (and certainly answer-worthy). I suspected there wouldn't be a clean formula for $\pi$, but at least now I know there's an algorithm. $\endgroup$
    – Jared Weinstein
    Commented Apr 25, 2012 at 16:33

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