Equality in $\mathbb F_q\left(\left(\frac1T\right)\right)$ Can one characterize the $a\in\mathbb F_q\left(\left(\frac1T\right)\right)$ such that $a(T+1)=a(T)$? Although this seems elementary, I did not manage to find a answer.
Thanks in advance for any help.
 A: Let $K$ be an arbitrary field of prime characteristic $p$.
Consider the linear operator $D:a(t)\mapsto a(t+1)-a(t)$ of $K(\!(1/t)\!)$. It is well-defined, as it maps $t^{-n}$ to $$(t+1)^{-n}=t^{-n}(1+t^{-1})^{-n}=t^{-n}(1-t^{-1}+t^{-2}-t^{-3}+\cdots)^{n}.$$
The question is to determine the kernel of $D$.

The kernel of $D$ is equal to the subalgebra $K(\!(1/(t^p-t))\!)$.

First, $D$ preserves $K[t]$. It preserves the obvious flag, mapping $t^n$ to a polynomial of degree $n$ if $p$ does not divide $n$, and of degree $<n$ if $p$ divides $n$. Therefore, the kernel $L_{\ge 0}$ of $D$ restricted to $K[t]$ has a basis $(P_n)_{n\ge 0}$ with $P_n$ of degree $np$; moreover for every sequence $(P_n)$ in this kernel of degree $pn$, the family $(P_n)_{n\ge 0}$ is a basis of $L_{\ge 0}$. Hence we can choose $P_n=(t^p-t)^n$.
In general, $D$ preserves $t^nK[\![1/t]\!]$ for every $n\in\mathbf{Z}$, and in particular preserves the decomposition $K[t]\oplus t^{-1}K[\![t^{-1}]\!]$.
Next, let $f\in K(\!(1/t)\!)$ be an arbitrary nonzero element of the kernel of $D$. For $n\in\mathbf{Z}$, decompose $(t^p-t)^nf$ in this decomposition:
$$(t^p-t)^nf(t)=Q_n(t^p-t)+R_n,\quad Q_n\in K[u],\;R_n\in t^{-1}K[\![t^{-1}]\!].$$
Hence $(t^p-t)^{n+1}f=Q_{n+1}(t^p-t)+R_{n+1}$, which dividing by $t^p-t$ yields
$$(t^p-t)^{n}f=\frac{Q_{n+1}(t^p-t)}{t^p-t}+\frac{R_{n+1}}{t^p-t}.$$
Write $Q_n(u)=Q_n(0)+uQ^*_n(u)$ with $Q_n^*\in K[u]$; then 
$$(t^p-t)^{n}f=Q_{n+1}^*(t^p-t)+\frac{Q_{n+1}(0)+R_{n+1}}{t^p-t}.$$
Hence $Q_{n+1}^*=Q_n$. That is, $$Q_n(u)=\frac{Q_{n+1}(u)-Q_{n+1}(0)}{u}.$$
It follows that there exists $Q(u)\in K(\!(1/u)\!)$ such that for all $n\in\mathbf{Z}$ (say $n\ge 0$ this is enough for us) we have $Q_n(u)=\pi(u^nQ(u))$, where $\pi$ is the projection to $K[u]$ in the decomposition $K(\!(1/u)\!)=K[u]\oplus u^{-1}K[\![1/u]\!]$. That is, $u^nQ(u)=Q_n(u)+R'_n(u)$ with $R'_n\in u^{-1}K[\![1/u]\!]$.
$$(t^p-t)^nf(t)=(t^p-t)^nQ(t^p-t)+R_n-R'_n.$$
Hence $f(t)-Q(t^p-t)\in (t^p-t)^{-n}K[\![1/t]\!]=t^{-pn}K[\![1/t]\!]$ for $n$ arbitrary large. This shows that $f(t)\in K(\!(1/(t^p-t))\!)$.
A: This is a field automorphism of order $p$, since $T+p=T$, so it fixes an index $p$ subfield. $\mathbb F_q (( \frac{1}{T^p - T } ))$ is an index $p$ subfield and is fixed, hence is the fixed field. This is the simplest example of a wildly ramified extension of local fields in equal characteristic, and is in particular an Artin-Schreier extension, generated over $\mathbb F_q((u))$ by the equation $T^p-T = u^{-1}$ where $u = \frac{1}{T^p - T}$.
