some rational functions over a field of characteristic 2 I would like to know what are the formal power series $$f(t) = \sum_a \omega_a t^{-a}$$ over an algebraicially closed field of characteristic 2, with two properties:  (1)  The series represents a rational function, i.e. the coefficients satisfy a linear recursion, and (2) $\omega_{2a} = \omega_a^2$ for $a \ge 0$.
One family of solutions is $\omega_a = p_a(u_1, \dots, u_r)$   where $p_a$ is the $a$-th power sum symmetric function in some finite subset of $F$,   $p_a = \sum_{i = 1}^r u_i^a$.
Are these (more or less) all the solutions?
 A: Kevin Buzzard gave the solution.  Here it is with a little more detail:
Our assumptions include $\omega_0 = \omega_0^2$.  Thus $\omega_0 \in  \{0, 1\}$.
The linear homogeneous recursion only kicks in eventually; say the $\omega_a$ for $a \ge N$  satisfy such a recursion.
Let $v_1, \dots, v_m$  be the distinct roots of the characteristic polynomial of the linear recursion.    Then there exist polynomials $h_1, \dots, h_m$ such that
$\omega_a = \sum_{i = 1} ^m  h_i(a) v_i^a $  for $a \ge N$.    Let $\alpha_i$ be the constant term of 
$h_i$ for each $i$.   Since the characteristic is $2$,  we have $h_i(2a) = \alpha_i$  for all $a$.
For $a \ge N$,
\begin{equation} 
 \sum_i  \alpha_i v_i^{4a} = \omega_{4 a}   =  \omega_{2a}^2 = \sum_i \alpha_i^2 v_i^{4a}.
\end{equation}
Because the characteristic of $F$ is $2$,  each element has a unique  $2^k$--th root for all $k \ge 1$;  in particular all the $v_i^4$ are distinct, so the displayed equation implies that $\alpha_i^2 = \alpha_i$ for all $i$, i.e. $\alpha_i \in \{0, 1\}$.    Let $u_1, \dots, u_d$ be the list of those $v_j$ such that $\alpha_j = 1$.  Then we have $\omega_{2a} = \sum_i u_i^{2a}$ for
$a \ge N$.   For an arbitrary $a \ge 1$,  chose $k$ such that $2^{k-1} a \ge N$.  Then
$\omega_a$ is the unique $2^k$--th root of $\omega_{2^k a} = \sum_i u_i^{2^k a}$, namely
$\omega_a = \sum_i u_i^a$. 
Thus we have $\omega_0 \in \{0, 1\}$ and $\omega_a = p_a(u_1, \dots, u_d)$ for $a \ge 1$.
THANKS, KEVIN ! 
