Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
Yeah, I think so. One can solve a general linear recurrence over an alg closed field: the general solution is that $\omega_a$ is a sum of things of the form $h(a).x^a$ with $h$ a polynomial and $x\in F$. Now your second assumption implies $h(a)^2=h(2a)$ for all integers $a\geq0$, but you are in char 2 so $h(2a)=h(0)=c$, the constant term, and so $h(a)$ is the unique (again as you're in char 2) square root of $c$ for all $a$, so $h$ may as well be replaced by a constant function $c$ satisfying $c=c^2$, and $c=1$ is the only interesting solution, giving you the solution you already spotted. –  Kevin Buzzard Oct 27 '10 at 22:02
add comment

1 Answer

up vote 0 down vote accepted

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 !

share|improve this answer
    
No worries. Make your answer community-wiki and accept it, and then this stops the problem re-appearing on the front page in 6 months time. –  Kevin Buzzard Oct 31 '10 at 8:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.