some rational functions over a field of characteristic 2 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T16:49:27Z http://mathoverflow.net/feeds/question/43879 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43879/some-rational-functions-over-a-field-of-characteristic-2 some rational functions over a field of characteristic 2 Fred Goodman 2010-10-27T21:31:47Z 2010-10-28T22:11:25Z <p>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 $a \ge 0$.</p> <p>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 $F$, $p_a = \sum_{i = 1}^r u_i^a$.</p> <p>Are these (more or less) all the solutions?</p> http://mathoverflow.net/questions/43879/some-rational-functions-over-a-field-of-characteristic-2/44026#44026 Answer by Fred Goodman for some rational functions over a field of characteristic 2 Fred Goodman 2010-10-28T22:11:25Z 2010-10-28T22:11:25Z <p>Kevin Buzzard gave the solution. Here it is with a little more detail:</p> <p>Our assumptions include $\omega_0 = \omega_0^2$. Thus <code>$\omega_0 \in \{0, 1\}$</code>.</p> <p>The linear homogeneous recursion only kicks in eventually; say the $\omega_a$ for $a \ge N$ satisfy such a recursion.</p> <p>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 $a$.<br> For $a \ge N$,<br> $$\sum_i \alpha_i v_i^{4a} = \omega_{4 a} = \omega_{2a}^2 = \sum_i \alpha_i^2 v_i^{4a}.$$ 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. <code>$\alpha_i \in \{0, 1\}$</code>. 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 $\omega_a = \sum_i u_i^a$. </p> <p>Thus we have <code>$\omega_0 \in \{0, 1\}$</code> and $\omega_a = p_a(u_1, \dots, u_d)$ for $a \ge 1$.</p> <p>THANKS, KEVIN ! </p>