P^n-1|Q^n-1
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From: blang on 30 Jun 2010 00:02 Yes, you are right for this special case.
From: master1729 on 30 Jun 2010 02:47 blang wrote : > Hi :) > > Let P and Q in IR[X] (with deg(P)>0 and deg(Q)>0) > such that : for all n in IN*, P^n-1|Q^n-1. > Is it true that we can find k in IN* such that Q=P^k > ? > > Thanks ! i dont think so.
From: master1729 on 30 Jun 2010 02:49 > blang wrote : > > > Hi :) > > > > Let P and Q in IR[X] (with deg(P)>0 and deg(Q)>0) > > such that : for all n in IN*, P^n-1|Q^n-1. > > Is it true that we can find k in IN* such that > Q=P^k > > ? > > > > Thanks ! > > i dont think so. apart from id(x) of course ...
From: blang on 2 Jul 2010 03:38 No idea ?
From: achille on 5 Jul 2010 13:12
On Jul 2, 7:38 pm, blang <bl...(a)laposte.net> wrote: > No idea ? Okay, I think the theorem is true for most (if not all) p(z). More precisely, I believe the theorem is true at least for those p(z) where p'(z) is non-zero for any complex number z that satisfies |p(z)| = 1. Following are my arguments. Perhaps someone else can double check whether these arguments are valid or not. Under the given assumption, the set |p(z)| = 1 is the disjoint union of finite many smooth closed curves (i.e. diffeomorphic to a bunch of circles). Let C be one of this smooth closed curve. It is easy to see/prove |q(z)| = 1 on C. Case 1: q'(z_*) = 0 for some z_* on C. for any z on C sufficiently close to z_*, we have: [1] q(z) / q(z_*) - 1 = O( ( z - z_* )^2 ) [2] p(z) / p(z_*) - 1 = p'(z_*)/p(z_*) ( z - z_* ) + O( ( z - z_*)^2 ) Let t_p, t_q \in [ 0, 1 ) such that p(z_*) = exp( 2 \pi i t_p) q(z_*) = exp( 2 \pi i t_q) let z_1, z_2, ... be any sequence on C such that p(z_i) = exp( 2 \pi i k_i/m_i ) where k_i, m_i are positive integers satisify m_i -> infinity and 0 < | k_i - m_i t_p | <= 1. We see [2] implies z_i - z_* = 2 \pi i (k_i/m_i - t_p) / (p'(z_*)/p(z_*)) + O((z_i- z_*)^2) => z_i - z_* = O(1/m_i). Since p(z_i)^(m_i) = 1 implies q(z_i)^(m_i) = 1. We can find integer l_i (up to multiple of m_i) such that p(z_i) = exp( 2 \pi i l_i/m_i ). Substitute this in [1], we see one can always fix the multiple of m_i in l_i so that l_i / m_i - t_q = O((z_i - z_*)^2) = O(1/m_i^2 ) => l_i - m_i t_q = O(1/m_i). [3]. Notice that unless t_q is an integer, it is always possible to choose an increasing integer sequence m_i such that the fractional part of m_i t_q stay close to 1/2. This will trigger a contradiction with [3]. Togather with the choice t_q \in [0,1), we must have t_q = 0 <=> q(z_*) = 1. Now [3] becomes l_i = O(1/m_i). Since l_i are integers, l_i = 0 for any sufficiently large i. This in turn implies q(z_i) = 1 for infinitely many z_i -> z_*. Since q(z) is a polynomial, q(z) = 1 identically. Case 2: q'(z) <> 0 on C, Let s \in [0, 1] -> z(s) \in C be a smooth parametrization of C with z'(s) <> 0 for all s. Since |p(z)| = 1 and p'(z) <> 0 on C, p'(z(s))/p(z(s)) z'(s) = i t_p(s) for some real smooth function t_p(s) on [0,1] which doesn't change sign. WOLOG, we will assume t_p(s) > 0. This implies the contour integral -i \int_C p'(z)/p(z) dz = \int^1_0 t_p(s) ds > 0. By Cauchy integral formula, this is equal to 2 \pi N_p where N_p is the number of root of p(z) in the interior of C. ie. the number of root of p(z) inside C is non-zero and the winding number of p(z(s)) as s moves from 0 to 1 is N_p > 0. Since |q(z)| = 1 and q'(z) <> 0 on C, we can define t_q(s) and N_q in a similar manner and the winding number of q(z(s)) as s moves from 0 to 1 is N_q > 0. Consider the rational function H(z) = q(z)^N_p / p(z)^N_q. Since |H(z)| = 1 on C, H(z(s)) = exp( i \theta(s) ) for some real smooth function \theta(s) on [0,1]. Now the winding number of H(z(s)) as s moves from 0 to 1 is 0. we can extend \theta(s) to a periodic function over R. This implies \theta(s) reaches extremum for some s \in [0,1]. This in turn implies there exists a point z_* on C such that H'(z_*) = 0. Once again, we have for z on C sufficiently close to z_*, [1]' H(z)/H(z_*) - 1 = O( (z-z_*)^2 ). Similar argument like case [1] (employing same definition for k_i and l_i) show that N_p( l_i / m_i - t_q ) - N_q ( k_i / m_i - t_p ) = O(1/m_i^2 ) => ( N_p l_i - N_q k_i ) - m_i ( N_p t_q - N_q t_p ) = O(1/ m_i) [3]' Once again, unless N_p t_q - N_q t_p is an integer, we can always find an increasing sequence m_i such that the fractional part of m_i ( N_p t_q - N_q t_p ) stays close to 1/2. This will trigger a contradiction with [3]'. Hence N_p t_q - N_q t_p = K <-- a integer! => H(z_*) = 1. Now [3]' becomes N_p l_i - N_q k_i - m_i K = O(1/m_i). Since the LHS is always an integer, for sufficiently large i, we will have N_p l_i - N_q k_i - m_i K = 0 => N_p l_i/m_i - N_q k_i/m_i = K a integer! => H(z_i) = 1 Once again, this implies H(z_i) = 1 for infinitely many z_i -> z_* and hence H(z) = 1 <=> q(z)^N_p = p(z)^N_q identically. Let N = gcd(N_p,N_q) and n_p = N_p/N, n_q = N_q/N. Recall C[z], the polynomial ring over C is an unique factorization domain. Factor p(z) and q(z) into irreducible factors over C[z], and by comparing factors between q(z)^N_p and p(z)^N_q, one immediately see that there exists polynomial f(z) \in C[z] and units c, d in C[z] such that: p(z) = c f(z)^n_p q(z) = d f(z)^n_q Since units in C[z] are just complex numbers, it is clear by rescaling f(z) appropriately, one can fix c to 1. Now choose integer m >=0, n_p > r >= 0 such that n_q = m n_p + r, then q(z) - 1 = d f(z)^n_p - 1 = d ( (p(z)-1) + 1 )^m f(z)^r - 1 and p(z) - 1 | q(z) - 1 => p(z) - 1 | d f(z)^r - 1 Since degree(d f(z)^r - 1 ) < degree(p(z)-1), this is possible only when d f(z)^r = 1. ie. q(z) = p(z)^m. |