PUMaC 2020 · 代数（A 组） · 第 6 题
PUMaC 2020 — Algebra (Division A) — Problem 6
题目详情
- Given integer n , let W be the set of complex numbers of the form re , where q is a rational n number so that qn ∈ Z and r is a real number. Suppose that p is a polynomial of degree ≥ 2 such that there exists a non-constant function f : W → C so that p ( f ( x )) p ( f ( y )) = f ( xy ) for n all x, y ∈ W . If p is the unique monic polynomial of lowest degree for which such an f exists n for n = 65, find p (10).
解析
- Given integer n , let W be the set of complex numbers of the form re , where q is a rational n number so that qn ∈ Z and r is a real number. Suppose that p is a polynomial of degree ≥ 2 such that there exists a non-constant function f : W → C so that p ( f ( x )) p ( f ( y )) = f ( xy ) for n 2 all x, y ∈ W . If p is the unique monic polynomial of lowest degree for which such an f exists n for n = 65, find p (10). Proposed by: Frank Lu Answer: 100009 Note: On the original algebra test, we had forgotten the phrase “ r is a real number.” Fix f (1) and p ( x ). 2 First, note that plugging in x = y = 1 yields that p ( f (1)) = f (1), and y = 1 yields that p ( f ( x )) p ( f (1)) = f ( x ). Hence, we see that the image of f is a root of the polynomial p ( u ) p ( f (1)) − u = 0 , which in par- 2 ticular means that f has a finite image. Furthermore, we thus see that p ( f ( x )) p ( f ( y )) p ( f (1)) = 2 f ( xy ) p ( f (1)) , which means that, in fact, that f ( x ) f ( y ) = f (1) f ( xy ) ∀ x, y ∈ W . If f (1) is n zero, then it follows that ∀ x ∈ W that f ( x ) = 0, so we consider when f (1) 6 = 0. Then, we see n that, letting g ( x ) = f ( x ) /f (1) that g ( x ) g ( y ) = g ( xy ) ∀ x, y ∈ W . n Since the image of g is finite, if there exists a value of x so that | g ( x ) | 6 = 1 , | g ( x ) | 6 = 0, then 2 g ( x ) , g ( x ) , . . . are all distinct, contradiction. Furthermore, g ( x ) = 0 for some x 6 = 0 means that g ( x ) g ( y/x ) = 0 = g ( y ) ∀ y ∈ W , so we take that we want | g ( x ) | = 1 for all x ∈ W . By n n a similar logic, we see that g ( x ) must be a root of unity, as again we will run into the issue where the image of g is infinite. We see that if p is a prime not dividing n , then g ( x ) can’t ever be a p th root of unity, since otherwise we could take the p th root of x to get another root (raising all of the roots to the p th power yields a permutation of the roots). Thus, we see that the minimal possible value for the degree of our polynomial is 5, which would then require it to have the 5th roots of unity as roots. 5 Thus, we see that p ( u ) p ( f (1)) − u = au − a for some complex number a , meaning that a a 5 5 p ( u ) = u + u − , which we can just write as p ( u ) = u + c ( u − 1) for some complex p ( f (1)) p ( f (1)) 5 constant c . By monic, we see that p ( x ) = x + x − 1 yields that p (10) = 100009 .