PUMaC 2023 · 个人决赛（A 组） · 第 3 题
PUMaC 2023 — Individual Finals (Division A) — Problem 3
题目详情
- Let f ( X ) be a monic irreducible polynomial over Z ; therefore, by Gauss’s Lemma, f is also 2 irreducible over Q (you may assume this). Moreover, assume f ( X ) | f ( X + n ) where n is an 2 integer such that n ̸ ∈ {− 1 , 0 , 1 } . Show that n ∤ f (0). 2
解析
- Let f ( X ) be a monic irreducible polynomial over Z ; therefore, by Gauss’s Lemma, f is also 2 irreducible over Q (you may assume this). Moreover, assume f ( X ) | f ( X + n ) where n is an 2 integer such that n ̸ ∈ {− 1 , 0 , 1 } . Show that n ∤ f (0). Proposed by Michael Cheng and Steven Wang 2 Solution: Let g ( X ) = X + p , so f ( X ) | f ( g ( X )). Note that if x is a (complex) root of f ( X ), 2 3 k then so is g ( x ). Therefore g ( x ) , g ( x ) , · · · are all roots of f , where g ( x ) = g ( · · · g ( x )) with g is applied k times. However, f has finitely many roots, so there must be some root x of f 0 k and some k ≥ 1 such that g ( x ) = x . This means that x is a root of the polynomial 0 0 0 k h ( X ) = g ( X ) − X. Irreducibility of f implies that f | h (this is a standard fact; we assume this for now and supply an elementary proof later), and thus f (0) | h (0). However, it is easy to see that 2 h (0) = n + ( n + · · · ) , 2 2 so n ∤ h (0) and n ∤ f (0). Now we prove the claim that we assumed in the proof above: Claim. Let α ∈ C be an algebraic number (i.e. α is the root of some rational polynomial). Then there is some unique irreducible monic polynomial f ( X ) ∈ Q [ X ] with f ( α ) = 0 ; moreover, for any polynomial h ( X ) ∈ Q [ X ] with h ( α ) = 0 , we have f ( X ) | h ( X ) in Q [ X ] . Proof. Let f ( X ) be a monic polynomial having α as a root with minimal degree. We claim that f satisfies the desired properties. Assume for contradiction that f is not irreducible, so f ( X ) = g ( X ) h ( X ) with g, h ∈ Q [ X ] and non-constant; moreover we may assume that g and h are both monic. Then α is a root of one of them, which contradicts the minimality of f . Now assume that h ∈ Q [ X ] with h ( α ) = 0. By the Euclidean algorithm (a.k.a. long division of polynomials), we can write h ( X ) = f ( X ) q ( X ) + r ( X ) , with q, r ∈ Q [ X ] and deg r < deg f . Now plugging in X = α gives r ( α ) = 0, which contradicts the minimality of f unless r ≡ 0, so f | h . The uniqueness of f thus follows. If g is another monic polynomial with the same prop- erties, then f | g and g | f , so f ( X ) = cg ( X ) for some constant c , but c = 1 since both are monic. In our case, f is the unique irreducible polynomial with x as a root, thus f | h in Q [ X ]; 0 that is h ( X ) = f ( X ) g ( X ) for some g ( X ) ∈ Q [ X ]. However, since f ∈ Z [ X ] and monic by assumption, and h ∈ Z [ X ] obviously, we actually have g ( X ) ∈ Z [ X ] (this follows from the Euclidean algorithm). 9