ON THE APPLICATIONS OF CYCLOTOMIC FIELDS IN INTRODUCTORY NUMBER THEORY 3 Now α prime to p , so if a$_{0}$ ≡ 0 ( p ), then a$_{0}$ ≡ 0 (1 − ζ$_{p}$ ), and so α ≡ 0 (1 − ζ$_{p}$ ), which is a contradiction. So a$_{0}$ ̸≡ 0 ( p ), and so a$_{1}$ − na$_{0}$ ≡ 0 has a unique solution k modulo p . Now (1 − ζ$_{p}$ ) | (1 − ζ 2 $_{p}$), and N ( $^{1}$− ζ 2 p $_{ζ$_{p}$}$) = N $^{(1}$− ζ 2 $_{p}$) N ζ$_{p}$ ) = 1, so (1 − ζ 2 $_{p}$) is associate to (1 − ζ$_{p}$ ). It follows that (1 − ζ$_{p}$ ) 2 | p , and so k is (still, since a$_{1}$ − na$_{0}$ ∈ Z ) the unique integral solution modulo p to a$_{1}$ − na$_{0}$ ≡ 0 (1 − ζ$_{p}$ ) $^{2}$. Then ζ k $_{p}$α ≡ a$_{0}$ (1 − ζ$_{p}$ ) $^{2}$, and therefore ζ k $_{p}$α is primary. □ Lemma 1. Let u be a unit in O $_{K}$. Then u u = ζ t for some t ∈ Z Proof. Write υ = u $_{u}$. Conjugation is a Galois automorphism on O K since ζ = ζ − 1 = ζ $^{p}$− $^{1}$. So u is also a unit, and so υ ∈ O $_{K}$. Now let σ$_{k}$ be the ( p − 1) Galois automorphisms on O K such that σ$_{k}$ ( ζ ) = ζ $^{k}$, k ∈ Z . Then for all 1 ≤ k ≤ ( p − 1), σ$_{k}$υ = σ$_{k}$u σ$_{k}$u = σ$_{k}$u σ$_{k}$u by the above remark. So | σ$_{k}$υ | = σ$_{k}$υσ$_{k}$υ = 1. So | σ$_{k}$υ | n = 1 for any n ∈ N . Now consider the polynomial f ( x ) = ∏ k =1 ( x − σ$_{k}$υ ). The coefficients of this polynomial are elementary symmetric polynomials in { σ$_{k}$υ : 1 ≤ k ≤ p − 1 } , and so are invariant by action by Gal ( K : Q ) = { σ$_{k}$υ : 1 ≤ k ≤ p − 1 } . So f ( x ) ∈ Z [ x ]. But then the coefficient of x k is s$_{(}$$_{p}$ − $^{1)}$− k where s$_{j}$ is the j th elementary symmetric polynomial. But by the previous paragraph, ∣ s$_{(}$$_{p}$ − $^{1)}$− k ∣ ≤ $^{p}$− $^{1}$− k ∑ j =1 | σ$_{k}$υ | k ≤ p − 1 − k . So there are finitely many possible such f ( x ) ∈ Z [ x ] since the coefficients are bounded. So there are finitely many possible roots since a polynomial of finite degree has a finite number of roots. But | σ$_{k}$υ $^{n}$| = 1 for any n ∈ N , so { υ n : n ∈ N } satisfy the same argument. So we must have υ n = υ n ′ for some n, n ′ ∈ Z . So υ $^{n}$−n ′ = 1, and it follows that υ is a root of unity in O $_{K}$. So by the basic properties of prime cyclotomic fields, we must have υ = ± ζ t for some t ∈ Z . Now consider congruence modulo λ = 1 − ζ . Then since 1 − ζ k 1 ζ = $^{k}$− 1 ∑ i =1 ζ i ∈ 〈 1 , ζ$_{p}$, ..., ζ $^{p}$− 2 p 〉 Z = O $_{K}$, ζ k ≡ 1 ( λ ) for all k ∈ Z . So since ζ k = ζ − k ≡ 1 ≡ ζ k ( λ ), α ≡ α ( λ ) for all α ∈ O $_{K}$. Namely, u ≡ u = ± ζ − $^{t}$u ≡ ± u ( λ ). So if υ = − ζ $^{t}$, u ≡ − u ( λ ) ⇒ 2 u ≡ 0 ( λ ) which is impossible since N ( λ ) = p ∤ N (2 u ) = 2 $^{p}$− 1 since p is odd. So υ = + ζ $^{t}$. □ Theorem 1. Let u be a unit in O $_{K}$. Then u is real ⇔ u is primary in O $_{K}$. Proof. Since O K = Z [ ζ$_{p}$ ] = 〈 1 , ζ$_{p}$, ..., ζ $^{p}$− 2 p 〉 $_{Z}$, we can write u as $^{p}$− 2 ∑ k =0 a$_{k}$ζ k for unique a$_{0}$, ..., a$_{p}$ − 2 ∈ Z . And so, noting that ζ $^{p}$− 1 = −$^{p}$ − 2 ∑ i =0 ζ $^{i}$, ζ − $^{t}$u = $^{p}$− 2 ∑ k =0 a$_{k}$ζ $^{k}$−t = $^{p}$− 2 k =0 ( a$_{k}$$_{+}$$_{t}$ − a$_{(}$$_{p}$ − 1)+ $_{t}$) ζ k where