6 KABALAN GASPARD equivalence of primary and real units in O K when K is a prime cyclotomic field, we can prove the result more rapidly. Lemma 2. Let i ̸≡ j ( p ) . Then the ideals I = ( x$_{0}$ + ζ $^{i}$y$_{0}$ ) and J = ( x$_{0}$ + ζ $^{j}$y$_{0}$ ) are relatively prime. Now N ( ζ i − ζ $^{j}$) = p since ( N ( ζ i − ζ $^{j}$)) 2 = $^{p}$− 1 ∏ k =1 ( ζ ik − ζ $^{jk}$) 2 = $^{p}$− 1 ∏ k =1 ( − ζ − k ( $^{j}$−i $^{)}$)(1 − ζ k ( $^{j}$−i $^{)}$) 2 = $^{p}$− 1 ∏ k =1 ( − ζ − $^{k}$)(1 − ζ $^{k}$) 2 = + ζ − p $^{p}$− 1 2 $^{p}$− 1 ∏ k =1 (1 − ζ $^{k}$) 2 = 1 · ( $^{p}$− 1 ∑ k =1 1 ) 2 = p $^{2}$. So N ( I + J ) | p . If N ( I + J ) = p , then since I ⊆ I + J , p = N ( I + J ) | N ( I ) = $^{p}$− 1 ∏ i =0 ( x$_{0}$ + ζ $^{i}$y$_{0}$ ) = x p 0 + y p 0 = z p $_{0}$. So since p is prime, p | z$_{0}$ ⇒ contradiction. So N ( I + J ) = 1, and therefore I + J = O $_{K}$. So I and J are coprime since P | I and P | J ⇒ P | I + J ⇒ P = O $_{K}$. □ Proof. Consider the ideal I + J . J contains the element − ( x$_{0}$ + ζ $^{j}$y$_{0}$ ), so x$_{0}$ + ζ $^{i}$y$_{0}$ − ( x$_{0}$ + ζ $^{j}$y$_{0}$ ) = ( ζ i − ζ $^{j}$) y$_{0}$ ∈ I + J . Likewise, since O K = Z [ ζ ], − ζ $^{j}$( x$_{0}$ + ζ $^{i}$y$_{0}$ ) = ζ $^{j}$x$_{0}$ + ζ i + $^{j}$y$_{0}$ ∈ I and ζ $^{i}$( x$_{0}$ + ζ $^{j}$y$_{0}$ ) = ζ $^{i}$x$_{0}$ + ζ i + $^{j}$y$_{0}$ ∈ J . So ζ $^{i}$x$_{0}$ + ζ i + $^{j}$y$_{0}$ − ζ $^{j}$( x$_{0}$ + ζ $^{i}$y$_{0}$ ) = ( ζ i − ζ $^{j}$) x$_{0}$ ∈ I + J . Now ( x$_{0}$, y$_{0}$ ) = 1 ⇒ there exist a, b ∈ Z such that ax$_{0}$ + by$_{0}$ = 1. So a ( ζ i − ζ $^{j}$) x$_{0}$ + b ( ζ i − ζ $^{j}$) y$_{0}$ = ( ζ i − ζ $^{j}$) ∈ I + J . Now x p $_{0}$+ y p 0 = z p 0 ⇒ $^{p}$− 1 ∏ i =0 ( x$_{0}$ + ζ $^{i}$y$_{0}$ ) = ( z$_{0}$ ) p as ideals. But { ( x$_{0}$ + ζ $^{i}$y$_{0}$ ) : 0 ≤ i ≤ p − 1 } are pairwise coprime. So by unique factorization of ideals, each of these ideals must be a p -th power. So in particular, taking i = 1, ( x$_{0}$ + ζy$_{0}$ ) = I p for some ideal I . So since ( x$_{0}$ + ζy$_{0}$ ) is principal, [ I ] has order dividing p in the ideal class group, but since p ∤ h , we must have that the order of [ I ] is 1. So I is principal. Let I = ( α ). Then ( x$_{0}$ + ζy$_{0}$ ) = ( α $^{p}$), and so x$_{0}$ + ζy$_{0}$ is associate to α $^{p}$. We write x$_{0}$ + ζy$_{0}$ = uα p where u is a unit in O $_{K}$. Then by 1 there exists a unique c modulo p such that ζ − $^{c}$u is primary. Let ζ − $^{c}$u = u$_{0}$ so that u = ζ $^{c}$u$_{0}$ where u$_{0}$ is primary. But u$_{0}$ is trivially a unit, and is therefore real by 1. So x$_{0}$ + ζy$_{0}$ = ζ $^{c}$u$_{0}$ α p where u$_{0}$ is real. Note that modulo p , α p ≡ ( $^{p}$− 2 ∑ i =0 a$_{i}$ζ i ) p ≡ $^{p}$− 2 ∑ i =0 a p $_{i}$ζ ip ≡ $^{p}$− 2 ∑ i =0 a p i ∈ Z ( p ). So α p ≡ α p ( p ). It follows that x$_{0}$ + ζy$_{0}$ = ζ $^{c}$u$_{0}$ α p ⇒ x$_{0}$ + ζy$_{0}$ ≡ ζ $^{c}$u$_{0}$ α p ( p ) ⇒ x$_{0}$ + ζy$_{0}$ ≡ ζ $^{c}$u$_{0}$ α p ( p ) ⇒ x$_{0}$ + ζ − $^{1}$y$_{0}$ ≡ ζ − $^{c}$u$_{0}$ α p ( p ). So we now have x$_{0}$ + ζy$_{0}$ ≡ ζ $^{c}$u$_{0}$ α p ( p ) ⇒ ζ − $^{c}$x$_{0}$ + ζ $^{1}$− $^{c}$y$_{0}$ ≡ u$_{0}$α p ( p ) and x$_{0}$ + ζ − $^{1}$y$_{0}$ ≡ ζ − $^{c}$u$_{0}$ α p ( p ) ⇒ ζ $^{c}$x$_{0}$ + ζ $^{c}$− $^{1}$y$_{0}$ ≡ u$_{0}$α p ( p ).