2 KABALAN GASPARD The definition of primary elements has historically been ambiguous in Number Theory. In [2], Dalawat shows that definitions of primary elements in O K even differ by country (” p -primary”, ” primaire ” and ” prim¨ ar ”) and, even though these definitions do form a chain of implications, they are not equivalent. We also note that it is not true that if p an arbitrary odd prime and µ prime in O $_{K}$, only one associate of µ is primary (for example, according to the above definition, both ± (4+3 ω ) are primary in the ring of integers of Q ( ω ) where ω = e 2 $_{π}$√ − 1 3 ). Proposition 1. Let α ∈ O K (not necessarily prime) and suppose α prime to p in O $_{K}$. Then there exists a k ∈ Z , unique (modulo p ), such that ζ k $_{p}$α is primary. Proof. Consider the ideal P = (1 − ζ$_{p}$ ) in O $_{K}$. Then the norm of the ideal N ( P ) = $^{p}$− ∏ i =1 (1 − ζ i $_{p}$) = p by the fact that Gal ( K : Q ) ≃ U ( Z /p Z ). So P is a prime ideal and is thus of degree 1. So by Dedekind’s Theorem in Algebraic Number Theory, any element of O K is the root of a monic polynomial of degree 1 in O $_{K}$/P . So in the particular case of α , α − a$_{0}$ = 0 in O $_{K}$/P for some a$_{0}$ ∈ Z . In other words, α ≡ a$_{0}$ (1 − ζ$_{p}$ ). So $^{α}$−a$_{0}$ $^{(1}$− ζ$_{p}$ ) ∈ O K and so, by the same argument, $^{α}$−a$_{0}$ $^{(1}$− ζ$_{p}$ ) ≡ a$_{1}$ (1 − ζ$_{p}$ ) for some a$_{1}$ ∈ Z . We stop repeating this here because multiplying the congruence by (1 − ζ$_{p}$ ), we now have a congruence modulo (1 − ζ$_{p}$ ) $^{2}$, which is what we want to consider. More precisely, we now have α − a$_{0}$ ≡ a$_{1}$ (1 − ζ$_{p}$ ) (1 − ζ$_{p}$ ) $^{2}$, so α ≡ a$_{0}$ + a$_{1}$ (1 − ζ$_{p}$ ) (1 − ζ$_{p}$ ) $^{2}$. We want to eliminate the (1 − ζ$_{p}$ ) term by multiplying both sides by ζ n p for some n ∈ Z . Notice that ζ$_{p}$ = (1 − (1 − ζ$_{p}$ )). So modulo (1 − ζ$_{p}$ ) $^{2}$, ζ n $_{p}$α ≡ ζ n $_{p}$a$_{0}$ + a$_{1}$ζ n $_{p}$(1 − ζ$_{p}$ ) ≡ a$_{0}$ (1 − (1 − ζ$_{p}$ )) n + a$_{1}$ (1 − ζ$_{p}$ )(1 − (1 − ζ$_{p}$ )) n ≡ a$_{0}$ (1 − n (1 − ζ$_{p}$ )) + a$_{1}$ (1 − ζ$_{p}$ )(1 − n (1 − ζ$_{p}$ )) since considering (1 − (1 − ζ$_{p}$ )) n as a polynomial in (1 − ζ$_{p}$ ), (1 − ζ$_{p}$ ) 2 divides (1 − ζ$_{p}$ ) i for i ≥ 2. So ζ n $_{p}$α ≡ a$_{0}$ + ( a$_{1}$ − na$_{0}$ )(1 − ζ$_{p}$ ) (1 − ζ$_{p}$ ) 2