A Note on Cyclotomic Integers

Abstract

In this note, we present a new proof that the ring $\mathbb{Z}[{\zeta }_n]$ is the full ring of integers in the cyclotomic field $\mathbb{Q}({\zeta }_n)$.

MSC:

• Primary 11R18

Acknowledgments

The author is grateful to two anonymous reviewers and to Professor Keith Conrad for their many helpful comments and suggestions. This paper is dedicated to the memory of Alan Thorndike, former professor of physics at the University of Puget Sound and a dear friend, teacher, and mentor.

Notes

1 It can be shown that the number p is a generator of the local maximal ideal in this case.

2 It is clear that ${\Phi }_n(X)$ divides $X^n-1$ in $\mathbb{Q}[X]$. The fact that ${\Phi }_n(X)$ must also divide $X^n-1$ in $\mathbb{Z}[X]$ is a consequence of the Gauss–Kronecker lemma (used in proving that the ring $\mathbb{Z}[X]$ is a unique factorization domain), or the fact that the ring $\mathbb{Z}$ is integrally closed.

3 If q > 2, then ${({\zeta }_q-1)}^2$ divides p in the ring $\mathbb{Z}[{\zeta }_n]$ and $G({\zeta }_n)H({\zeta }_n)=({\zeta }_q-1)(1+\text{multiple} \text{of} ({\zeta }_q-1))$. In that case $G({\zeta }_n)$ also divides ${\zeta }_q-1$ in the local ring of $\mathbb{Z}[{\zeta }_n]$ at $\wp$. In other words, $G({\zeta }_n)$ and ${\zeta }_q-1$ are associates in that local ring, and the local maximal ideal can also be generated by $G({\zeta }_n)$.

As a practical matter, we can assume that q > 2 because if q = 2 and $n=2k$, where k is odd, then the field $\mathbb{Q}({\zeta }_n)$ and the ring $\mathbb{Z}[{\zeta }_n]$ are exactly the same as the field $\mathbb{Q}({\zeta }_k)$ and the ring $\mathbb{Z}[{\zeta }_k]$. So we can exclude the case when 2 divides n exactly to the power 1.