I cannot find a good reference for the proof that the ring of integers in a cyclotomic field $\mathbb{Q}(\zeta_n)$ is $\mathbb{Z}[\zeta_n]$. The proof I usually find does an induction on the number of prime factors of $n$, coupled with a lengthy and somewhat computational proof in the case where $n$ is the power of a prime.

Do you know a quicker and possibly more conceptual approach?

  • $\begingroup$ I have rolled back to the previous version, since LaTeX is rendered in the title too. Is there any reason why it was removed? $\endgroup$ Mar 6, 2010 at 17:04
  • $\begingroup$ Ok, I have reverted to the version without LaTeX, since that is not rendered on the home page. $\endgroup$ Mar 6, 2010 at 17:16

3 Answers 3


The ring of integers $R_n$ of ${\mathbf Q}(\zeta_n)$ contains ${\mathbf Z}[\zeta_n]$ as a subring with finite index. To show the containment of rings is an equality, it suffices to show the inclusion ${\mathbf Z}[\zeta_n] \rightarrow R_n$ becomes an isomorphism after tensoring with ${\mathbf Z}_p$ for all $p$, and this basically boils down to showing the ring of integers of ${\mathbf Q}_p(\zeta_n)$ is ${\mathbf Z}_p[\zeta_n]$ for all $p$. Now you have to know something about how to compute rings of integers in unramified and totally ramified extensions of local fields. I'm leaving off some details here, admittedly, and since I used the terrible word "compute" maybe this isn't an answer you are looking for.

You didn't tell us whether you were okay with the inductive argument but disliked (apparently) the prime-power case or you were unhappy with both aspects.

  • $\begingroup$ I'm fine with the inductive argument; it is a simple reduction step using the fact that $\mathbb{Q}(\zeta_{mn})$ is the composite of $\mathbb{Q}(\zeta_{m})$ and $\mathbb{Q}(\zeta_{n})$ if $m$ and $n$ are relatively prime. But the standard proof for the prime power case is not very illuminating. Your answer and that of Arno make me understand better what's going on. $\endgroup$ Mar 7, 2010 at 0:06
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    $\begingroup$ Suppose K = Q(a) where a is the root of a poly. in Z[x] that's Eisenstein with respect to p. Using the Eisenstein condition, the natural map Z[a] --> O_K/pO_K has kernel pZ[a], so Z[a] and pO_K meet in pZ[a]. Therefore the group O_K/Z[a] has no element of order p, so [O_K:Z[a]] is not divisible by p. At the same time this index is a factor of disc(f), so if you are in the case that disc(f) is a p-power (up to sign), then [O_K:Z[a]] is both not divisible by p and is a power of p, so the index is 1: O_K = Z[a]. When a = zeta_{p^r} - 1 this happens, so O_K = Z[zeta_{p^r}-1} = Z[zeta_{p^r}]. $\endgroup$
    – KConrad
    Mar 7, 2010 at 1:21
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    $\begingroup$ Very nice. In the course I am teaching on local fields, we just did the structure theory of unramified and totally ramified extensions, so this will make a great homework problem! $\endgroup$ Mar 7, 2010 at 3:18

In case of a prime power it is not so hard. Denote $R$ the ring of integers of ${\mathbf Q}(\zeta_n)$. First note that $X^n -1$ is separable mod $p$ when $p$ does not divide $n$. This implies $R [1/p] = {\mathbf Z}[\zeta_n, 1/p]$. To check that $R = {\mathbf Z}[\zeta_n]$ it then suffices to show that the local rings of ${\mathbf Z}[\zeta_n]$ at all primes above $p$ are DVRs. But you have the explicit element $\lambda = 1 - \zeta$ that you defined above and its norm equals $p$. This implies that there is only one prime above $p$ and that this ideal is generated by $\lambda$. In particular the local ring at this prime is regular and hence a DVR.

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    $\begingroup$ after a look in Langs alg number theory, I think the "taking-tensor-products-to-reduce-to-prime-powers" argument is the easiest. It is really quick: take a look on p 68 prop 17, you only need some basic properties of the different of a number field. $\endgroup$
    – mnr
    Mar 8, 2010 at 9:33

I don't know if this is what you're looking for, but I like the proof in Neukirch's Algebraic Number Theory Section I.10. Conceptually the idea (for $n$ a prime power) is that if the discriminant is a prime power and there is an element $\lambda$ whose norm is that prime, then $\mathbf{Z}[\lambda]$ should be the ring of integers. Here, $\lambda = 1- \zeta$ so $\mathbf{Z}[\lambda] = \mathbf{Z}[\zeta]$.

Then for general $n$ you use the fact that $\mathbf{Q}(\zeta_n)$ is the compositum of $\mathbf{Q}(\zeta_{\ell^r})$, and $\zeta_{\ell^r}$ is a power of $\zeta_n$.

  • $\begingroup$ I fear this is the "computational" proof I already know. By the way, is your second sentence an actual theorem? This may help streamlining the proof. $\endgroup$ Mar 6, 2010 at 16:49
  • $\begingroup$ I don't know that it's a significant enough phrase to call a theorem, but it's a good way to find the ring of integers when you have a totally ramified prime in $\mathbf{Z}[\lambda]$. $\endgroup$
    – stankewicz
    Mar 6, 2010 at 16:58
  • $\begingroup$ @stankewicz: Ok, but apart from what we call it, is that true? Namely: if $\lambda$ is an algebraic integer in a number field with norm $p$ and if the discriminant is a power of $p$, then the ring of intgers is $\mathbb{Z}[\lambda]$. $\endgroup$ Mar 6, 2010 at 17:07
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    $\begingroup$ @Ferretti: If I were to state it as a theorem, it would be this: Let $K$ be an algebraic number field whose discriminant is a power of $p$ a prime number which is totally ramified in $K$. If $\mathbf{Z}[\lambda]$ is an order of $K$ in which $p$ totally ramifies then $\mathbf{Z}[\lambda]$ is the ring of integers. $\endgroup$
    – stankewicz
    Mar 6, 2010 at 17:23
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    $\begingroup$ @Andrea Ferreti: The discriminant of $\mathbf Z[\lambda]$ is the discriminant of the ring of integers $R$ times the $d$th power of the index of $\mathbf Z[\lambda]$ in $R$, where $d$ is the degree of the extension. So if the discriminant of $\mathbf Z[\lambda]$ has no $d$th powers, one must have $\mathbf Z[\lambda]=R$. $\endgroup$
    – ACL
    May 27, 2010 at 15:27

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