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Let $p \neq 2$ be a prime and $\zeta_p$ be a (nontrivial) root of unity. It is well known that $(1-\zeta_p)^{p-1}$ is divisible by $p$ inside $\mathbb{Z}[\zeta_p]$, i.e., there exists $a \in \mathbb{Z}[\zeta_p]$ such that $(1-\zeta_p)^{p-1}= pa$ inside $\mathbb{Z}[\zeta_p]$.

(Remark. As @KCd remarked this ring is in general not a factorial ring, as I initially wrongly assumed.)

(sketch/idea: calculation performed modulo $p$ gives inside $\mathbb{F}_p[X]$ the factorization $X^p-1=(X-1)^p=(X-1)(X-1)^{p-1}$; now inside $X:= \zeta_p$ and use that $\mathbb{F}_p[\zeta_p]$ is a domain, implying $(\zeta_p-1)^{p-1}$ is zero inside $\mathbb{F}_p[\zeta_p]$, so it lifts to an element in the principal ideal $p\mathbb{Z}[\zeta_p]$)

Question: Is it possible to calculate $a \in \mathbb{Z}[\zeta_p]$ satisfying $(\zeta_p-1)^{p-1}=pa$ explicitly?

user26857
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user267839
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    The term "factorial ring" means unique factorization domain, and $\mathbf Z[\zeta_p]$ for prime $p$ is not factorial when $p \geq 23$. It is a Dedekind domain for all primes $p$. – KCd Apr 13 '24 at 19:39
  • @KCd: Thanks, for that subtile point, I wasn't aware of it until now! Could you recommend a source discussing this issue on (non)factoriarity of $\mathbb{Z}[\zeta_p]$ in dependence of $p$? – user267839 Apr 16 '24 at 09:47
  • @Addendum: That's also very interesting, because Dedekind domains are regular, but only local regular rings are factorial (Auslander Buchsbaum Thm), so this ring witnesses that locality is crucial to get factoriarity from regularity. – user267839 Apr 16 '24 at 09:53
  • @user267839 it's been a while since I had a copy of the text so I can't remember the chapter number, but Washington's Introduction to Cyclotomic Fields covers the cases when $\mathbb{Q}(\zeta_n)$ has class number 1, but it is quite far into the book. The Brauer-Siegel theorem shows that there are only finitely many such fields, and then there is a chapter dedicated to determining exactly which ones have class number 1. – Warren Moore Apr 16 '24 at 10:02
  • @WarrenMoore: Right, thanks. Class number one implies a priori only principal ideal domain, but for Dedekind domains the latter is equivalent to beeing unique factorization domain. – user267839 Apr 16 '24 at 10:13

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Yes. For $0 < i < p$, write $g_i = \frac{\zeta^i - 1}{\zeta - 1}$. Then $g_i \in \mathbb{Z}[\zeta_p]$ (and not just the field of fractions) since $(x^i - 1)/(x-1) = 1 + x + \dots + x^{i-1}$. Further, $g_i$ is a unit in $\mathbb{Z}[\zeta_p]$, because $$ \frac{\zeta - 1}{\zeta^i - 1} = \frac{\zeta^M - 1}{\zeta^i - 1}, $$ for any integer $M$ that is $1$ mod $p$, and if we pick $M$ to be divisible by $i$, we can again use the "geometric series trick" to expand.

Now we have $g_i (\zeta - 1) = \zeta^i - 1$. Multiply this for all values of $i$, and we get $$ (\zeta - 1)^{p-1} \prod_{i=1}^{p-1} g_i = \prod_{i=1}^{p-1} (\zeta^i - 1). $$

Now, since the primitive $p$th roots of unity (together with $1$) are the roots of $x^p-1$, the right-hand side is the polynomial $(x^p-1)/(x-1) = 1 + x + x^2 + \dots + x^{p-1}$, evaluated at $x=1$, which is $p$.

We deduce that $$ (\zeta - 1)^{p-1}/p = \prod_{i=1}^{p-1} g_i^{-1}. $$

user26857
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hunter
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