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Let $\zeta=e^{\frac{2\pi i}{p^n}}$, $p$ is an odd prime. Is $\mathbb{Z}[\zeta]$ a UFD?

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    $\begingroup$ That would make Fermat's Last Theorem a lot easier to prove! $\endgroup$
    – Derek Holt
    Commented Sep 5, 2013 at 8:38

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The ring of integers $\mathbb{Z}[\zeta_p]$ is an UFD if and only if the class number of $\mathbb{Q}[\zeta_p]$ is $1$. This is the case if and only if $p\le 19$. For bigger primes $p$ the class numbers $h(p)$ grow rapidly:

$$ h(23)=3, \; h(29)=8, \; h(31)=9, \ldots ,h(89)=13379363737, $$

$$ h(137)=646901570175200968153, etc. $$

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Some more can be said here, though. For instance, writing ${\bf{Q}}(\zeta_{p^n})$ to denote the extension of ${\bf{Q}}$ obtained by adjoining a primitive $p^n$-th root of unity $\zeta_{p^n}$ (for $p > 2$ and $n \geq 1$), one can consider the analogous question for the maximal totally real subfields ${\bf{Q}}_n = {\bf{Q}}(\zeta_{p^{n+1}})^+$ of ${\bf{Q}}(\zeta_{p^{n+1}})$, i.e. the degree-$p^n$ extensions of ${\bf{Q}}$ contained in the cyclotomic ${\bf{Z}}_p$-extension of ${\bf{Q}}$. The following intriguing results are then known. Let $h_n$ denote the class number of the ring of integers of the cyclotomic extension ${\bf{Q}}_n$. Let $e_n = \operatorname{ord}_p(h_n)$ denote the exponent of $p$. Iwasawa proved that there exist integers $\lambda, \mu$, and $\nu$, independent of $n$, such that $e_n = \lambda n + \mu p^n + \nu$ for all $n$ sufficiently large. Ferrero and Washington later proved that $\mu = 0$ in this setting. There are some hard open conjectures (e.g. due to Greenberg) about this sort of behaviour of class numbers in more general settings, though very little seems to be known or understood at present ...

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