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This is a question related to Fermat's last theorem. Let $p\geq5$ be a prime number, and let $\zeta$ be a primitive $p$th root of unity. Consider the Fermat's last theorem: \begin{equation} z^p = (x+y)(x+y\zeta)\cdots(x+y\zeta^{p-1}) \end{equation} where $(x,y,z)$ are pairwise coprime in $\mathbb{Z}$. How to prove that the terms on the right-hand side are coprime in $\mathbb{Z}[\zeta]$. Does this only apply to the "first case" of the theorem(i.e. $p\nmid xyz$)?

Some background: if we have this conclusion and together if we assume that $\mathbb{Z}[\zeta]$ is a UFD, this would be a starting point to prove the theorem. But the assumption $\mathbb{Z}[\zeta]$ is a UFD is not true in general. More detailed discussions can be found in section 1.5 of chapter 1 of Matthew Baker's lecture note http://people.math.gatech.edu/~mbaker/pdf/ANTBook.pdf

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    $\begingroup$ I reckon they are coprime iff $p\nmid z$. $\endgroup$ – Angina Seng Jun 24 at 17:59
  • $\begingroup$ Where is the intuition coming from? $\endgroup$ – FanFanKa Jun 24 at 18:12
  • $\begingroup$ Intuition? What's that? $\endgroup$ – Angina Seng Jun 24 at 18:27
  • $\begingroup$ I mean where $p\nmid z$ comes from? $\endgroup$ – FanFanKa Jun 24 at 18:44
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Let $\omega_p=e^{2\pi i/p}$. We have the following factorization of $x^p+y^p=z^p$ in $\mathbb{Z}[\omega_p]$, $$(x+y)(x+\omega_py)\ldots(x+\omega_p^{p-1}y)=z^p$$ Let $\pi$ be some irreducible element in $\mathbb{Z}[\omega_p]$ such that $\pi\mid(x+\omega_py)$. Let $\pi\mid(x+\omega_p^jy)$ for some $j\neq1$. Let $\mathrm{WLOG}$ $j>1$. Then $\pi\mid(x+\omega_py-x-\omega_p^jy)=y\omega_p(1-\omega_p^{j-1})$. Therefore the fact that $\omega_p^p=1$ we get, $$\pi\mid y\omega_p(1-\omega_p^{j-1})\omega_p^{p-1}\prod\limits_{\substack{i\neq(j-1)\\ 1\leq i\leq(p-1)}}(1-\omega_p^i)=y\omega_p^p\prod_{i=1}^{p-1}(1-\omega_p^i)=yp$$ Since $\mathbb{Z}[\omega_p]$ is assumed to be a $\mathrm{UFD}$, $\pi$ is a prime. Hence $\pi\mid(x+\omega_py)\mid z^p$ implies $\pi\mid z$. Since $z,yp$ are assumed to be relatively prime, $\exists$ $m,n\in\mathbb{Z}$ such that $zm+nyp=1$. This implies $\pi\mid1$ and hence $\pi$ is a unit. This is a contradiction since $\pi$ is irreducible. Similarly, we can show that no two distinct elements on the right-hand side product share the same irreducible factor. Hence they are pairwise relatively prime.

@AnginaSeng is correct, it holds for the case $p\nmid z$ and $x,y,z$ are relatively prime.

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    $\begingroup$ What is Ex. $16$? $\endgroup$ – saulspatz Jun 24 at 18:24
  • $\begingroup$ Thanks for the answer. I am also confused by Ex. 16 and also seems $\mathbb{Z}[\omega_p]$ is UFD only when $p\leq19$. $\endgroup$ – FanFanKa Jun 24 at 18:43
  • $\begingroup$ Sorry, I was also writing my analysis homework in Overleaf, there I have copied one phrase, so mistakenly I pasted it here while writing. Sorry to everyone. $\endgroup$ – Shubhrajit Bhattacharya Jun 24 at 19:04
  • $\begingroup$ I think no need to assume $\mathbb{Z}[\omega_p]$ to be a UFD. Since from $z, yp$ are relatively prime, we have $z^p, yp$ are relatively prime. We still can conclude $\pi | 1$, which is a contradiction. $\endgroup$ – FanFanKa Jun 24 at 23:48

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