coprime terms in the factorization of $x^p + y^p$

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: $$$$z^p = (x+y)(x+y\zeta)\cdots(x+y\zeta^{p-1})$$$$ where $$(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

• I reckon they are coprime iff $p\nmid z$. – Angina Seng Jun 24 at 17:59
• Where is the intuition coming from? – FanFanKa Jun 24 at 18:12
• Intuition? What's that? – Angina Seng Jun 24 at 18:27
• I mean where $p\nmid z$ comes from? – FanFanKa Jun 24 at 18:44

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.
• What is Ex. $16$? – saulspatz Jun 24 at 18:24
• Thanks for the answer. I am also confused by Ex. 16 and also seems $\mathbb{Z}[\omega_p]$ is UFD only when $p\leq19$. – FanFanKa Jun 24 at 18:43
• 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. – FanFanKa Jun 24 at 23:48