Lack of Unique Factorization in $\mathbb{Z}[\zeta_n]$ and Fermat's Last Theorem

Question

Given that $x^n + y^n =z^n$, $n \ge 3$ and $x,y,z \in \mathbb{N}$. Which can be rewritten as $$x^n = (z-y)(z- \zeta_{n}y) \dots(z-\zeta^{n-1}_{n}y)$$

I saw a remark that if, in general, $\mathbb{Z}[\zeta_n]$ has the unique factorization property, FLT could be proved by factoring $x$ and the various $(z-\zeta^{k}_{n}y)$, $k = 1,\dots,n-1$.

I can see that unique factorization being applicable in $\mathbb{N}$, then $x^n = p_{1}^n\dots p_{m}^n$ for the appropriate prime factors.

And if unique factorization were applicable, each term on the RHS of the equation above could also be factored.

And this as far as I can get.

So I would appreciate help to see how unique factorization (if valid) would have proved the theorem, and how this attempted proof actually fails in the absence of unique factorization.

(I would especially appreciate any answers that do not venture into prime ideals, since I feel this question precedes their advent.)

Answer 1

This proof can be found in many texts on algebraic number theory; see here for references. Basically we can write each $z-\zeta_n^iy=\alpha_i^n$ as an $n$-th power and then apply rather lengthy divisibility arguments in the ring $\mathbb{Z}[\zeta_n]$. This is just not true if the ring is not factorial. So this proof fails for all $n$ where $\mathbb{Z}[\zeta_n]$ is not a factorial ring, i.e., does not have class number $1$. In particular, it fails for all primes $p>19$, and for all $n>90$.