# Common factors of the ideals $(x - \zeta_p^k)$, $x \in \mathbb Z$, in $\mathbb Z[\zeta_p]$

I'm trying to understand a proof of the following Lemma (regarding Catalan's conjecture):

Lemma: Let $x\in \mathbb{Z}$, $2<q\not=p>2$ prime, $G:=\text{Gal}(\mathbb{Q}(\zeta_p):\mathbb{Q})$, $x\equiv 1\pmod{p}$ and $\lvert x\lvert >q^{p-1}+q$. Then the map $\phi:\{\theta\in\mathbb{Z}:(x-\zeta_p)^\theta\in\mathbb{Q}(\zeta_p)^{*q}\}\rightarrow\mathbb{Q}(\zeta_p)$, $\$ $\phi(\theta)=\alpha$ such that $\alpha^q=(x-\zeta)^\theta$ is injective.

I don't understand the following step in the proof:

The ideals $(x-\sigma(\zeta_p))_{\sigma \in G}$ in $\mathbb{Z}[\zeta_p]$ have at most the factor $(1-\zeta_p)$ in common. Since $x\equiv 1\pmod{p}$ the ideals do have this factor in common.

Could you please explain to me why these two statements are true?

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$\LaTeX$ tip: instead of (\text{mod }p), use \pmod{p}, which gives appropriate spacing. – Arturo Magidin Jun 27 '12 at 20:15
to be honest I'm not entirely sure wheater the ideals are in $\mathbb{Q}(\zeta_p)$ or $\mathbb{Z}[\zeta_p]$ and it isn't clear from the context. – Julian Jun 27 '12 at 20:48
@Julian Ideals in fields are not so interesting, so I think it's the latter. – Dylan Moreland Jun 27 '12 at 20:52

In the world of ideals gcd is the sum. Pick any two ideals $(x-\sigma(\zeta_p))$ and $(x-\sigma'(\zeta_p))$. Then $\sigma(\zeta_p)-\sigma'(\zeta_p)$ will be in the sum ideal. This is a difference of two distinct powers of $\zeta_p$, so generates the same ideal as one of the $1-\zeta_p^k$, $0<k<p$. All of these generate the same prime ideal as $(1-\zeta_p)$, because the quotient $$\frac{1-\zeta_p^k}{1-\zeta_p}=\frac{1-\zeta_p^k}{1-(\zeta_p^k)^{k'}}=\left(\frac{1-(\zeta_p^k)^{k'}}{1-\zeta_p^k}\right)^{-1}$$ is manifestly a unit in the ring $\mathbb{Z}[\zeta_p]$ (here $kk'\equiv1\pmod p$). Because the ideal $(1-\zeta_p)$ is a non-zero prime ideal, hence maximal, this gives the first claim you asked about.
The second claim follows from the well known fact that up to a unit factor (i.e. as a principal ideal) $(p)$ is the power $(1-\zeta_p)^{p-1}$. If $x=mp+1$, $m$ a rational integer, then $(1-\zeta_p)$ divides both $mp$ and $(1-\sigma(\zeta_p))$, and therefore also $x-\sigma(\zeta_p)$.