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?

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 $x-\sigma(\zeta_p)$.
