TITLE: Any element of $\mathbf{Z}[\xi]$ is congruent to an integer modulo $(1-\xi)^2$ if multiplied by a suitable power of $\xi$
QUESTION [10 upvotes]: I'm currently reading Kummer's famous paper on Fermat's Last Theorem (if anyone wants the link, I'll post it, but the paper is in German). There's the following statement in there, which should be "very easy to prove": 

Let $\xi$ be a primitive $p$-th root of unity with $p$ an odd prime. Any element $x \in \mathbf{Z}[\xi]$ can be multiplied with a power $\xi^r$ such that it is congruent to an integer mod $(1-\xi)^2$. 

I've tried to prove it, but I need help. I thought one could write up a basis $1, (1-\xi)^2, ..., (1-\xi)^{p-1}$, but this is wrong, as has been shown in an answer to an earlier question of mine. 
Thanks!
Added Bounty-question: I went through the paper again and saw that I missed the hypothesis that $x$ does not lie in the ideal $(1 - \xi)$. It didn't really help me proving it though. Can anyone find a proof under this stronger condition?

REPLY [2 votes]: With your additional hypothesis that $x$ does not lie in the prime ideal $(1- \zeta)$, the question becomes much easier, because of its now "local" nature. To simplify notations, put $\pi=1 - \zeta $  throughout and consider the ring $A = \mathbf Z_p[\zeta]$, where $\mathbf Z_p$ denotes the $p$-adic integers. The group  $U$ of units (= invertible elements) of $A$ is endowed with a descending filtration $U > U_1 > ... > U_n > ...$, where $U_n = 1+ (\pi^{n})$ . The residual field  $k = A/(\pi)$ is isomorphic to $\mathbf F_p$ because $p$ is totally ramified, and the residue map $A \to \mathbf F_p$ induced by $\zeta \to 1$ gives rise to an isomorphism $U/U_1 \cong \mathbf F_p^{*}$ (multiplicative group), whereas for $n\ge 1$ the map $u \to u - 1$ gives rise to $U_n/U_n+1 \cong (\pi^{n})/(\pi^{n+1})\cong \mathbf F_p$ (additive group) (see e.g. Cassels-Fröhlich, chap. 1, propos. 4). One gets in particular an exact sequence  $0 \to \mathbf F_p \to U/U_2 \to \mathbf F_p^{*} \to 1$, which is split because the two extreme terms have coprime orders. Given the explicit isomorphisms above, this shows that $U/U_2 = <\zeta>.\mathbf Z_p$ , hence for any $u\in U$ , a certain $u.\zeta^{r}$ will be congruent to $a$ mod $(\pi^{2})$, with $a \in\mathbf Z_p$. Coming back to our $x \in \mathbf Z[\zeta], \notin (\pi)$, and using the density of $\mathbf Z$ in $\mathbf Z_p$,  a certain $x.\zeta^{r}$ will be will be congruent to $b$ mod $(\pi^{2})$, with $b \in\mathbf Z$ as desired.
The $p$-adics were unknown at Kummer's time, but the above proof can be adapted to be (probably)more in Kummer's spirit. Consider the ring $B = \mathbf Z[\zeta]= \mathbf Z[\pi] $. Every element $x \in B$ can be uniquely written under the form $x = a_0 + a_1\pi + ... + a_{p - 1}\pi^{p-1}$, with $a_n \in \mathbf Z$.   Analogously to above, the natural exact sequence  $ 0 \to (\pi)/(\pi^{2}) \to B/(\pi^{2}) \to B/(\pi) \cong \mathbf F_p \to 0$ gives rise to another exact sequence  $ 0 \to (\pi)/(\pi^{2})\cong \mathbf F_p \to (B/(\pi^{2}))^{*} \to (B/(\pi))^{*} \cong \mathbf F_p^{*} \to 0$, where $(.)^*$ denotes the units.  The hypothesis that $x \notin (\pi)$ means that $a_0$ is not a multiple of $p$, hence by the definition of the residue map, that $x$ mod $(\pi)$ $\in (B/\pi)^{*}$. Analogously to above, one can conclude that a certain $x.\zeta^{r}$ will be will be congruent to $b$ mod $(\pi^{2})$, with $b \in\mathbf Z$ as desired.
NB: This second proof is just a formalization of the one given by @Starfall.