Any element of $\mathbf{Z}[\xi]$ is congruent to an integer modulo $(1-\xi)^2$ if multiplied by a suitable power of $\xi$

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?

Solution 1:

This is false - $ 1 - \xi $ can never be multiplied by a power of $ \xi $ to be an integer modulo $ (1 - \xi)^2 $. Let $ \mathfrak p = (1 - \xi) $ throughout the post.

To see this, note that $ p \equiv 0 \pmod{\mathfrak p^2} $ since the ideal $ (p) $ totally ramifies as $ (p) = \mathfrak p^{p-1} $, so if $ \xi^k (1 - \xi) $ is an integer modulo $ \mathfrak p^2 $, without loss of generality it can be taken to be in the set $ \{ 0, 1, \ldots, p-1 \} $. It cannot be $ 0 $, because that would imply $ 1 - \xi \in \mathfrak p^2 $ and $ \mathfrak p = \mathfrak p^2 $, but $ 1 - \xi $ is not a unit: it has norm $ \pm p $. It cannot be anything else in this set, because $ \mathbf Z \cap \mathfrak p = p\mathbf Z $, therefore all of the nonzero integers in the set are invertible modulo $ \mathfrak p $, and thus modulo $ \mathfrak p^2 $. Thus, such a congruence would imply that $ 1 - \xi $ is invertible modulo $ \mathfrak p^2 $, which is absurd, since the ideals $ (1 - \xi) = \mathfrak p $ and $ \mathfrak p^2 $ are not coprime.

Solution 2:

Perhaps you dropped some hypothesis in your statement, which should be: "Any unit $u$ in $\mathbf Z[\zeta]$ can be multiplied, etc." By Dirichlet's unit theorem (or by direct computation), $u$ is of the form $\zeta^r .v$, where $v$ is a unit in $\mathbf Z[\zeta +\zeta ^{-1}]$ (= the ring of integers of the maximal totally real subfield). But $(1-\zeta)(1-\zeta ^{-1}) = 2 -(\zeta +\zeta ^{-1})$, and $(1-\zeta)/(1-\zeta ^{-1})$ is a unit, hence taking quotients mod $(1-\zeta)(1-\zeta ^{-1})$ shows that every element of $\mathbf Z[\zeta +\zeta ^{-1}]$ is congruent to a rational number mod $(1-\zeta)^2$ QED

I guess that this property is a preliminary in the proof of the so called Kummer lemma, in which case you could fine more information in Washington's book, §5.6.