Why is such an ideal ambiguous? Suppose I have an $R$-ideal $I$ with 
$$I=(1-\zeta)^n XR$$
with $R=\mathbb{Z}[\zeta]$, $\zeta$ a primitive $p$-th root, $X$ an ideal in the integral closure of $\mathbb{Z}$ in $\mathbb{Q}[\zeta + \zeta^{-1}]$ (would it be $\mathbb{Z}[\zeta + \zeta^{-1}]$ ?) and the exponent $n$ is $0$ or $1$. Why is such an ideal ambiguous? I can calculate it just for the case where $X$ is ambiguous. But it should work in general.
Thanks for help!
p.s. maybe anyone even knows where I could find the proof for the other direction too? (Each ambiguous $R$-ideal can be written in the form above)
 A: Apologies if this is wrong, but it seems to me that what you're trying to prove is not true...
In general when $n=0$ $I$ is ambiguous iff $X$ is ambiguous.
Indeed when it is a subring of $R$, the integral closure $C$ of $\mathbb Z$ in $\mathbb Q[\zeta+\zeta^{-1}]$ is the set of fixpoints of conjugation in $R$ and conjugation commutes with all automorphisms, so $\sigma(X)\subseteq XR$ implies $\sigma(X)\subseteq XC=X$ therefore $X$ is ambiguous.
For a more concrete example, let $p=5$ and $n=0$, then $C$ is indeed a subring of $R$ because $C=\mathbb Z[\zeta+\zeta^{-1}]$.
Let $X = (\zeta+\zeta^{-1}-3)C$.
If $I$ is ambiguous, then given the automorphism $\sigma$ that maps $\zeta$ to $\zeta^2$, we should have
$$\sigma(\zeta+\zeta^{-1}-3)=\zeta^2+\zeta^{-2}-3=-\zeta-\zeta^{-1}-4\in XR$$
and therefore $-\zeta-\zeta^{-1}-4 = (\zeta+\zeta^{-1}-3)r$ for some $r\in R$.
However in $\mathbb Q[\zeta]$ this has the only solution $7/11(\zeta+\zeta^{-1}) + 17/11$, which does not belong to $R$.
Therefore $I$ cannot be ambiguous.
