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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)

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What do you mean by ambiguous? – tomasz Dec 12 '14 at 9:21

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.

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