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Suppose $a,m\in\mathbb Z_{\ge2}$. Let's consider the ring $A=\mathbb Z[(1+\alpha)/2]$, where $\alpha^2=1-4a^m$, and the ideal $I=(a,(1+\alpha)/2)$, we need to show that $I^n$ is non-principal when $0<n<m$.

It's easy to show by induction that $I^n=(a^n,(1+\alpha)/2)$, and we have $A/a^nA\cong\mathbb Z/a^n\mathbb Z\times\mathbb Z/a^n\mathbb Z$, $A/I^n\cong\mathbb Z/a^n\mathbb Z$. Note that $I^m=((1+\alpha)/2)$, which is principal.

Since the case is a bit more general than computations in specific quadratic fields, maybe there's more general methods to determine whether an ideal is principal.

Any help is welcome. Thanks!

Yai0Phah
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  • I think that here you are expected to rely on specific properties of imaginary quadratic fields. Most notably the norm. See this related question, where the asker is using the norm to exclude the possibility that the ideal could be principal. Not sure that the technique matches your needs exactly, but it sure rang a bell. – Jyrki Lahtonen Nov 10 '15 at 09:31
  • @JyrkiLahtonen Thanks. The norm of $a^n$ or $(1+\alpha)/2$ is a power of $a$, not necessarily a prime or primary, sadly. Additional note: it's a starred exercise in a mid-term examination of algebra mostly at undergraduate level (i.e, that of M.Artin's Algebra, say), and I failed to work it out during the exam. – Yai0Phah Nov 10 '15 at 09:52
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    The element of minimal norm in $A \setminus Z$ is $(1+\alpha)/2$. So all you need to do is verify that the powers of I are not generated by integers. –  Nov 12 '15 at 21:25
  • @franzlemmermeyer Therefore a fortiori $(1+\alpha)/2$ is irreducible, and a principal ideal containing an irreducible element should be an ideal generated by that, or the unity ideal, right? Thanks! – Yai0Phah Nov 13 '15 at 09:06

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