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How can one show that every nonzero element $x$ of the ring $\mathbb{Z}[\sqrt{35}]$ is contained in finitely many ideals? It is obvious in case of $x$ being invertible, but a general case is out of my sight. Is something special about the number $35$ (except it is composite)? The ring is not UFD ($35=5\cdot 7=\sqrt{35}\cdot\sqrt{35}$), and so neither it is PID, thus the standard factorization argument does not work here. However, this ring is Noetherian -- maybe it would be helpful somehow?

I will appreciate any hints. TIA.

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    Yes, thank you. I've just noticed after sending the comment that as ideals $(a+b\sqrt{35})=\mathbb{Z}[\sqrt{35}](a+\sqrt{35}b)$, which is different than that what$I$wrote. Thank you for your help!2012-06-07

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Hint $\ $ If ideal $\rm\:I\:$ contains $\rm\:\alpha\ne 0\:$ then $\rm\:I\:$ contains its norm $\rm\: 0\ne n = \alpha\alpha'\in \mathbb Z.\:$ Therefore

$\rm\:I = (n,\,\beta_1,\,\beta_2,\,\ldots\,)\ \Rightarrow\ I = (n,\,\beta_1\,mod\: n,\,\beta_2\,mod\:n,\,\ldots\,)$

But there are only finitely many $\rm\:\! \beta_i\,mod\: n,\:$ so only finitely many such reduced generating sets.