# Every nonzero $x\in\mathbb{Z}[\sqrt{35}]$ belongs to finitely many ideals

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.

-
Can you show that the ideal generated by $x$ has a finite index in the ring (as an abelian group)? A finite ring obviously has finitely many ideals. Apply the correspondence principle. –  Jyrki Lahtonen Jun 7 '12 at 11:08
@JyrkiLahtonen: Ok, I understand the second part of the hint. But is there any nice and short method for proving that $[\mathbb{Z}[\sqrt{35}]:(a+b\sqrt{35})]<\infty$? I have thought about presenting $\mathbb{Z}[\sqrt{35}]$ as $\mathbb{Z}\times\mathbb{Z}$ (those two are isomorphic as abelian groups) and dividing by the subgroup induced by an ideal. Right way? –  Damian Sobota Jun 7 '12 at 11:45
Go, go, go!${{}}$ –  Jyrki Lahtonen Jun 7 '12 at 11:51
@JyrkiLahtonen: Every ideal $(a+b\sqrt{35})$ seen as a subgroup of $\mathbb{Z}\times\mathbb{Z}$ is a subgroup generated by the pair $(a,b)$, so $(a+b\sqrt{35})=\mathbb{Z}a+\mathbb{Z}b$. We get that $(\mathbb{Z}\times\mathbb{Z})/(\langle a\rangle\times\langle b\rangle)\cong(\mathbb{Z}/a\mathbb{Z})\times(\mathbb{Z}/b\mathbb{Z})$. Thus the only problem left is the case when $a=0$ xor $b=0$. Is that correct what I have just said? –  Damian Sobota Jun 7 '12 at 12:16
No. That's mostly wrong. The subgroup generated by $(a,b)$ only contains pairs $(ka,kb)$ with $k$ any integer, and this always has infinite index in $\mathbb{Z}\times\mathbb{Z}$. But you are not interested in subgroups. You're interested in ideals. The ideal generated by $x=a+b\sqrt{35}$ also contains the element $\sqrt{35}x=(35b)+a\sqrt{35}$. So you will be looking at the subgroup $I$ generated by $(a,b)$ and $(35b,a)$. –  Jyrki Lahtonen Jun 7 '12 at 12:23
show 1 more comment

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.