Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.