In some cases, it is quite straightforward to prove that a specific ideal cannot be principal. For example, in the ring of integers of $\mathbb{Q}(\sqrt{-5})$, the ideal $(2,1+\sqrt{-5})$ is not principal, by taking norms (since this is one of the ideals in the factorization of 2).

However, in that case, we used that the norm of a generating element would have to equal $\pm 2$.

Now, let $K=\mathbb{Q}(\sqrt{-39})$ and let $I=(2,\alpha-1)$ (where $\alpha$ the root of $x^2-x+10$, the minimal polynomial of $\sqrt{-39}$). I want to show that this ideal is not principal. (specifically, it is the square of one on the primes in the factorization of $2\mathcal{O}_K$).

Any suggestions?


  • $\begingroup$ What is $\alpha$? $\endgroup$ – Adrián Barquero Apr 12 '11 at 17:18
  • $\begingroup$ Ops, I forgot. I've added what $\alpha$ is now. Thanks. $\endgroup$ – Fredrik Meyer Apr 12 '11 at 17:22
  • $\begingroup$ How does your original argument not work? I suppose you need to be a little more careful since the ring of integers is $\mathbb{Z}\left[\frac{1+\alpha}{2}\right]$, but you can still take norms and argue that the norm of a prospective generator would have to divide $N(2) = 4$ and $N(\alpha-1) = 40$. Then search for elements $(a + b\alpha)/2$, where $a \equiv b$ modulo $2$, such that $N((a + b\alpha)/2) = (a^2+40b^2)/4$ divides $4$. I think you need to check norms up to 16. $\endgroup$ – Michael Chen Apr 12 '11 at 17:44
  • $\begingroup$ I don't think you're correct when you say $(2,\alpha-1)=Q^2$ for some prime $Q$ dividing 2 - if $2{\cal{O}}_K=Q^2P_1^{e_1}\cdots P_r^{e_r}$, then $$e(Q/2)f(Q/2)+\sum_{i=1}^re(P_i/2)f(P_i/2)=2f(Q/2)+\sum_{i=1}^r e_if(P_i/2)=2$$ forces all $e_i=0$, i.e. $Q^2=2{\cal{O}}_K$. $\endgroup$ – Zev Chonoles Apr 12 '11 at 17:53
  • $\begingroup$ @Michael Chen: Note that we've developed some conflicting notation in the question and comments. Fredrik Meyer says that $\alpha$ is a root of $x^2-x+10$ (which is not, as he also writes, a minimal polynomial of $\sqrt{-39}$). So, we may take $\alpha = \frac{1+\sqrt{-39}}{2}$. Whichever root we take, the ring of integers is $\mathbb{Z}[\alpha]$ and the norm of $\alpha-1$ is $10$. (Is my interpretation correct, Fredrik Meyer?) $\endgroup$ – Jonas Kibelbek Apr 12 '11 at 17:55

$\rm (\beta)\: =\: (2,\:\alpha-1)\ \Rightarrow\ (\beta\:\beta')\: =\: (2)\ \Rightarrow\Leftarrow\ $ via $ \rm\ (2,\:\alpha-1)\ (2,\:\alpha'-1)\: =\: (4,\:10,\:2\alpha-2,\:2\alpha'-2)\: =\: (2)$

Simpler, avoiding (conjugate) ideals: $\rm\ \beta\ |\ 2,\:\alpha-1\ \Rightarrow\ N(\beta)\ |\ N(2),\:N(\alpha-1)\:,\:$ i.e. $\rm\:\beta\beta'\ |\ (4,10)= 2$

  • $\begingroup$ I'm not sure if I understand your answer. What are $\beta^\prime$ and $\alpha^\prime$? $\endgroup$ – Fredrik Meyer Apr 12 '11 at 23:10
  • $\begingroup$ @Fred: Conjugates. Multiply the first equation by its conjugate to get the second equation. I appended a simpler form avoiding conjugate ideals. $\endgroup$ – Bill Dubuque Apr 13 '11 at 0:14
  • $\begingroup$ So the argument is this?: if $(2,\alpha-1)$ is principal, it would divide $2$ which is impossible since it is a power of a factor of two. $\endgroup$ – Fredrik Meyer Apr 13 '11 at 0:17
  • $\begingroup$ @Fredrik: Could you clarify that last comment? I was thinking that, Bill's answer implies $N(\beta) = 2$, and you can show that there is no element with norm 2: $N\left(\frac{a+b\sqrt{-39}}{2}\right) = \frac{a^2+39b^2}{4} = 2$. $\endgroup$ – Michael Chen May 7 '11 at 15:23

I think that again an approach by contradiction using norms does work. Since $\alpha$ is a root of $x^2 - x + 10$ then

$$\alpha = \frac{1 \pm \sqrt{-39}}{2}$$

So for example if you pick the negative sign then you want to show that the ideal $$I = \left \langle 2, \frac{1 - \sqrt{-39}}{2} - 1 \right \rangle = \left \langle 2, \frac{1 + \sqrt{-39}}{2} \right \rangle$$

is not principal. So if you suppose it is principal then it may be of the form $I = \langle a + b\sqrt{-39} \rangle$ for $a, b \in \mathbb{Z}$ or $I = \left \langle \frac{a + b\sqrt{-39}}{2} \right \rangle$ with $a, b \in \mathbb{Z}$.

Then in the first case by taking norms you get $(a^2 + 39b^2) | 2$ because it divides $\mathrm{\textbf{N}}(2) = 4 $ and $\mathrm{\textbf{N}} \left ( \frac{1 + \sqrt{-39}}{2} \right ) = 10$. This case is impossible because the corresponding diophantine equation has no solutions in integers.

And well, in the other case the only difference is that you get

$$\frac{a^2}{4} + \frac{39b^2}{4} | 2 \implies a^2 + 39b^2 | 8$$

And again a case by case analysis shows that this is not possible. So the ideal is not principal.


Let me just say that in principle you should be able to decide whether any ideal $I$ in an imaginary quadratic number ring $R = \mathbb{Z}[\sqrt{-d}]$ is principal or not by looking at norms. Indeed, because the unit group of $R$ is finite (usually just $\pm 1$ in fact), there are going to be at most finitely many elements $\alpha \in R$ with $N(\alpha) = N(I)$ and you could write a short piece of code to enumerate them all. Assuming that $I = \langle \beta_1,\ldots,\beta_n \rangle$ (we can arrange for $n = 2$, but never mind that) we just check whether $\beta_i/\alpha \in R$ for all $i$. If so, then $\langle \alpha \rangle \supset I$, and since they have the same norm they must be equal.

Added: My answer above is unnecessarily cautious. Of course for any $\alpha \in R$ and any unit $u$ of $R$, the principal ideals $\langle \alpha \rangle$ and $\langle u \alpha \rangle$ are equal. Thus the above argument works whenever you have a ring $R$ for which you can algorithmically determine all elements $\alpha$ with $\# R/\langle \alpha \rangle = N$ for any $N \in \mathbb{Z}^+$. All rings of integers of number fields have this property. Of course there are probably much more efficient algorithms than this: unfortunately my knowledge of the algorithmic aspects of algebraic number theory is very poor.


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