Show that a specific ideal is not principal

Question

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?

Thanks.

Answer 1

$\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. $\rm\:\beta\beta'\ |\ (4,10)= 2$

Answer 2

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.

Answer 3

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.