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.
Best Answer
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.