Let $P(x)\in\mathbb{Z}[x]$ be monic and irreducible over $\mathbb{Q}[x]$, and let $\theta$ be a root of $P(x)$. Let $K = \{a + b\theta\} \subseteq \mathbb{Z}[\theta]$. When is it the case that there are infinitely many $\alpha\in K$ s.t. $\text{Nm}(\alpha) = 1$?


closed as off-topic by Venkataramana, Andreas Blass, Karl Schwede, Stefan Kohl, S. Carnahan Sep 27 '14 at 2:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Andreas Blass, Karl Schwede, Stefan Kohl, S. Carnahan
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    $\begingroup$ The notation $K$ is a bit unfortunate (since it will connote a field to many readers) for something that in general won't even be a ring. $\endgroup$ – GNiklasch Sep 26 '14 at 16:59
  • 1
    $\begingroup$ If $\deg(P) \geq 3$ there will only be finitely many by Thue's theorem. If $\deg(P) = 2$, it will depend on whether or not $\mathbb{Q}[\theta]$ is imaginary quadratic or real quadratic. $\endgroup$ – Jeremy Rouse Sep 26 '14 at 17:02

If and only if $\mathbb{Q}(\theta)$ is a real quadratic field.

In the imaginary quadratic case, and when $P$ has degree $1$, there are only finitely many units in the ring of integers of the field. When the degree of $P$ is at least $3$, the norm condition amounts to a Thue equation. In the real quadratic case, some power of the fundamental unit will be congruent to $1$ modulo the conductor ideal of the order and will thus be an element of the order along with its powers.


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