Are the $\{\alpha \in \mathbb{Q}(\theta) | Tr(\alpha\cdot\mathbb{Z}[\theta]) \subset \mathbb{Z}\}$ the algebraic integers of $\mathbb{Q}(\theta)$?

Question

Are the $\{\alpha \in \mathbb{Q}(\theta) \,\,|\,\, Tr(\alpha\cdot\mathbb{Z}[\theta]) \subset \mathbb{Z}\}$ the algebraic integers of $\mathbb{Q}(\theta)$?

$\theta$ is a complex root of a monic irreducible polynomial $f$; $Tr$ denotes the Trace.

Answer 1

Not in general. In the case where $\Bbb Z[\theta]$ is the ring of integers the $\alpha$ with Tr$(\alpha \Bbb Z[\theta])\subseteq\Bbb Z$ form a fractional ideal of the ring of integers, the inverse different that contains the ring of integers. Unless $\Bbb Q(\theta)=\Bbb Q$ the inverse different is non-trivial.

As an example, consider $\theta=\sqrt2$. Then Tr$(\alpha\Bbb Z[\sqrt2] \subseteq\Bbb Z[\sqrt 2])$ iff $\alpha\in\frac14\sqrt 2\Bbb Z[\sqrt2]$ etc.

Answer 2

This is never true, unless the field is the rational numbers.

The set of numbers with your property is a fractional ideal which (by definition) is the inverse different. The inverse norm of thus ideal is the discriminant of the field. A theorem of Minkowski says that the discriminant of a non-trivial field has absolute value strictly greater than one.