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

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.

• If you replace $\theta$ with $n\theta$, the thing on the left changes, while the thing on the right doesn't – mercio May 23 '17 at 19:39

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.
• What I'm trying to understand is why $f'(\theta)\mathcal{O}_{\mathbb{Q}(\theta)} = \mathbb{Z}[\theta]$. I found the above statement in Weiss, "Algebraic Number Theory", Proposition 3-7-14, but maybe i did not understand his notation. – ilmarchese May 23 '17 at 18:37
• When $\mathcal{O}_K=\Bbb Z[\theta]$ with $f$ the minimum polynomial of $\theta$, then $f'(\theta)$ generates the different of $K$. – Lord Shark the Unknown May 23 '17 at 18:40
• Unfortunately I don't know what a different is. I'm studying the General Number Field Sieve algorithm, and to extend the search for squares in $\mathbb{Q}(\theta)$, the statement of my first comment is used. It's desiderable for me to understand why, but with my lack of knowledge, maybe for the moment it is sufficient to be sure that it is true. – ilmarchese May 23 '17 at 18:50