Quadratic extension has trace and norm algebraic integers.

Question

Let $K = \mathbb{Q}(\sqrt{d})$ be a non-trivial extension of $\mathbb{Q}$. Why are $Tr_{K / \mathbb{Q}}(x) , N_{K / \mathbb{Q}}(x)$ algebraic integers for all $x \in \mathcal{O}_K$? (Here is $Tr_{K / \mathbb{Q}}, N_{K / \mathbb{Q}}$ denote the trace and norm)

Answer 1

I actually make a mistake, it should be $x \in \mathcal{O}_{K}$, in which case the claim follows. This is because $Tr_{K / \mathbb{Q}}(x) = x + \sigma(x)$ and $N_{K / \mathbb{Q}}(x) = x \cdot \sigma(x)$, where $\sigma(a + b \sqrt{d}) = a - b\sqrt{d}$. Since $\sigma \in Aut(K / \mathbb{Q})$, $x \in \mathcal{O}_{K} \implies \sigma(x) \in \mathcal{O}_{K}$, $\mathcal{O}_{K}$ being a ring implies $Tr_{K / \mathbb{Q}}(x), N_{K / \mathbb{Q}}(x) \in \mathcal{O}_{K}$.

Answer 2

Actually, the trace and the norm of an algebraic integer are ordinary integers.

Indeed, the trace and the norm of an algebraic integer $\alpha$ in a finite extension $L/\mathbb Q$ are the trace and the determinant of the map $L \to L$ given by $x \mapsto \alpha x$. By taking an integral basis of $\mathcal{O}_L$, the corresponding matrix has integer entries and so has integer trace and determinant.