Trace of an algebraic integer is an integer?

Question

Let $F$ be a number field and let $\alpha \in F$.

If $\alpha \in \mathcal{O}_F$, then it is known that $N(\alpha) \in \mathbb{Z}$.

I was wondering if something similar can be said about the trace? I know that the trace of any element in $F$ is in $\mathbb{Q}$, but I was wondering if that element is in the ring of integers, is it then an integer?

If $\alpha \in \mathcal{O}_F$, is $tr(\alpha) \in \mathbb{Z}$?

Answer 1

Yes, absolutely. The trace $tr(\alpha)$ is the trace of the $\mathbb Q$-linear map $$\begin{align*} F & \to F \\ \beta & \mapsto \alpha \beta \end{align*}$$ If we express this linear map in an integral basis, the matrix representation has entries in $\mathcal O_F$, so that $tr(\alpha) \in \mathcal O_F \cap \mathbb Q = \mathbb Z$.

For a statement about general field extensions you can look here: Relative trace and algebraic integers

Answer 2

An algebraic number $\alpha \in F$ is integral iff its minimal polynomial has integral coefficients. Let $m(T) = T^n + c_1 T^{n-1} + \cdots + c_n$ be its minimal polynomial. Letting $d = [F:\mathbb{Q}(\alpha)]$, then $\DeclareMathOperator{\tr}{tr} \tr(\alpha) = -c_1 d$ and $\operatorname{N}(\alpha) = (\pm c_n)^d$.

If we denote the roots of $m$ by $\alpha = \alpha_1, \ldots, \alpha_n$, then $c_i = (-1)^is_i(\alpha_1, \ldots, \alpha_n)$, the where $s_i$ is the $i^\text{th}$ elementary symmetric function. So more generally, if $\alpha$ is an algebraic integer, then every elementary symmetric function in its Galois conjugates is in $\mathbb{Z}$.