2

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 $tr(\alpha) \in \mathbb{Z}$?

Bart Michels
  • 26,985
  • 6
  • 59
  • 123
yktd
  • 133
  • 6

2 Answers2

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

Bart Michels
  • 26,985
  • 6
  • 59
  • 123
  • (you should make clear it is the trace of the $\mathbb{Q}$-linear map, when restricted to $O_F$ it is the trace of the (free) $\mathbb{Z}$-module map) – reuns Mar 02 '19 at 17:28
  • Thanks, edited. Although one could, I never view it as a $\mathbb Z$-linear map, I just take a basis such that the $\mathbb Q$-linear map has a matrix with integral entries. – Bart Michels Mar 02 '19 at 17:31
  • 1
    There is a $\Bbb{Q}$-basis of $F$ which is also a $\Bbb{Z}$-basis of $O_F$ (ie. $O_F$ is a free $\Bbb{Z}$-module and $F = O_F \otimes_\Bbb{Z} \Bbb{Q}$) but when looking at $Tr_{F/K}$ then $O_F$ isn't free $O_K$-module in general so we'd need to replace $O_F$ by a finite index free $O_K$-module $M$ containing $\alpha$ and show the trace on $M$ is the same as the one on $F$ – reuns Mar 02 '19 at 17:36
1

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 $\mathbb{Z}$.

Viktor Vaughn
  • 20,897
  • This is not correct. Say in $\mathbb{Q}(\sqrt{2})$. The norm of $2$ is $4$. However, the minimal polynomial of $2$ is $T-2$. – Upc Jan 09 '20 at 21:01
  • @Upc Yes, if $\alpha$ does not generate $F$ you have to multiply by the degree $[F : \mathbb{Q}(\alpha)]$. Thanks for the comment; I've edited to address it. – Viktor Vaughn Jan 10 '20 at 02:38