What is the trace of an integer of algebraic degree $1$?

Question

Potentially dumb question here. At least I'm fairly sure it's not a duplicate.

When it comes to quadratic integers, I think I know pretty well what the trace and norm are. Given a quadratic integer $a + b \sqrt{d}$ (the various restrictions on $a$ and $b$ and $d$ being stipulated to, as lawyers say), the minimal polynomial is $x^2 - 2a + (a^2 - db^2)$, the trace is $-2a$ and the norm is $a^2 - db^2$.

For algebraic integers of higher degree, meaning $k > 2$, I think I still understand that the norm is: in the polynomial $x^k + \alpha_{k - 1} x^{k - 1} + \ldots + N$, where $\alpha_k = 1$ implicitly, $N \in \textbf{Z}$, $N$ is the norm. But what is the trace?

I thought I would figure it out by figuring out the trace and norm of algebraic integers of degree $1$. Which is to say the numbers of $\textbf{Z}$. Given an integer $n$ of algebraic degree $1$, the minimal polynomial is $x - n$ and the only solution is $x = n$.

So the norm is $-n$ and the trace is $0$? Is this right or wrong?

Answer 1

For an algebraic number $\alpha$, let $L=\Bbb Q(\alpha)$ be the extension it generates, and let $p(x) = \sum_{i=0}^k a_i x^i$ with $a_k=1$ its normed minimal polynomial (so $k = deg(p)$). Then the trace and norm of $x$ are given by

$Tr_{L\vert \Bbb Q}(\alpha)= \bf{\color{red}{-}} $$a_{k-1}$

$N_{L\vert \Bbb Q}(\alpha)= (-1)^{k}\cdot a_0$.

So up to $\pm$, the trace is $a_{k-1}$ and the norm is $a_0$, but one has to be careful about the sign.

Note that there are at least two other definitions of the trace resp. norm which are very much worth learning: 1) as trace resp. determinant of the multiplication with $x$ viewed as a linear operator on the $\Bbb Q$-vector space $L$; 2) as sum resp. product of all $\sigma(\alpha)$, where $\sigma$ runs over the different embeddings of $\alpha$ into an algebraic closure of $\Bbb Q$. Note that 2) is particularly handy if the extension is Galois, and is related to the highlighted one by Vieta's formulae and the observation that

$$p(x) = \prod_{\sigma}(x-\sigma(\alpha)).$$

Note in particular how if $k=1$, of course we have $Tr(\alpha) = N(\alpha) = \alpha$.

In the quadratic case, for $\alpha= a+b\sqrt d$, all above formulae actually give -- and I recommend doing these calculations yourself as an enlightening exercise -- that $N(\alpha) = a^2-b^2d$ and $Tr(\alpha)=2a$ (your trace has the wrong sign).

Note that this has not much to do with the numbers being integral, and generalises to arbitrary field extensions. For a good first look, I recommend these and these lecture notes. The highlighted formula above are to be found in there in Cor. 5.10 resp. Section 2.1.3.

Answer 2

If you let $\sigma _1, \sigma _2, ...\sigma _d$ be the field embeddings K$ \rightarrow \mathbb{C}$, then the $\mathrm{Norm}(a)= \sigma _1(a)\sigma _2(a) ...\sigma _d(a)$, while $\mathrm{Trace}(a)= \sigma _1(a)+\sigma _2(a)+ \cdots+\sigma _d(a)$.

When dealing with a degree 1 extension, then there is only one field embedding, namely the identity.

So the norm and trace of $a$, an algebraic integer in $\mathbb{Q}$, is equal to $a$.