Embeddings of pure cubic field in complex field I know that the complex embeddings (purely real included) for quadratic field $\mathbb{Q}[\sqrt{m}]$ where $m$ is square free integer, are 


*

*$a+b\sqrt{m} \mapsto a+b\sqrt{m}$

*$a+b\sqrt{m} \mapsto a-b\sqrt{m}$
So, norm of $a+b\sqrt{m}$ is $a^2-mb^2$.
Motivated by this , I want to calculate norm of $a+\sqrt[3]{n}$ in $\mathbb{Q}[\sqrt[3]{n}]$ where $n$ is positive cubefree integer. I am able to calculate the norm to be $a^3+n$ using the fact that it's equal to the negative of constant term of the minimal polynomial.  But I don't get same answer when I assume embeddings to be


*

*$a+\sqrt[3]{n} + 0\sqrt[3]{n^2} \mapsto a+\sqrt[3]{n} + 0\sqrt[3]{n^2}$

*$a+\sqrt[3]{n} + 0\sqrt[3]{n^2} \mapsto a-\sqrt[3]{n} + 0\sqrt[3]{n^2}$

*$a+\sqrt[3]{n} + 0\sqrt[3]{n^2} \mapsto a+\sqrt[3]{n} - 0\sqrt[3]{n^2}$
So what are the correct conjugation maps for pure cubic field case?
 A: The conjugates of $\sqrt[3]{n}$ are $\omega\sqrt[3]{n}$ and $\omega^2\sqrt[3]{n}$, where $\omega$ is a primitive cube root of unity. Therefore the norm of $a+\sqrt[3]{n}$ is
$$ (a+\sqrt[3]{n})(a+\omega\sqrt[3]{n})(a+\omega^2\sqrt[3]{n})=a^3+n$$
using the fact that $1+\omega+\omega^2=0$.
Edit: the three complex embeddings of $\mathbb{Q}(\sqrt[3]{n})$ are
$$ a+b\sqrt[3]{n}+c\sqrt[3]{n^2}\mapsto a+b\sqrt[3]{n}+c\sqrt[3]{n^2} $$
$$ a+b\sqrt[3]{n}+c\sqrt[3]{n^2}\mapsto a+b\omega \sqrt[3]{n}+c\omega^2\sqrt[3]{n^2} $$
$$ a+b\sqrt[3]{n}+c\sqrt[3]{n^2}\mapsto a+b\omega^2\sqrt[3]{n}+\omega \sqrt[3]{n^2}$$
Here $a,b,c\in\mathbb{Q}$ and $\sqrt[3]{n}$ is the real cube root of $n$.
A: The norm can be computed without leaving $\mathbb Q$.
The norm of $\alpha \in \mathbb{Q}[\sqrt[3]{n}]$ is the determinant of the linear map $x \mapsto \alpha x$.
Taking the basis $1, \sqrt[3]{n}, \sqrt[3]{n^2}$, the matrix of this map for $\alpha=a+\sqrt[3]{n}$ is
$$
\pmatrix{ a & 0 & n \\ 1 & a & 0 \\ 0 & 1 & a }
$$
whose determinat is $a^3+n$.
