Embeddings of pure cubic field in complex field

Question

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?

Answer 1

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 $n$.

Answer 2

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 $a^3+n$.