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?


2 Answers 2


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 $n$.

  • $\begingroup$ So we can use complex conjugates to calculate norm though they don't exist in number field? $\endgroup$ Jul 14, 2016 at 17:39
  • $\begingroup$ Sure. As you said, the norm is equal to the negative of the constant coefficient of the minimal polynomial, and if you factor the minimal polynomial in some splitting field, you'll see that the norm is the product of the conjugates. $\endgroup$ Jul 14, 2016 at 17:42
  • $\begingroup$ can you please write the embeddings explicitly? I think that you have just restated my way of solving. $\endgroup$ Jul 14, 2016 at 17:44
  • $\begingroup$ The three complex embeddings of $\mathbb{Q}(\sqrt[3]{n})$ are $\sqrt[3]{n}\mapsto \sqrt[3]{n}$, $\sqrt[3]{n}\mapsto \omega\sqrt[3]{n}$, and $\sqrt[3]{n}\mapsto \omega^2\sqrt[3]{n}$. $\endgroup$ Jul 14, 2016 at 17:45
  • $\begingroup$ this is not the way we write embeddings. It should be like "identity map" and so on, as stated for quadratic field case above. Am I wrong? $\endgroup$ Jul 14, 2016 at 17:49

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 $a^3+n$.

  • $\begingroup$ Aha! Thanks for reminding me this approach (ex. 17, ch 2, Marcus) $\endgroup$ Jul 14, 2016 at 18:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .