Number Theory: Calculating Norm.

Question

Let $K_{1}=\mathbb{Q}(\sqrt[3]{6})$ and $K_{2}=\mathbb{Q}(\sqrt[3]{12})$. The goal is to show that ring of integers in $K_{1}$, $K_{2}$ is spanned by $\{1,\sqrt[3]{6},\sqrt[3]{36}\}$, respectively $\{1,\sqrt[3]{12},\sqrt[3]{18}\}$. In the beginning of the proof I calculated both discriminants correctly to be $-972$ using the following formula: $disc = \det([Tr]_{i,j})$. In a further part of the proof the following norms are being calculated.

For $K_{1}$: Let $$w=(a+ b\sqrt[3]{6}+c\sqrt[3]{36})/p.$$ I understand that:

$N(w) = a^{3}/p^{3} N(1) + b^{3}/p^{3} N(\sqrt[3]{6})+ c^{3}/p^{3}N(\sqrt[3]{36})$.

Similar to the calculation of the discriminant I get the right result by using: $N(\alpha) = \det(\begin{bmatrix} \det_{i,j} \end{bmatrix})$, by just replacing the trace in the discriminant formula with the determinant.

($N(w)= 3/4 b^{3} + 9/4 c^{3}$, we assumed $a = 0, p=2$.)

However I should get the same result (same assumptions) for $K_{2}$ namely $N(w')$ with $w'= (a+ b\sqrt[3]{12}+c\sqrt[3]{18})/p$.

Why does my method for the norm not work for $K_{2}$? What method works?

Answer 1

You deal with two pure cubic fields $\mathbf Q(\sqrt[3]m)$, with $m =$ a cubic free integer, which are of different types: $6$ is square free, but not $12$. Actually, integral bases of pure cubic fields are completely known. For convenience, write $\alpha=\sqrt[3]m$. Two cases occur:

1. If $m$ is square free, an integral basis is $\{1, \alpha, \alpha ^2\}$ if $m\neq \pm 1$ mod $9$, and $\{1, \alpha, (\alpha ^2 \pm \alpha +1)/3\}$ if $m\equiv \pm 1$ mod $9$.

2. If $m$ is not square free, write $m=hk^2$, with $h$ and $k$ square free and coprime. Then an integral basis is $\{1, \alpha, \alpha ^2 /k\}$ if $m\neq \pm 1$ mod $9$, and $\{1, \alpha,(\alpha ^2 \pm k^2\alpha+k^2)/3k\}$ if $m\equiv \pm 1$ mod $9$.

See D. Marcus, Number Fields, end of chap. $2$ and exercise $41$.