A Question on Ideals in Algebraic Number Fields

Question

So I am not sure if the answer to this question is a trivial one. It would be very nice if it were totally true.

Let $K$ be a number field over $\mathbb{Q}$ and let $N_{K/\mathbb{Q}}$ be the norm mapping.

Let us make a small definition for simplicity:

For an ideal $I \subset \mathbb{O}_K$ let $S_I = \{N_{K/\mathbb{Q}}(x) : x \in I\} \subset \mathbb{Z}$

Consider two ideals $\alpha, \beta \subset \mathbb{O}_K$

Is it then always the case that for any $\sigma \in Gal(K/\mathbb{Q})$ that: $S_{\alpha \cdot \beta} = S_{\alpha \cdot \sigma(\beta)}$ ?

Answer 1

Only if $\sigma$ acts trivially on the class group (e.g. for imaginary quadratic $K$, only if the class group is $2$-torsion).

Indeed, suppose the ideal class $[\beta]$ of $\beta$ is inverse to the ideal class $[\alpha]$ of $\alpha$. Then $\alpha \cdot \beta$ is principal. We have

$$ N(\alpha \cdot \beta)=N(\alpha) N(\beta) = N(\alpha) N(\sigma(\beta))=N(\alpha\cdot \sigma(\beta)).$$

Since $\alpha \cdot \beta $ is principal, $N(\alpha\cdot \beta)\in S_{\alpha\cdot \beta}$ being the norm of a generator of the ideal. But as long as $[\sigma(\beta)]\neq [\beta]$, the ideal $\alpha \cdot \beta$ is not principal, so $N(\alpha\cdot \sigma(\beta))\notin S_{\alpha\cdot\sigma(\beta)}$ since any element of an ideal whose norm matches the norm of the ideal must be a generator.

Thus $S_{\alpha \cdot \beta} \neq S_{\alpha \cdot \sigma(\beta)}$.