Product of all embbedings of a fractional ideal is principal

Question

Let $F$ be a number field of degree $n$ and $\mathfrak a \subset F$ a fractional ideal. How do you show that $$\prod_{k=1}^n \sigma_k(\mathfrak a)$$ is in the trivial ideal class with a rational generator, hence of the form $a \mathcal O _F$ for some $a\in \mathbb Q$? Here, $\sigma_1 , \dots, \sigma_n$ are the $n$ embeddings of $F$ into $\overline{ \mathbb Q}$.

Answer 1

Let $F/K$ be an extension of number fields of degree $n$, assumed first to be normal. Associated to $F/K$ there are two norm maps defined on ideals of $O_F$ :

1) The algebraic norm $\nu$, defined by $\nu (\mathcal I)=\prod \sigma (\mathcal I)$, where the product bears over the automorphisms $\sigma\in Gal(F/K)$

2) The arithmetic norm $N$ defined by $N \mathcal I$ = the ideal of $O_K$ generated by the elements $N_{F/K} (x)$ for $x\in \mathcal I$, where $N_{F/K}$ is the usual norm $F^* \to K^*$

and

3) An extension map $\epsilon$ defined on ideals $\mathcal J$ of $O_K$ by $\epsilon (\mathcal J)$ = the ideal of $O_F$ generated by the natural inclusion $\mathcal J \subset O_F$ .

From the definitions one can easily check that $\nu = \epsilon .N$ , which gives what you want when $K=\mathbf Q$.

When $F/K$ is not Galois, the arithmetic norm $\nu$ is still defined, this time with $\sigma$ running through the embeddings of $F$ into an algebraic closure of $\mathbf Q$ , but the image $\nu (\mathcal I)$ has no reason to be inside $O_F$.