TITLE: On the absolute norm of an ideal
QUESTION [6 upvotes]: Let $K$ be a number field, with number ring $\mathscr{O}_K$. Let $\mathfrak{a}$ be an ideal in $\mathscr{O}_K$ and let $\mathfrak{N}({\mathfrak{a}})$ denote the absolute norm of $\mathfrak{a}$. How can be proved that $\mathfrak{N}(\mathfrak{a})$ is an element of $\mathfrak{a}$?

REPLY [3 votes]: Here's an alternative approach.
Factor $\mathfrak{a}$ as $\displaystyle \prod_i \mathfrak{p}_i^{e_i}$. Then, we know that 
$$\|\mathfrak{a}\|=\prod_i\|\mathfrak{p}_i\|^{e_i}$$
Now, it's well known that $\|\mathfrak{p}_i\|=p^{f(\mathfrak{p}_i\mid p)}$ if $\mathfrak{p}_i\cap \mathbb{Z}=(p)$ (this is almost tautological). In particular, we see that $\|\mathfrak{p}_i\|\in \mathfrak{p}_i$. So,
$$\|\mathfrak{a}\|=\prod_i \|\mathfrak{p}_i\|^{e_i}\in \prod_i \mathfrak{p}_i^{e_i}=\mathfrak{a}$$