# On the absolute norm of an ideal [closed]

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 $\mathfrak{a}$?

• I think this question shouldn't be closed - the question is perfectly clear, and it's one of the basic things in algebraic number theory. Also there is no comment whatsoever about why it got closed (especially 3 years after it was asked). So I vote for reopening. – user8268 Dec 16 '19 at 0:36

Since $\mathfrak N(\mathfrak a)=|\mathscr O_K/\mathfrak a|$ we have $\mathfrak N(\mathfrak a)\times 1=0$ in $\mathscr O_K/\mathfrak a$, i.e. $\mathfrak N(\mathfrak a)\in\mathfrak a$.

• Sorry, I don't get why $\mathfrak N(\mathfrak a)\times 1=0$?? – Leonardo May 5 '16 at 20:40
• @Leonardo: the order of an element of a group (in this case of $1\in\mathscr O_K/\mathfrak a$, additive group) divides the order of the group – user8268 May 5 '16 at 21:27
• why does that mean $\mathfrak N(\mathfrak a)\in\mathfrak a$. – 王李远 May 27 '18 at 9:55

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,

$$\|\mathfrak{a}\|=\prod_i \|\mathfrak{p}_i\|^{e_i}\in \prod_i \mathfrak{p}_i^{e_i}=\mathfrak{a}$$