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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 $\mathfrak{a}$?

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  • $\begingroup$ 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. $\endgroup$ – user8268 Dec 16 '19 at 0:36
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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. $\mathfrak N(\mathfrak a)\in\mathfrak a$.

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  • $\begingroup$ Sorry, I don't get why $\mathfrak N(\mathfrak a)\times 1=0$?? $\endgroup$ – Leonardo May 5 '16 at 20:40
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    $\begingroup$ @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 $\endgroup$ – user8268 May 5 '16 at 21:27
  • $\begingroup$ why does that mean $\mathfrak N(\mathfrak a)\in\mathfrak a$. $\endgroup$ – 王李远 May 27 '18 at 9:55
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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,

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

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