On the absolute norm of an ideal [closed]

Solution 1:

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. $\mathfrak N(\mathfrak a)\in\mathfrak a$.

Solution 2:

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,

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