Attaining the norm of an ideal in a number field by the norm of an element Let $K$ be a number field of degree $n$ and $\mathfrak{a}$ be an ideal in its ring of integers $\mathcal{O}_K$. We can consider:


*

*The norm $N(\mathfrak{a})$ of $\mathfrak{a}$.

*The norms $N(x)$ of the elements $x\in\mathfrak{a}$.


It is well known that:


*

*For all $x\in\mathfrak{a}$, $N(\mathfrak{a})|N(x)$, so $\lvert N(x) \rvert \ge N(\mathfrak{a})$

*$N(\mathfrak{a})\in\mathfrak{a}$

*By point 2., $\mathfrak{a}$ contains an element of norm $N(\mathfrak{a})^n$.


But does there exist an element $x\in\mathfrak{a}$ such that precisely
$$\lvert N(x) \rvert =N(\mathfrak{a})\ ?$$
 A: This happens if and only if $\mathfrak a$ is a principal ideal. One direction is easy if you know that $N(\mathfrak (x)) = \lvert N(x) \rvert$ for all $x \in \mathcal O_K$ (left side is the ideal norm of the principal ideal). For the other direction let $x \in \mathfrak a$ with $\vert N(x) \rvert = N(\mathfrak a)$. Then the index $[ \mathfrak a : (x) ]$ is equal to $\frac{N(\mathfrak a)}{\lvert N(x) \rvert} = 1$ (some isomorphism theorem tells you $\mathcal O_K / \mathfrak (x) \cong (\mathcal O_K / \mathfrak a)/(\mathfrak a / (x))$ . Hence $(x) = \mathfrak a$.
A: We have $\langle N(x) \rangle = N(\langle x \rangle)$, so a counterexample would have to come from a non-UFD.
The usual example of such a thing is $\mathbb{Z}[\sqrt{-5}]$, so let's try it out. The prime ideal lying over $2$ is $\langle 2, 1 + \sqrt{-5} \rangle$, and it has norm $\langle 2 \rangle$.
However, $N(a + b \sqrt{-5}) = a^2 + 5 b^2$, which clearly can never equal $2$ or $-2$.
A: If and only if $\mathfrak{a}$ is principal. Note that the norm of the ideal generated by $x$ is the same as the field norm of $x$. Thus 
$$
N(x) = N(\mathfrak{a}) \Longleftrightarrow N(x\mathcal{O}_K) = N(\mathfrak{a})\\
$$
But since $x \in \mathfrak{a}$, we have $x\mathcal{O}_K \subseteq \mathfrak{a}$. Since the norms are equal, this forces $x\mathcal{O}_K = \mathfrak{a}$.
