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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:

  1. For all $x\in\mathfrak{a}$, $N(\mathfrak{a})|N(x)$, so $\lvert N(x) \rvert \ge N(\mathfrak{a})$
  2. $N(\mathfrak{a})\in\mathfrak{a}$
  3. 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 $$\lvert N(x) \rvert =N(\mathfrak{a})\ ?$$

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Incidentally, the norm of an ideal is also an ideal. (although when we take norms down to $\mathbb{Z}$, we can equate ideals of $\mathbb{Z}$ with non-negative elements of $\mathbb{Z}$) –  Hurkyl Dec 22 '12 at 16:20
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3 Answers

up vote 7 down vote accepted

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 $(x) = \mathfrak a$.

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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 $-2$.

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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 $x\mathcal{O}_K = \mathfrak{a}$.

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