[Math] Attaining the norm of an ideal in a number field by the norm of an element

algebraic-number-theoryidealsnormed-spaces

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
$$\lvert N(x) \rvert =N(\mathfrak{a})\ ?$$

Best Answer

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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