## In an algebraic integer ring, the norm of an ideal principally generated by a rational integer is the absolute value of the norm of its generator

Let $K$ be an algebraic extension of $\mathbb{Q}$ of degree $n$, and let $\mathcal{O}$ be the ring of integers in $K$. Let $k \in \mathbb{Z}$. Show that the norm $N((k))$ is precisely $|N(k)|$. (The first norm is the norm of $(k)$ as an ideal, while the second norm is the norm of $k$ as an element.)

We can assume without loss of generality that $k$ is positive.

Recall that the conjugates of $k$ over $K$ are all the same (Theorem 5.10 in TAN), so that $N(k) = k^n$.

Let $\{\omega_i\}$ be an integral basis for $K$. Suppose $A$ is an ideal containing $k$. By Theorem 9.3 in TAN, there exists an element $\alpha = \sum c_i\omega_i$ such that $A = (k,\alpha)$. Note that we may assume that each $c_i$ is in $[0,k)$. In particular, $N((k)) \leq k^n$. We claim that this $\alpha$ is in fact unique. To that end, suppose $\sum c_i\omega_i \equiv \sum d_i\omega_i$ mod $(k)$ for some $0 \leq c_i, d_i < k$. Then we have $\sum c_i\omega_i - \sum d_i \omega_i = \sum ke_i\omega_i$ for some $e_i$. Since $\{\omega_i\}$ is an integral basis, we have $c_i \equiv d_i$ mod $k$ for each $i$, and thus $c_i = d_i$. So in fact $N((k)) = k^n = N(k)$, as desired.