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.

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