An upper bound on the number of ideals in an algebraic integer ring which contain a fixed rational integer

Let K = \mathbb{Q}(\theta) be a finite extension of \mathbb{Q} of degree n, and let \mathcal{O} be the ring of integers in K. Let k be a rational integer. Prove that k is contained in at most k^n ideals in \mathcal{O}.

[Disclaimer: For this exercise, I am going to refer to a theorem which is proved four pages later in the text. This does not introduce a cycle in the logic. I do not like to do this, but I wasn’t able to come up with a different proof. My best idea was to try to diagonalize a matrix over \mathbb{Z}, or to take c_i minimal such that c_i\omega_i \in A, and use the reduction mod k trick. If anyone sees how to prove this using only the machinery developed through chapter 8 in TAN, I’d greatly appreciate hearing about it. :)]

Let k \in \mathcal{O} be a nonzero rational integer, and suppose A is an ideal containing k. By Theorem 9.3, there exists an element \alpha such that A = (k,\alpha).

Now let \{\omega_i\}_{i=1}^n be an integral basis for K, and say \alpha = \sum_i a_i\omega_i. Via the division algorithm, we have a_i = q_ik + r_i for some 0 \leq r_i < k, so that \alpha = k\gamma + \sum_i r_i\omega_i, where \gamma = \sum_i q_i\omega_i. Now A = (k, \sum_i r_i\omega_i).

However, note that there are only k^n distinct choices for the r_i. Thus A is one of at most k^n ideals in \mathcal{O}.

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