## 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 $\mathcal{O}$.