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

Let be a finite extension of of degree , and let be the ring of integers in . Let be a rational integer. Prove that is contained in at most ideals in .

[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 , or to take minimal such that , and use the reduction mod 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 be a nonzero rational integer, and suppose is an ideal containing . By Theorem 9.3, there exists an element such that .

Now let be an integral basis for , and say . Via the division algorithm, we have for some , so that , where . Now .

However, note that there are only distinct choices for the . Thus is one of at most ideals in .

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