Interplay between integral bases in algebraic number fields and bases of ideals of integers

Let K be an algebraic number field and let B = \{\omega_i\}_{i=1}^n be an integral basis for K. Let \alpha \in K be a nonzero algebraic integer. Show that \alpha B = \{\alpha\omega_i\}_{i=1}^n is a basis of an ideal I of integers in K if and only if I = (\alpha).


Suppose first that \alpha B is a basis of an ideal I– that is, (\alpha B) = I. Certainly, since \alpha B \subseteq (\alpha), we have I \subseteq (\alpha). Now B is an integral basis for K, so that in particular there exist rational integers t_i such that \sum t_i\omega_i = 1. Then \sum t_i \alpha \omega_i = \alpha, and we have \alpha \in I. Hence I = (\alpha).

Conversely, suppose I = (\alpha). Certainly (\alpha B)_\mathbb{Z} \subseteq (\alpha), where (\alpha B)_\mathbb{Z} denotes the set of all \mathbb{Z}-linear combinations of elements from \alpha B. Again because B is an integral basis for K, we have \alpha \in (\alpha B)_\mathbb{Z}. Hence (\alpha B)_\mathbb{Z} = I. Next we claim that \alpha B is \mathbb{Z}-linearly independent. To that end, suppose \sum t_i \alpha \omega_i = 0 for some t_i \in \mathbb{Z}; then since the ring of integers in K is an integral domain, \sum t_i \omega_i = 0. Since B is an integral basis for K, we have t_i = 0. Thus \alpha B is \mathbb{Z}-linearly independent, and so is a basis for I = (\alpha).

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