## 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 $I = (\alpha)$.