## If A is an ideal of integers in an algebraic number field and if bA is contained in A, then b is an integer

Let $K$ be an algebraic number field with ring of integers $\mathcal{O}$, and let $A \subseteq \mathcal{O}$ be an ideal. Suppose $\beta \in K$ such that $\beta A \subseteq A$. Prove that $\beta \in \mathcal{O}$.

Let $\{\omega_i\}$ be a basis for $A$ over $\mathbb{Z}$. (This exists by Theorem 7.10). In particular, for each $i$, there exist rational integers $c_{i,j}$ such that $\beta\omega_i = \sum_i c_{i,j}\omega_j$. Rearranging, we have $0 = \sum_j (c_{i,j} - \delta_{i,j}\beta)\omega_j$ for each $i$, where $\delta_{i,j}$ is the Kronecker delta. In particular, $[\omega_1 \cdots \omega_n]^\mathsf{T}$ is a nontrivial solution to the matrix equation $Mx = 0$, where $M = [c_{i,j} - \delta_{i,j}\beta]$. Thus $\mathsf{det}(M) = 0$. On the other hand, by the Leibniz expansion for determinants, $\mathsf{det}(M)$ is a polynomial in $\beta$ having coefficients in $\mathbb{Z}$. Thus $\beta$ is an algebraic integer, and more specifically $\beta \in \mathcal{O}$.