## In the set of ideals of a ring of algebraic integers in some fixed finite extension of QQ, if AB = A then B = (1)

Let $A,B \subseteq \mathcal{O}_K$ be ideals in the ring of integers in $K = \mathbb{Q}(\theta)$, with $A \neq 0$. Show that if $AB = A$ then $B = (1)$, using only the existence of a basis for $A$.

Let $T = \{\omega_i\}_{i=1}^n$ be a basis for $A$ over $\mathbb{Z}$. ($T$ exists by Theorem 7.10 since $A \neq 0$.) Since $AB = A$, we have $\omega_i = \sum_j \gamma_{i,j} \omega_j$ for each $i$, where $\gamma_{i,j} \in B$, using the definition of the ideal product and collecting terms in $\omega_j$ as necessary. Rearranging, we have $0 = \sum_j (\gamma_{i,j} - \delta_{i,j}) \omega_j$ for each $i$, where $\delta_{i,j}$ is the Kronecker delta. That is, $[\omega_1\ \cdots\ \omega_n]^\mathsf{T}$ is a nontrivial solution to the matrix equation $Mx = 0$, where $M = [\gamma_{i,j} - \delta_{i,j}]$. Thinking of $\mathcal{O}$ as embedded in the field $K$, we have $\mathsf{det}(M) = 0$. On the other hand, via the Leibniz expansion of $\mathsf{det}(M)$, we have $\mathsf{det}(M) = \beta - 1$ for some $\beta \in B$. So $-1 \in B$, and we have $B = (1)$.