## In an algebraic integer ring, multiplication of ideal classes is well defined

Let $\mathcal{O}$ be the ring of integers in an algebraic number field $K$. Recall that we defined a relation $\sim$ on the set of ideals in $\mathcal{O}$ as follows: $A \sim B$ if and only if there exist nonzero $\alpha,\beta \in \mathcal{O}$ such that $(\alpha)A = (\beta)B$. Show that if $A_1 \sim B_1$ and $A_2 \sim B_2$, then $A_1A_2 \sim B_1B_2$.

We have $\alpha_1,\alpha_2,\beta_1,\beta_2$ such that $(\alpha_1)A_1 = (\beta_1)B_1$ and $(\alpha_2)A_2 = (\beta_2)B_2$. Then $(\alpha_1\alpha_2)A_1A_2 = (\beta_1\beta_2)B_1B_2$, so that $A_1A_2 \sim B_1B_2$.