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.

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