The least common multiple of two ideals in an algebraic integer ring

Let K be an algebraic number field with ring of integers \mathcal{O}. Let A and B be ideals in \mathcal{O}. Suppose A and B factor into prime ideals as A = \prod P_i^{k_i} and B = \prod P_i^{\ell_i}, where k_1,\ell_i \geq 0. Prove that A \cap B = \prod P_i^{\max(k_i,\ell_i)}. Prove also that (A \cap B)(A+B) = AB.

Note that A \cap B is an ideal, and so factors as a product of prime ideals as A \cap B = \prod P_i^{t_i} for some t_i \geq 0. Since A \cap B \subseteq A, we have \prod P_i^{t_i} \subseteq \prod P_i^{k_i}. In particular, t_i \geq k_i for each i. Similarly, since A \cap B \subseteq B we have t_i \geq \ell_i for each i. Thus t_i \geq \max(k_i,\ell_i) for each i. Conversely, we have \prod P_i^{\max(k_i,\ell_i)} \subseteq A,B, so that \prod P_i^{\max(k_i,\ell_i)} \subseteq A \cap B. Thus A \cap B = \prod P_i^{\max(k_i,\ell_i)}, as desired.

Using the previous result, we have (A \cap B)(A + B) = \prod P_i^{\max(k_i,\ell_i)} \cdot \prod P_i^{\min(k_i,\ell_i)} = \prod P_i^{\max(k_i,\ell_i) + \min(k_i,\ell_i)} = \prod P_i^{k_i+\ell_i} = \prod P_i^{k_i} \cdot \prod P_i^{\ell_i} = AB.

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