## 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 $(A \cap B)(A + B) = \prod P_i^{\max(k_i,\ell_i)} \cdot \prod P_i^{\min(k_i,\ell_i)}$ $= \prod P_i^{\max(k_i,\ell_i) + \min(k_i,\ell_i)}$ $= \prod P_i^{k_i+\ell_i}$ $= \prod P_i^{k_i} \cdot \prod P_i^{\ell_i}$ $= AB$.