Ideal lying over an ideal shows up in prime decomposition

Question

Say we have a number ring $R$ of a number field $K.$ We know that $R$ has unique prime ideal decomposition. My question is as follows. Let $I \subset J$ be a containment of ideals. Then let $I = P_1 \ldots P_r$ be its prime ideal decomposition. Then can we write $I = JP_q \ldots P_r?$ I think yes because every prime ideal lying over $J$ also lies over $I.$

Answer 1

Yes, and this holds in any Dedekind domain $R$.

If you decompose your ideals as products over the non-zero prime ideals of $R$, you have $$I = \prod_P P^{n_P} \quad\text{and}\quad J = \prod_P P^{m_P},$$ where $n_P, m_P \geq 0$ are integers and are zero for all but finitely many prime ideals $P \subset R$. Then $$I \subseteq J \iff \forall P, n_P \geq m_P.$$ Thus we may write $$I = J \prod_P P^{n_P -m_P},$$ since $n_P -m_P \geq 0$.