Relatively prime ideals in Dedekind domains definition Let $R$ be a Dedekind domain and let $I$ and $J$ be two ideals such that $I+J=R$; that is, $I$ and $J$ are relatively prime. 
Now, since $R$ is a Dedekind domain, $I$ and $J$ have unique (up to order of factors) decompositions into products of prime ideals.
My question is: Is $I$ and $J$ being relatively prime via the definition in the first sentence above equivalent to saying their prime factorizations have no common factor? 
 A: If $I$ and $J$ have the same prime ideal factor, say $\mathfrak{p}$, then $I+J\subseteq \mathfrak{p}$.
On the other hand if $I+J\neq R$, then there exists a prime (maximal) ideal $\mathfrak{p}$ of $R$ such that $I+J\subseteq \mathfrak{p}$. In particular $I\subseteq \mathfrak{p}$ and $J\subseteq \mathfrak{p}$. This implies that $\mathfrak{p}$ is a common factor of both $I$ and $J$ (see Lemma).
Lemma. Let $R$ be a Dedeking ring. If $I = \prod_{i=1}^k\mathfrak{q}_i^{n_i}$ is a factorization onto product of powers of distinct prime ideals and $I\subseteq \mathfrak{p}$, then $\mathfrak{q}_{i_0}=\mathfrak{p}$ for some $i_0$.
Proof. Suppose first that for each $i$ there exists $x_i\in \mathfrak{q}_i^{n_i}$ such that $x_i\not \in \mathfrak{p}$. Then
$$x_1x_2...x_k \in \prod_{i=1}^k\mathfrak{q}_i^{n_i} = I$$
but $x_1x_2...x_k\not \in \mathfrak{p}$. This is a contradiction with $I\subseteq \mathfrak{p}$. Hence there exists $i_0$ such that $q_{i_0}^{n_{i_0}}\subseteq \mathfrak{p}$. Taking radicals we have
$$\mathfrak{q}_{i_0} = \sqrt{\mathfrak{q}_{i_0}^{n_{i_0}}} \subseteq \sqrt{\mathfrak{p}} = \mathfrak{p}$$
Since all nonzero prime ideals in a Dedeking ring are maximal, we derive that $\mathfrak{q}_{i_0} = \mathfrak{p}$.  
