$\gcd(I, J) = I + J$ and $\mathrm{lcm}(I, J) = I \cap J$ Following is a statement in Marcus Number Fields which I am not able to prove.
Let $R$ be a Dedekind domain and $I, J$ be non-trivial ideals in it.
Then by unique factorization of ideals in a Dedekind domain, $I = \mathfrak{p_1}^{r_1}\ldots \mathfrak{p_n}^{r_n}$ and $J = \mathfrak{p_1}^{s_1} \ldots \mathfrak{p_n}^{s_n}$ where $r_i \geq 0$ and $s_i \geq 0$. We define
$$\gcd(I,J) = \mathfrak{p_1}^{\min \lbrace r_1,s_1 \rbrace} \ldots \mathfrak{p_n}^{\min \lbrace r_n, s_n \rbrace}.$$
Now I want to show $\gcd(I, J) = I + J$. It is obvious that $I + J \subset \gcd(I, J)$,but I have no clue about how to proceed for proving the other inclusion.
 A: You can easily reduce to the case $\min\{r_i,s_i\}=0$, that means there are no common prime ideals in the prime factorization of $I$ and $J$, by considering the ideals $I \operatorname{gcd}(I,J)^{-1}$ and $J \operatorname{gcd}(I,J)^{-1}$.
But in this case, we have $\operatorname{gcd}(I,J)=R$ by definition, so it all comes down to this well known statement:
If $I_1, \dotsc, I_n$ are pairwise coprime ideals of a ring, then the two ideals
$I_1^{r_i} \dotsc I_m^{r^m}$ and $I_{m+1}^{r_{m+1}} \dotsc I_n^{r^n}$
are also coprime for any choice of $1 \leq m \leq n-1$.
There is a very nice proof: First of all note that we can assume all $r_i=1$ due to $1 \in \sqrt{I} \Longleftrightarrow 1 \in I$.
Then we proof the following statement: If $J_1, \dotsc, J_n$ are pairwise coprime, then $J_1$ is coprime to the product $J_2 \dotsb J_n$. This is a one-liner:
$$R = (J_1+J_2)(J_1+J_3)\dotsb (J_1+J_n) \subset J_1 + J_2 \dotsb J_n$$
Use this statement to the above situation and deduce that the ideals $I_1 \dotsb I_m, I_{m+1}, \dotsc, I_n$ are pairwise coprime. Use it once more to deduce that $I_1 \dotsb I_m$ and $I_{m+1} \dotsb I_n$ are coprime.
A: In Marcus,we have already proved the cancellation theorem,of which we will make extensive use and the fact that $I \mid J$ iff $IC=J$ for some ideal $C$. 
Since $I \subset I+J \Rightarrow I=(I+J)C$. For any prime $\mathfrak{q}$ occuring in decompositon of $I+J$,it must be one of the $\mathfrak{p_i}$ as in the question.Hence $\mathfrak{p_1}^{r_1}\ldots \mathfrak{p_n}^{r_n}=(\mathfrak{p_1}^{t_1}\ldots \mathfrak{p_n}^{t_n})C$.If $t_i>r_i$ for some $i$, then $t_i=r_i+l_i$ where $l_i\geq1$.Then cancelling respecting prime powers,we get $\mathfrak{p_1}^{r_1}\ldots\mathfrak{p_{i-1}}^{r_{i-1}}\mathfrak{p_{i+1}}^{r_{i+1}}\ldots \mathfrak{p_n}^{r_n}\subset \mathfrak{p_i}$ which is not possible. Hence $t_i\leq r_i$.Similarly for $J$ we get $t_i\leq s_i$.Consequently, $t_i\leq \min\lbrace r_i,s_i \rbrace$ and hence $\gcd(I,J) \subset I+J$. 
