The relation between the product of two ideals and the intersection of these two ideals in a Dedekind domain.

Question

Let $R$ be a Dedekind domain, $A,B$ be two ideals of $R$. Show that $AB=(A+B)(A\cap B)$.

I know that $AB=I(A\cap B)$ for some ideal $I$ in $R$ because $AB\subseteq A\cap B$. But why $I$ is necessarily $A+B$?

Answer 1

Since $R$ is a Dedekind domain, every ideal can be written as a product of prime ideals. Therefore there are prime ideals $\mathfrak{p}_1,\dots,\mathfrak{p}_n$ such that $$ A=\mathfrak{p}_1^{a_1}\cdots \mathfrak{p}_n^{a_n}$$ and $$ B=\mathfrak{p}_1^{b_1}\cdots \mathfrak{p}_n^{b_n}$$ with $a_i,b_i\geq 0$.

To prove that $AB=(A+B)(A\cap B)$, it's enough to observe that $$ A+B=\mathfrak{p}_1^{\min\{a_1,b_1\}}\cdots \mathfrak{p}_n^{\min\{a_n,b_n\}}$$ and $$ A\cap B=\mathfrak{p}_1^{\max\{a_1,b_1\}}\cdots \mathfrak{p}_n^{\max\{a_n,b_n\}}$$ while $$ AB=\mathfrak{p}_1^{a_1+b_1}\cdots \mathfrak{p}_n^{a_n+b_n}$$

It's worth noting that this is essentially the way that one proves $mn=\gcd(m,n)\mathrm{lcm}(m,n)$ for natural numbers $m$ and $n$.