# Does $I(J\cap K)=IJ\cap IK$ hold in a Dedekind ring?

For ideals in any ring, we have the relation $I(J\cap K)\subseteq IJ\cap IK$. Do we actually have equality if we are in a Dedekind domain?

I've been looking around for a reference, but haven't found anything. I haven't been able to construct a counterexample (if there is one).

Hint:

$$\mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_m^{e_m}\cap\mathfrak{p}_1^{f_1}\cdots\mathfrak{p}_m^{f_m}=\mathfrak{p}_1^{\max\{e_1,f_1\}}\cdots\mathfrak{p}_m^{\max\{e_m,f_m\}}$$

and

$$a+\max\{b,c\}=\max\{a+b,a+c\}$$

if $a,b,c\geqslant 0$.

• Thanks, this mostly clears it up for me, but why is $$\mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_m^{e_m}\cap\mathfrak{p}_1^{f_1}\cdots\mathfrak{p}_m^{f_m}\subseteq\mathfrak{p}_1^{\max\{e_1,f_1\}}\cdots\mathfrak{p}_m^{\max\{e_m,f_m\}}?$$ I see the other inclusion. – Katrina Xiao Apr 8 '14 at 7:58
• Sorry, I don't know why it isn't compiling correctly. – Katrina Xiao Apr 8 '14 at 8:00
• @KatrinaXiao This is just because dividing and containment are the same thing for Dedekind domains. Thus, the intersection of two ideals is the ideal maximal among containment. So, $I\subseteq K$ and $I\subseteq J$ if and only if $I\subseteq K\cap J$. So, this says that $K\mid I$ and $J\mid I$ if and only if $K\cap J\mid I$. – Alex Youcis Apr 8 '14 at 8:07
• Hmm, that's what made me feel funny, since the larger the exponents of the prime ideals, the smaller the whole ideal is, which seemed weird since I'm looking for the ideal maximal among containments. – Katrina Xiao Apr 8 '14 at 8:10
• @KatrinaXiao I don't quite follow. Containment and divisibility are reversed. So, the ideal minimal with respect to being divisible by both $K$ and $J$ should be the ideal maximal to containment in both $I$ and $J$. – Alex Youcis Apr 8 '14 at 8:11