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If $R$ is a Dedekind domain and $I\subset R$ is a non-zero ideal then by the Noetherian property of $R$, I can show that there are distinct non-zero prime ideals $P_1,...,P_r$ s.t. $P_1^{a_1}\cdots P_r^{a_r}\subset I$ with $a_1,...,a_r\geq 0$.

I'd be grateful for an answer to the following:

If $P$ is a prime ideal in $R$ s.t. $P\notin \{P_1,...,P_r\}$ and $I\subset P$ then is there some $P_i\subset P$?

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  • $\begingroup$ If you assume that a Dedekind domain is an integral noetherian domain where unique factorization of non-zero ideals into prime powers exists, then your statement is trivial. So the correct question would be : What is your definition of a Dedekind domain? (There are multiple, and they are all equivalent, after some work ; the problem here is that you are doing this work!) $\endgroup$ – Patrick Da Silva Jun 14 '15 at 0:38
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You can show that the ordinary definition of primes in commutative rings is equivalent to this one:

$P$ is prime if whenever $A,B$ are ideals such that $AB\subseteq P$, then $A\subseteq P$ or $B\subseteq P$.

By applying this to the product of powers if primes in your factorization, you get an affirmative proof.

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