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Let $R$ be a commutative, Dedekind (and therefore Noetherian) ring with $1$. Let $I$ be a non-prime ideal of $R$, and let $a,b$ be elements of $R$ such that $a\not\in I,b\not\in I$ but $ab\in I$. Let $\cal P$ be the set of prime ideals appearing in the Dedekind factorization of $(I,a)$ or $(I,b)$. Then $\cal P$ is finite, and I ask : is it always true that some ideal in $\cal P$ must appear in the Dedekind factorization of $I$ ?

If true, this would provide a theoretical "algorithm" to compute the Dedekind factorization of any ideal of such a ring $R$.

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"To divide is to contain."

Thus, if $P$ divides $(I,a)$ then $P$ contains $(I,a)$ and so $P$ contains $I$. This means that $P$ divides $I$.

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  • $\begingroup$ Thank you for your answer. It is perhaps worthwile to mention that "to divide is to contain" is proved at math.stackexchange.com/questions/103692/… $\endgroup$ Oct 17 '14 at 17:06
  • $\begingroup$ @EwanDelanoy, I wonder what role is played by $b$ ... $\endgroup$
    – lhf
    Oct 17 '14 at 17:08
  • $\begingroup$ No role at all, it just appears naturally when you try to use the non-prime character of $I$ $\endgroup$ Oct 17 '14 at 17:09

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