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I'm to prove that every proper ideal is a product of maximal ideals which are uniquely determined up to order. I have no idea even how to start in the proof to solve this question :( May anybody help me ?

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Hint: Every PID is a UFD. Does that help you get started?

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  • $\begingroup$ yes that's it !! UFD means elements are irreducable or product of irreducables > and an element is irreducable iff the ideal generated by it is maximal among principal ideals , but in PID all ideals are principal . We are done !! what a wonderful hint ! thank u :) $\endgroup$ – Enas Jan 16 '14 at 18:21
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Hints:

In such a ring, $(ab)=(a)(b)$, and the nonzero maximal ideals are the same thing as the nonzero prime ideals. Do you know what the prime ideals of PID's look like?

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  • $\begingroup$ i guess the ideals generated by prime numbers, or irreducable since in PID ( irreducabel iff prime ) $\endgroup$ – Enas Jan 16 '14 at 18:32
  • $\begingroup$ @Enas Close: nonzero prime elements. $\endgroup$ – rschwieb Jan 16 '14 at 18:40
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Every PID is a trivial example of Dedekind domain. In a Dedekind domain we can factor every proper ideal as a product of prime (maximal) power ideal uniquely up to rearrangements of those prime ideals. Proof of this is easy and can be seen in any book of Algebraic Number Theory.

Though I have not given a concrete proof of your question but I tried to give you some more information about this question i.e. this result is not true only in case of PID but is also true for some special type of domains.

I hope the information presented by me will be helpful to you.

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