In a Dedekind Domain we have unique decomposition of ideals into product of prime ideals. My question out of curiosity is: Is there a ring satisfying this which is not a Dedekind Domain?
Rings where ideals decompose into product of primes
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$\begingroup$
abstract-algebra
ring-theory
maximal-and-prime-ideals
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0A domain is Dedekind iff every nonzero proper ideal is a product of prime ideals. Equivalently with maximal ideals. Are you asking about generalizations to rings (vs, domains)? – 2017-02-04
2 Answers
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Something that looks like a Dedekind Domain but is not a domain.
In particular every $\mathbb{Z}/n\mathbb{Z}$ with $n$ square-free should work, by the Chinese remainder theorem.
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Any integral domain with unique ideal factorization is already a Dedekind domain, see here. For rings with zero divisors, this is not true in general - consider $\mathbb{Z}/6$ for example.
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1It would be better to give a specific link to the result that you cite, rather than a generic Google search. – 2017-02-04
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0@BillDubuque a good pointer is https://en.wikipedia.org/wiki/Primary_decomposition (noetherian rings) ? – 2017-02-04
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0@BillDubuque Yes, of course, I am sorry. Cut and paste went wrong. – 2017-02-04