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Does there exist a name for the class of commutative rings with identity that satisfy the following:

For any 2 ideals $I_1,I_2$ of R,we have : $I_1 I_2= (I_1\cap I_2)(I_1+I_2) $

I would also like to see an example of a ring not satisfying the above property.

Thank you

Amr
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  • http://math.stackexchange.com/questions/158533/is-there-a-distributive-law-for-ideals and http://mathoverflow.net/questions/49259/when-is-the-product-of-two-ideals-equal-to-their-intersection answer your question. – Derek Allums Jul 05 '16 at 13:49
  • Any noetherian UFD, which is not a PID, does not satisfy this identity. Can anyone give a noetherian ring (no PID), which satisfies this identity for any two ideals? – MooS Jul 05 '16 at 13:54
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    @MooS The question seems local, so shouldn't it be true for Dedekind domains? – Mohan Jul 05 '16 at 14:10
  • Oh yes, if you start an argument with 'Any UFD, which is not a PID' and search for a counterexample, you should always think of Dedekind domains... – MooS Jul 05 '16 at 14:19
  • Next Question would be: Given a noetherian domain, which satisfies the identity. Is it a Dedekind domain? This should rephrase as: Given a local noetherian domain, which satisfies the identity: Is it a DVR? If not: Does it suffice to add one the two missing properties (one-dimensional, normal). – MooS Jul 05 '16 at 14:33

2 Answers2

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If $R$ is a noetherian domain with that property $R$ is called Dedekind domain, see Larsen and McCarthy' Multiplicative Theory of Ideals, theorem 6.20. However, see D. D. Anderson's homepage, he has a paper with a similar title for arbitrary commutative rings.

user26857
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E.R
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  • He has published so many papers with titles that would fit in this context. Can you name the paper to speed up my search? :) – MooS Jul 05 '16 at 14:51
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    See for example ' D. D. Anderson, On the ideal equation $I(B\cap C) = IB\cap IC$, Canadian Mathematical Bulletin 26 (1983), 331-332.' that is a useful paper. – E.R Jul 05 '16 at 16:33
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In a (noetherian) factorial ring, which is not a PID, we can always find $f,g$, such that $\operatorname{gcd}(f,g) =1$ but the ideal $(f,g)$ is not the unit-ideal. Then we have

$$((f)+(g))((f) \cap (g))=(f,g)(fg) \subsetneq (fg).$$

For instance, take $k[x,y]$ and the ideals $(x)$ and $(y)$.

MooS
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