I am looking for an example of prime ideal $P$ such that $\bigcap_{n=1}^{\infty}P^n$ is not prime.
In a Prüfer domain such an intersection is always a prime ideal.
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Tusif Ahmed
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3A local ring that is not an integral domain? In that case the intersection is the zero-ideal. – awllower Mar 07 '16 at 14:25
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3Thank you. But can you give me such example in integral domains? – Tusif Ahmed Mar 07 '16 at 14:27
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2Sorry, but I find out that the model in my mind is a discrete valuation ring, which is an integral domain... Sorry again. – awllower Mar 07 '16 at 14:28
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2you mean the intersection of all powers of maximal ideals? may be that ideal is idempotent... awllower – Tusif Ahmed Mar 07 '16 at 14:29
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2Yes, that is possible, but you demand an example of a non-integral domain, so let me try to find another. – awllower Mar 07 '16 at 14:30
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1By Krull intersection theorem, this is not possible in a Noetherian local integral domain. – awllower Mar 07 '16 at 14:55
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1So one have to find such example on non-Noetherian domains – Tusif Ahmed Mar 07 '16 at 14:57
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Related: http://math.stackexchange.com/questions/1688325/an-example-of-prime-ideal-p-in-an-integral-domain-such-that-bigcap-n-1-i – user26857 Aug 30 '16 at 15:11
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Sorry for commenting on such an old question, but why is such an intersection a prime ideal for Prufer domain ? – Jun 25 '18 at 05:35
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This intersection is prime for Prufer domains. For proof see Theorem 23.3 of R. Gilmer, Multiplicative ideal theory, Marcel Dekker, New York, 1972. – Tusif Ahmed Jul 18 '18 at 16:38
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Let $(R,m)$ be a (commutative Noetherian) local ring which is not a domain. By Krull's intersection theorem, $\bigcap_{n=1}^{\infty} m^n=0$ is not a prime.
One can use appropriate quotient of local domain to have local ring which is not a domain: $R=K[[X]]/(X^t)$
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