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I am looking for an example of prime ideal $P$ in an integral domain such that the ideal $\bigcap_{n=1}^{\infty}P^n$ is not a prime ideal.

This is a followup to this question where the ring was not assumed to be an integral domain.

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Tusif Ahmed
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    What's an example where it's not the zero ideal? – Gerry Myerson Mar 08 '16 at 12:26
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    In order to find such example one should have to find prime ideal $P$ such that $P$ is not idempotent and $\bigcap_{n=0}^{\infty}P^n\neq (0)$. So in finitely generated case we only need to find $P$ such that $\bigcap_{n=0}^{\infty}P^n\neq (0)$. Because of curse in integral domain there is no idempotent finitely generated ideal. – Tusif Ahmed Mar 08 '16 at 13:01
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    For a non-zero intersection, take $D$ the valuation ring of the Hahn series field $\mathbb{Q}[[T^{\mathbb{Z}^2}]]$ with the lexicographical order on $\mathbb{Z}^2$, and $P = (T^{(0,1)})$. Then $\bigcap P^n = (T^{(1,0)})$ which is non-zero (but prime...). – Captain Lama Mar 08 '16 at 13:02
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    Thank you Captain Lama. In order to understand your example frist I have to understand the Hahn series field $K[[T^\Gamma]]$. – Tusif Ahmed Mar 08 '16 at 13:16
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    Captain Lema but I need prime $P$ in an integral domain such that the intersection $\bigcap_{n=1}^{\infty}P^n$ is not prime. In your example $\bigcap P^n$ is nonzero prime. – Tusif Ahmed Mar 08 '16 at 13:24
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    @TusifAhmed Captain Lama wanted to show that your claim "in finitely generated case we only need to find $P$ such that $\cap_{n=0}^\infty P^n≠(0)$" is not correct, the intersection being a non-zero prime ideal. His example also answers the first comment. – user26857 Mar 08 '16 at 15:45
  • Okay I understand Thank you. – Tusif Ahmed Mar 08 '16 at 17:58
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    There are examples of local integral domains $(R,m)$ such that $m\supsetneq m^2=m^3=\cdots$. – user26857 Jun 26 '18 at 17:45

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Let $k$ be any field. Set $$R = \bigcup_{n=1}^{\infty} k\left[x,\ y,\ x^{1/n} y^{1/n} \right].$$ Each one of the terms in the union is a domain, so the rising union is also a domain. Let $P$ be the ideal $\langle x, y, x y, x^{1/2} y^{1/2}, x^{1/3} y^{1/3}, x^{1/4} y^{1/4}, \cdots \rangle$. Then $R/P = k$, so $P$ is prime (and even maximal).

We have $P^k = \langle x^k , y^k, x y, x^{1/2} y^{1/2}, x^{1/3} y^{1/3}, x^{1/4} y^{1/4}, \cdots \rangle$ and $\bigcap_{k=1}^{\infty} P^k = \langle x y, x^{1/2} y^{1/2}, x^{1/3} y^{1/3}, x^{1/4} y^{1/4}, \cdots \rangle$. So $R/\bigcap_{k=1}^{\infty} P^k = k[x,y]/(xy)$ which is not a domain, and we see that $\bigcap_{k=1}^{\infty} P^k$ is not prime.

David E Speyer
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    Note that the rings $k[x,y,x^{1/n}y^{1/n}]$ do not quite form an increasing union: for instance, $k[x,y,x^{1/2}y^{1/2}]$ is not contained in $k[x,y,x^{1/3}y^{1/3}]$. But there is a cofinal subsequence which is increasing, e.g. those of the form $k[x,y,x^{1/n!}y^{1/n!}]$. – Eric Wofsey Sep 27 '18 at 05:16