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I was thinking of the fact that, if you take an infinite number of prime ideals $p_i\mathbb{Z}$ in the ring $\mathbb{Z}$, then their intersection is zero. What kind of commutative rings satisfy that the intersection of an infinite number of prime ideals is zero?

Is this true for any integral domain?

Marco Flores
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  • In addition to the answers below, I'll remark that if you change your statement to "the intersection of ALL prime ideals is zero", then this is equivalent to saying that the ring is reduced. But you probably knew this already! – Kenny Wong May 14 '17 at 17:46
  • yep (: I regret the way I stated the question though, because now I want to know if the thing holds for principal ideal domains. But it's my turn to do some thinking I guess – Marco Flores May 14 '17 at 17:49
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    The prime ideals in a PID are precisely the principal ideals of the form $(p_i)$, where $p_i$ is an irreducible element. If an element $x$ is contained inside infinitely many prime ideals, then it has infinitely many irreducible divisors. Since every PID is a UFD, this is impossible, unless $x = 0$. – Kenny Wong May 14 '17 at 17:58

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Take $R=\mathbb{C}[x,y]$, polynomial ring in two variables. Consider maximal ideals $P_a=(x-a,y)$ where $a\in\mathbb{C}$. Then $\cap_a P_a=(y)\neq 0$.

Mohan
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This is not true for any integral domain. Consider a polynomial ring in countably infinitely many variables: $k[x_1,x_2,\dots]$, and consider the ideals $I_1 = (x_1,x_2,\dots)$, $I_2 = (x_1, x_3, x_4,\dots)$, and so on. In other words, $I_k$ is generated by $x_1$ and $x_i$ for all $i>k$. Then the intersection of all $I_k$'s is just $(x_1)$.

Andrew
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In the ring of integers the fundamental theorem of arithmetics holds. This is the reason that the intersection as mentioned is the zero ideal.

Wuestenfux
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