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For which integral domains $R$ (not filed) the ring $R[x_1, \ldots, x_n]$ satisfies descending chain condition for radical ideals? I am not expert in Ring Theory and I need an answer to construct some examples in a different area of algebra. This question may have a trivial answer and so I apologize if it is meaningless or quite trivial.

P.S. Is there an infinite ring without nonzero nilpotent element having the above property?

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    $\begingroup$ If $R$ is a domain and contains the integers, then the sequence $(x) \supset (x) \cap (x-1) \supset (x) \cap (x-1) \cap (x-2) \supset \cdots$ is a nonterminating descending chain of radical ideals. $\endgroup$
    – Graham Leuschke
    Commented Feb 15, 2014 at 20:24
  • $\begingroup$ @Graham Leuschke: Thank you, your example shows that there is no such an infinite domain. $\endgroup$
    – Sh.M1972
    Commented Feb 15, 2014 at 20:41
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    $\begingroup$ The same proof shows that there is no such domain for $n\geq 2$, right? Because $R[x_1,\dots,x_n]=R[x_1][x_2,\dots,x_n]$, and $R[x_1]$ is always an infinite domain. Then clearly for $n=1$ finite domains do have this property, answering the question. $\endgroup$
    – Will Sawin
    Commented Feb 15, 2014 at 20:51
  • $\begingroup$ Thank you for comments. What happens if we use commutative rings without nonzero nilpotent elements instead of domains? $\endgroup$
    – Sh.M1972
    Commented Feb 15, 2014 at 21:02
  • $\begingroup$ Some quotient of $R[x_1]$ is an infinite domain, providing the desired example. (A chain of ideals in $Q[x_2,\dots,x_n]$ can be lifted to $R[x_1][x_2,\dots,x_n]$, where $Q$ is a domain which is an infinite quotient of $R[x_1]$.) So, it remains to consider the case $n=1$, which is not rich either. $\endgroup$
    – Sasha Anan'in
    Commented Feb 15, 2014 at 21:40

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