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A regular local ring is a domain. But in general, a regular ring is not domain, so you can find regular rings with nilpotent elements.

I am unable to construct an example of (A, I) as

A is a regular ring, I is a nilpotent ideal and A/I is regular

Any help would be appreciated

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1 Answer 1

up vote 6 down vote accepted

There is no. A regular ring is reduced :

  • a ring is reduced if and only if all its localization at prime ideals are reduced ;
  • the localization at a prime ideal of a regular ring is a local regular ring (definition) hence reduced.

However, a regular ring can contains zero-divisors. For example the ring $k[x]\times k[y]$, representing the disjoint union of two affine lines, is not a domain but regular.

And by the way, if $I$ is a nilpotent ideal, $A/I$ needs not contain nilpotent element, and in fact, it would have less nilpotent elements than $A$.

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"the localization at a prime ideal of a local ring is regular (definition) hence reduced." there's a typo here: you mean that localization of a regular ring is still regular. –  S123 Feb 28 '12 at 14:07
    
Corrected, thank you ! –  Lierre Feb 28 '12 at 15:16
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