Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In a commutative ring R, if I is a primary ideal of R. Is R/I is local ring. Its my understanding i want to know wheater its right or not. kindly help me

share|cite|improve this question
All you can say about the quotient ring by a primary ideal is that $I$ is primary iff $R/I\ne 0$ and all zero-divisors of $R/I$ are nilpotent. – Ehsan M. Kermani Jan 31 '13 at 7:39
up vote 0 down vote accepted

Every prime ideal is primary. But is $R/P$ a local ring for a given prime ideal $P$ for any ring $R$? If it were, then every domain would be a local ring, which is certainly not true.

share|cite|improve this answer
thanks Rankeys but when primary ideal is not prime then it forms local ring? – Visual Patner Jan 31 '13 at 6:36
No. As a hint, look at powers of a prime ideal. These are in general primary ideals which are not prime. – Rankeya Jan 31 '13 at 6:40
rings of integer Z/(p^n) is a local ring. it has only maximal ideal generated by (p). I am confuse because i didnt study in literature that factor ring of primary ideal is local but it happen when i am taking the examples.... – Visual Patner Jan 31 '13 at 6:57
Okay. Then take the example $(x^2)$ in $\mathbb{C}[x,y]$. – Rankeya Jan 31 '13 at 6:59

In a Dedekind domain I believe it is true that when you quotient out by a primary ideal you get a local ring: For every primary ideal is a prime power $p^n$ and when one does $A/p^n$, the only prime ideal in here is $p/p^n$.

Also it is not true that the quotient of a primary ideal is always a local ring: Take $A = \Bbb{Z}[x]$ and $I = (x)$; $I$ is primary and $A/I \cong \Bbb{Z}$ which is not a local ring.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.