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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

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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
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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.

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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
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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.

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