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We know that every prime ideal is primary ideal. But can we say, every primary ideal is a power of prime ideal? if it is not correct a counterexample.

Thanks.

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I am a little tired right now so I can't figure out an example, but even though the example with $\mathbb Z$ seems convincing (which is probably your intuition behind it, I believe), I don't think it's true ; perhaps you should just check this out. en.wikipedia.org/wiki/Primary_ideal There's an example there where a primary ideal is not always a power of its radical. –  Patrick Da Silva Dec 22 '11 at 10:03
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$(X,Y^2)\subset K[X,Y]$, where $K$ is a field. –  Pierre-Yves Gaillard Dec 22 '11 at 10:38
    
@Pierre; I was searching for an easier counterexample. –  Vahid Dec 22 '11 at 15:11
    
So, you knew the answer. –  Pierre-Yves Gaillard Dec 22 '11 at 15:19
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Dear @Matt: If $K=A\times B$ where $A$ and $B$ are nonzero rings, then the image of $(1,0)\in K$ in $K[X,Y]/(X,Y^2)$ is a non-nilpotent zero divisor. (More generally, if $(0)$ is not primary in $K$, then $(X,Y^2)$ is not primary in $K[X,Y]$.) ($+1$ to your answer!) –  Pierre-Yves Gaillard Jun 16 '12 at 16:34

1 Answer 1

From Atiyah-MacDonald, page 51, Example 2):

"Let $A=k[x,y], \mathfrak q = (x,y^2)$. Then $A / \mathfrak q \cong k[y]/(y^2)$, in which the zero-divisors are all the multiples of $y$, hence are nilpotent. Hence $\mathfrak q$ is primary, and its radical $\mathfrak p$ is $(x,y)$. We have $\mathfrak p^2 \subset \mathfrak q \subset \mathfrak p$ (strict inclusions), so that a primary ideal is not necessarily a prime power."

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Even though this example was already given in the comments I am posting it as an answer to reduce the number of questions without answers. –  Rudy the Reindeer Jun 16 '12 at 15:44
    
Sorry I don't get it. how can you conclude from the fact $p^{2} \subset q \subset p$ that $q$ is not the power of prime ideal. I think at most you could only conclude that it is not the square of a prime ideal. Please explain for me. Thanks... –  le duc quang Sep 22 '13 at 8:09
    
@leducquang If $p^2 \subsetneq q \subsetneq p$ then neither $q=p^2$ nor $q=p$ but there are no powers of $p$ between $p$ and $p^2$ so therefore $q$ is not equal to a power of $p$. –  Rudy the Reindeer Sep 23 '13 at 5:51
    
Why $q$ must be a power of $p$ and not be a power of some other ideal? I don't get it. Can you make it clearer? –  le duc quang Sep 23 '13 at 15:36
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@leducquang Can you give me a list of all prime ideals containing $q$? –  Rudy the Reindeer Sep 25 '13 at 6:06

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