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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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    Dear @Pierre-YvesGaillard Thank you very much for both, your reply and the plus one!2012-06-17

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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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    @leducquang Can you give me a list of all prime ideals containing $q$?2013-09-25