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I want to find an example of ideal $Q$ such that $\sqrt{Q}$ is prime, but $Q$ is not primary.

It is clear that our domain would not be PID because $\sqrt{Q}$ should not be maximal ideal.

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Take $Q=(xy,y^2)\subseteq k[x,y]$. Then $\sqrt{Q}=(xy,y)=(y)$, which is prime, and $Q$ is not primary because $xy\in Q$, $y\notin Q$, but $x^n\notin Q$ for all $n\ge1$.

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    $\begingroup$ I love this example $Q$ because it satisfies the "naive simplification" of the definition of "primary" that a lot of people want to make, but isn't actually primary. $\endgroup$ – rschwieb Nov 23 '16 at 17:01

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