Example of ideal which is not primary, but its radical is prime.

Question

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.

thanks for your good answer – user371596 Nov 23 '16 at 11:25 — user371596, Nov 23 '16 at 11:25

score 6 · Answer 1 · answered Nov 23 '16 at 08:13

6

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 $n\ge1$.

answered Nov 23 '16 at 08:13

Alex Mathers

18,509

3

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. – rschwieb Nov 23 '16 at 17:01

Example of ideal which is not primary, but its radical is prime.

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