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Let $A$ be a ring and consider three prime ideals in $A$ with inclusions $P\subset Q\subset P'$. Suppose that $P,P'\in \operatorname{Ass}(A)$. Must then also $Q\in \operatorname{Ass}(A)$?

user26857
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evgeniamerkulova
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1 Answers1

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It seems not. E.g., let $k$ be a field, $A=k[x,y,z]/(z^2,xz,yz)$ (plane with embeddeded point). The associated primes are $(z)$ and $(x,y,z)$, but the intermediate ideal $(x,z)$ is not.

John Brevik
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  • Thanks for your nice answer: +1. I see that, since $(z)=Ann(x)=Ann(y)$ (moreover it is a minimal prime) and $(x,y,z)=Ann (z)$ in $A=k[x,y]\oplus k\cdot z$, these prime ideals are indeed associated. But why is the prime ideal $(x,z)$ not associated? – evgeniamerkulova Feb 19 '18 at 12:27
  • Ah, I see now: It is because we have the primary decomposition $(0)=(x,y)∩(z)$. The only associated primes of $R$ are then $\sqrt (z)=(z)$ and $\sqrt {(x,y)}=(x,y,z)$ . Since $(x,z)$ is not one of those two, it is not an associated prime. – evgeniamerkulova Feb 19 '18 at 23:02