# An example of a ring $R$ with the property that for each $P=Ann_R(r)\in {\rm Min}(R)$ we have $Ann_R(P)=Rr$.

I'm looking for an example of a commutative (preferably local) ring $R$ such that ${\rm dim}R>0$ and $R$ has the property that for each $P=Ann_R(r)\in {\rm Min}(R)$ we have $Ann_R(P)=Rr$.

This question is a follow-up of my previous question

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## 1 Answer

I believe the graded ring $k[x,y]/(xy)$ satisfies this property. The minimal primes are just $(x)$, $(y)$ and the annihilators are just the other ideal. The dimension is 1. This is not however local.

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You can localize at the ideal $(x,y)$ to get a local example. – Graham Leuschke May 13 '13 at 14:07
More generally, any reduced Gorenstein local ring of dimension one will satisfy this double-annihilator property. – Graham Leuschke May 13 '13 at 15:31
Thanx. Could you give a hint on how to prove it? – QED May 13 '13 at 17:22
Consider $R=k[x,y]/(xy)$. Then R is just the vector space spanned by $(1,x,x^2,...,y,y^2,...)$ because any mixed terms of $xy$ will be eliminated. Thus we can calculate the annihilator by considering products of things in this basis. $x^i*y^j=0$ where as all other products just give a power of a variable. Hence the annihilators are proven. We note that $(x)$ and $(y)$ are the minimal primes. Then $(x) \subset (x,y) \subset R$ is a maximal chain of primes so the dimension is 1. – Stephen Sturgeon May 13 '13 at 18:18