1
$\begingroup$

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

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ You can localize at the ideal $(x,y)$ to get a local example. $\endgroup$ May 13, 2013 at 14:07
  • 1
    $\begingroup$ More generally, any reduced Gorenstein local ring of dimension one will satisfy this double-annihilator property. $\endgroup$ May 13, 2013 at 15:31
  • $\begingroup$ Thanx. Could you give a hint on how to prove it? $\endgroup$
    – QED
    May 13, 2013 at 17:22
  • 1
    $\begingroup$ 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. $\endgroup$ May 13, 2013 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.