Suppose that $R$ is a (local) ring and $r\in R$. When do the equations $Ann_R(Ann_R(r))=Rr$ or $\sqrt{Ann_R(Ann_R(r))}=\sqrt{Rr}$ hold? I already know that it holds for Artinian Gorenstein rings (due to an exercise in Bruns-Herzog) and it seems to be true for $R=\Bbb {Z}/n\Bbb {Z}$. The question is more interesting when we also assume that $Ann_R(r)\in Ass(R)$
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1$\begingroup$ Quasi-Frobenius Rings are precisely the (Artinian) rings that Ann(ann(I))=I, ann(Ann(T))=T , Ann= right annihilator, ann=left annihilator. $\endgroup$– Maximiliano ValleCommented May 7, 2013 at 12:00
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1$\begingroup$ $R=\Bbb {Z}/n\Bbb {Z}$ is itself an artinian Gorenstein ring. $\endgroup$– Graham LeuschkeCommented May 7, 2013 at 13:19
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$\begingroup$ math.stackexchange.com/questions/384374/… $\endgroup$– user26857Commented May 11, 2013 at 14:35
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3 Answers
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For what it's worth, it suffices for $(r)$ to be an interesection of minimal primes; in fact, more generally if $J$ is any intersection of minimal primes then $J=Annih(Annih(J))$. (This requires only that $R$ be commutative and noetherian; you don't need local.)
This is Lemma 2.17 in an old paper of mine called "Patching Modules of Finite Projective Dimension", where the stated hypothesis is "Let $R$ be any (commutative) ring", but it now seems to me that the proof requires $R$ to be noetherian.
answered May 7, 2013 at 12:29
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$\begingroup$ Thanks for the informative answer. But I'm curious about the conditions on $r\in R$ that implies the equations. Don't you think that $Ann_R(r)\in Ass(R)$ can help somehow? $\endgroup$– QEDCommented May 7, 2013 at 19:04
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$\begingroup$ Is there some analog of this for non-Noetherian rings? $\endgroup$ Commented Jul 27, 2014 at 17:25
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Long time ago, I proved, but never published, that if $R$ is generically Gorenstein, meaning that all $R_P$ are Gorenstein with $P$ an associated (necessarily) minimal prime, then $\operatorname{Ann}(\operatorname{Ann}(I))=I$, for any ideal $I$ with non-zero annihilator.
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1$\begingroup$ Interesting. Is it possible to access the details of the proof? $\endgroup$– QEDCommented Aug 22, 2013 at 11:09
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Let $l(X)$ denote a left annihilator, and $r(X)$ denote a right annihilator.
A ring is called right P-injective if $l(r(a))=Ra$ for all $a\in R$. You can find this condition discussed in detail in Nicholson and Yousif's Quasi-Frobenius rings on page 96.
Rings which are right self-injective are right $P$-injective (and so that's why $\Bbb Z/n\Bbb Z$ has the property for every $n>1$.)
Also, all von Neumann regular rings are left and right $P$ injective, and these can be noncommutative, nonNoetherian and non-self-injective.
