When does "second annihilator" of a (principal) ideal equal the ideal itself , ie $Ann_R(Ann_R(r))=Rr$?

Question

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 $Ann_R(r)\in Ass(R)$

Answer 1

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.

Answer 2

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.

Answer 3

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.