On Gorenstein ring of dimension zero

Question

Let $R$ be an Artinian local ring. Then $R$ is a Gorenstein ring (i.e., $R$ is an injective $R$-module) iff for any ideal $I$ of $R$, Ann$($Ann$(I))=I$. Why? (We call $R$ Gorenstein if injective dimension of $R$ is finite.)

Answer 1

This is a consequence of a much more general equivalence of conditions defining quasi-Frobenius rings:

Theorem 15.1 (T.Y.Lam, Lectures on modules and rings pg 409)

For any ring $R$, the following are equivalent:

(1) $R$ is right Noetherian and right self-injective
(4) $R$ is two-sided Artinian and satisfies the following conditions:
-(a) $l.ann(r.ann(L))=L$ for every left ideal $L$
-(b) $r.ann(l.ann(T))=T$ for every right ideal $T$

The $l.ann$ means "left annihlator" and the $r.ann$ means "right annihilator," but of course in your case the both amount to $Ann(Ann(-))$

You can find a complete proof (it's a little involved) in the book cited, an I think probably in Anderson and Fuller's book, and most definitely in Nicholson and Yousif's book.

I bet that the extra assumptions you have about commutativity and localnes will allow you to make some simplifications.

Another fact to keep in mind is this one:

Among commutative Artinian local rings, the Gorenstein ones are exactly the ones with a (nonzero) minimal ideal contained in all other ideals. Perhaps you will find a second path using this equivalence.