If $R$ is generically Gorenstein, then $\operatorname{Ann}(\operatorname{Ann}(I))=I$ for every ideal $I$ with non-zero annihilator

Question

Let $R$ be a Noetherian ring such that $R_P$ is a Gorenstein ring (https://en.wikipedia.org/wiki/Gorenstein_ring) for every minimal prime ideal $P$ of $R$.

Is it true that $\operatorname{Ann}_R (\operatorname{Ann}_R(I))=I$ for every ideal $I$ of $R$ with $\operatorname{Ann}_R(I)\ne 0$ ?

Answer 1

I don't think so. Take $R = k[x,y]/(xy)$, where $k$ is a field. This ring is Gorenstein, so it is locally Gorenstein.

Take $I = x^2R$. Then $ann_R(I) = yR$ and $ann_R yR = xR$.