Example of a commutative, local, dual ring with nilradical $N$ such that $ann(N)\nsubseteq N$ For an ideal $I\lhd R$ in a commutative ring $R$, let $ann(I)$ denote the annihilator of $\{x\in R\mid xI=\{0\}\}$. A commutative ring $R$ is said to be a dual ring if for every ideal $I$ of $R$, $ann(ann(I))=I$.

What are examples of commutative, local, dual rings with nonzero Krull dimension  whose nilradical $N$ satisfies $ann(N)\nsubseteq N$? (Or is this not possible?!)

I'd be surprised if examples weren't possible.
A great deal of commutative, local, dual rings that I've encountered (many of them trace back to Hajarnavis and Norton's articles) turn out to be zero dimensional, so their Jacobson radical (the maximal ideal) is nil, and $ann(N)\subseteq N$ trivially (as long as $N$ is nonzero, which it is in these examples. Otherwise we are looking at a field.)
The only examples I've seen that aren't $0$-(Krull) dimensional are based on a construction which uses a valuation domain $D$, its field of fractions $Q$, and the $D$ module $M=Q/D$ in a trivial extension $R=D(+)M$ whose ideals are linearly ordered. The problem with this construction is that $0(+)M=N$ is the nilradical, $N$ is a faithful $D$ module, and $N^2=\{0\}$, which implies that $ann(N)=N$ (in $R$.)
Other than that construction, I'm not sure how to produce local, dual rings with nonzero Krull dimension.
(NB: to me, local means "has a unique maximal ideal," no Noetherian assumption.)

If you happen to be following the question, the news is that a solution was given on mathOverflow by Keith Kearnes.
He showed that one can actually prove a lemma that in such a ring, each prime contains its annihilator, which was a pleasant surprise. Armed with this lemma, it is easy to prove that the nilradical contains its own annihilator.
