*I [asked this question on math.stackexchange](https://math.stackexchange.com/q/1288445/29335) a month ago with no progress, even after a bounty. I hope to eliminate one if the other receives a satisfactory answer.*




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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 $R$, $ann(ann(I))=I$.



> 
> I am looking for an example (if one is possible) of a commutative, local, dual ring with Krull dimension greater than $0$ such that the nilradical $N$ satisfies $ann(N)\nsubseteq N$.
> 
> 
> 


During my search in the literature, the only dual rings I found which weren'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 $R$.)


I would be grateful for an example, or else some leads on easy methods to construct local dual rings that might lead to an example.




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Additions:


* it's worth noting that no example would be Noetherian: it's well known Noetherian dual rings are quasi-Frobenius, hence Artinian (and thus zero-dimensional).
* UPDATE: as is apparent now, no such ring exists! See the accepted solution. It is a pleasant surprise...



Non-field example of a commutative, local, dual ring with nilradical $N$ such that $ann(N)\nsubseteq N$
There is no such ring.


To lessen my typing, let me introduce some abbreviations.
C = commutative,
L = local,
D = dual,
K = Krull dimension greater than $0$,
$A(I)$ := $\textrm{Ann}(I)$ for $I\lhd R$,
N = $\textrm{Nil}(R)$.


**Theorem.** If $R$ is a CLDK ring, then $A(N)\subseteq N$.


The proof requires the following


**Lemma.** If $R$ is a CLD ring that is not a field
and $\mathfrak p\lhd R$ is a
prime ideal, then $A(\mathfrak p)\subseteq \mathfrak p$.


**Proof of Lemma.**
If $\mathfrak p$ is the maximal ideal, then $A(\mathfrak p)$
is a minimal ideal. Since $R$ is not a field,
this yields $R\neq A(\mathfrak p)$,
so $A(\mathfrak p)\subseteq \mathfrak p$ by locallness of $R$.
For the rest of the proof we consider only the case where
$\mathfrak p$ is nonmaximal.


As is well known,
if $\mathfrak p$ is a prime ideal, then it is $\cap$-irreducible.
(I.e., it is finitely meet irreducible.)
Reason: if $\mathfrak p = I\cap J$ and $\mathfrak p < I, J$,
then we contradict primeness by $IJ\subseteq I\cap J = \mathfrak p$.


**Claim.** If $\mathfrak p$ is nonmaximal, then it is *not*
$\bigcap$-irreducible. (I.e., $\mathfrak p$ is not infinitely meet irreducible.) Said another way, $\mathfrak p$ *will* equal the
complete intersection of all ideals that are properly above $\mathfrak p$.


**Proof of Claim.** Else $\mathfrak p^\*:=\bigcap\_{\mathfrak p < I} I$ is the smallest ideal
strictly above $\mathfrak p$. Then $\mathfrak p^\*/\mathfrak p$ is the smallest nonzero ideal
in the domain $R/\mathfrak p$. Notice that this domain is not a field, since
$\mathfrak p$ was nonmaximal, so by its minimality $\mathfrak p^\*/\mathfrak p$ is a proper ideal
of $R/\mathfrak p$.


$R/\mathfrak p$
is a domain and $0 < \mathfrak p^\*/\mathfrak p\cdot \mathfrak p^\*/\mathfrak p\leq \mathfrak p^\*/\mathfrak p$,
forcing $\mathfrak p^\*/\mathfrak p = (\mathfrak p^\*/\mathfrak p)^2$. Now minimal idempotent ideals in commutative rings, like
$\mathfrak p^\*/\mathfrak p$, are generated by idempotents, meaning
$\mathfrak p^\*/\mathfrak p = (e)$ for some $e\in R/\mathfrak p$ satisfying $e^2=e$. The
element $e$ cannot be zero,
because $\mathfrak p^\*/\mathfrak p$
is not zero, and it cannot be $1$, since $\mathfrak p^\*/\mathfrak p$
is not $R/\mathfrak p$. Thus the domain $R/\mathfrak p$ has a proper idempotent $e$, which is absurd.
($e(1-e) = 0\neq e, (1-e)$ contradicts the definition of a domain.). This proves the claim. \\


Now back to the proof of the lemma. By the claim, if
$\mathfrak p$ is not maximal, then it equals the
intersection of all the ideals that properly contain it.
Applying the lattice anti-isomorphism $I\mapsto A(I)$, which must preserve
the complete lattice operations of the ideal lattice of $R$,
we obtain that $A(\mathfrak p)$ is the join $\bigvee\_{\mathfrak p < I} A(I)$ of all ideals of the
form $A(I)$ where $\mathfrak p < I$. For any such $I$ we have
$$
I\cdot A(I) = 0\subseteq \mathfrak p,
$$
and $I \not\subseteq \mathfrak p$, so $A(I)\subseteq \mathfrak p$.
Hence the join $\bigvee\_{\mathfrak p < I} A(I) = A(\mathfrak p)$
is also contained
in $\mathfrak p$. \\


Now


**Proof of Theorem.**
Suppose $R$ is a CLDK ring and that $\mathfrak p$ is a minimal prime
of $R$. If $\mathfrak q$ is a different minimal prime, then
$$
\mathfrak p \cdot A(\mathfrak p) = 0 \subseteq \mathfrak q,
$$
so either $\mathfrak p\subseteq \mathfrak q$ or
$A(\mathfrak p)\subseteq \mathfrak q$. The former cannot happen,
since $\mathfrak q$ is assumed to be minimal and different from $\mathfrak p$,
so we must have the latter. To reiterate: if $\mathfrak p$ is a minimal prime of $R$, then $A(\mathfrak p)$ is contained in every minimal prime different from $\mathfrak p$.


By the lemma, $A(\mathfrak p)\subseteq \mathfrak p$ as well,
so $A(\mathfrak p)$ is contained in all of the minimal primes. In other
words, if $\mathfrak p$ is a minimal prime, then
$$
A(\mathfrak p)\subseteq \bigcap\_{\mathfrak q \textrm{min}} \mathfrak q
= N.
$$
Dualizing this yields $A(N)\subseteq \mathfrak p$ for every minimal prime $\mathfrak p$. It follows from this that
$$
A(N)\subseteq \bigcap\_{\mathfrak p \textrm{min}} \mathfrak p
= N. 
$$
\\


