I asked this question on math.stackexchange a month ago with no progress, even after a bounty. I hope to eliminate one if the other receives a satisfactory answer.

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.


  • 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...

  • $\begingroup$ Such examples will be difficult, considering that if you further assume that the ring is Noetherian, then dual rings are precisely zero dimensional Gorenstein rings. $\endgroup$
    – Mohan
    Commented Jun 26, 2015 at 21:42
  • $\begingroup$ a.k.a. the quasi-Frobenius rings, yes, I'm aware of those. The search for an example has been even more obscure than the time I spent looking for a right-not-left coHopfian ring (and finally found one in the literature.) $\endgroup$
    – rschwieb
    Commented Jul 16, 2015 at 15:42
  • $\begingroup$ May I ask where you found these examples with Krull dimension greater than 0? I would be interested to see the proof that this $R = D \oplus M$ is a dual ring. $\endgroup$ Commented Jul 16, 2015 at 16:07
  • $\begingroup$ @MannyReyes Always a pleasure :) In Nicholson and Yousif's book Quasi-Frobenius rings it is example 6.6 on page 133 ("Clark example"). It's also $P$-injective, simple-injective but not self-injective. I thought for some time about generalizing it to higher Krull dimension. I was thinking that it might work for any valuation ring, and so you could get deeper chains of primes. Feel free to drop me an email again if you think there'll be a lot to talk about :) $\endgroup$
    – rschwieb
    Commented Jul 16, 2015 at 18:14

1 Answer 1


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 $\mathfrak p$. \\


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 $$ A(N)\subseteq \bigcap_{\mathfrak p \textrm{min}} \mathfrak p = N. $$ \\

  • $\begingroup$ Nice! I had noticed that I only needed to prove primes contained their annihilator, but I didn't apply the same analysis to the prime ideal itself, which clearly gives us that. The part about a domain not having a minimal ideal can be abbreviated a lot, but it certainly looks complete! $\endgroup$
    – rschwieb
    Commented Jul 18, 2015 at 11:39
  • $\begingroup$ @rschwieb it would be fair to accept the answer and let Keith Kearnes have the bounty. $\endgroup$
    – user70876
    Commented Jul 22, 2015 at 13:09
  • $\begingroup$ Dear @LennyExeter : No need to be impatient: a glance at my network profile should tell you such prodding is unnecessary. Regards $\endgroup$
    – rschwieb
    Commented Jul 22, 2015 at 13:53
  • $\begingroup$ Dear @rschwieb - apologies! $\endgroup$
    – user70876
    Commented Jul 22, 2015 at 14:17

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