11
$\begingroup$

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.)

$\endgroup$
5
  • $\begingroup$ I might be missing something: isn't it true that any field is a local and dual ring? Then fields would form a family of examples. $\endgroup$
    – Pierre-Guy Plamondon
    May 21, 2015 at 12:20
  • $\begingroup$ @Pierre-GuyPlamondon What you said is true, but a field has Krull dimension zero, so it doesn't satisfy the requirement of having nonzero Krull dimension. $\endgroup$
    – rschwieb
    May 21, 2015 at 13:30
  • $\begingroup$ That is true. I don't see another example at the moment. Perhaps the condition on the Krull dimension should be added in the statement of your question? $\endgroup$
    – Pierre-Guy Plamondon
    May 21, 2015 at 13:34
  • $\begingroup$ @Pierre-GuyPlamondon Oh, I see what you mean. I actually forgot that fields are exceptional in my discussion since the annihilator of N isn't proper. You're right that I have to eliminate them manually. Among nonfields, Krull dimension > 0 is necessary, and that is what I had been thinking of. Thanks for catching that before it went too far. In questions with a lot of hypotheses, it's very annoying to have left out a detail like that. $\endgroup$
    – rschwieb
    May 21, 2015 at 14:03
  • $\begingroup$ For those following: I've crossposted it to mathoverflow. I'm currently trying a bounty on that version of the question. $\endgroup$
    – rschwieb
    Jul 16, 2015 at 15:46

1 Answer 1

Reset to default
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .