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Let $P$ be a poset and denote by $\operatorname{Hom}(P, \mathbb N)$ the set of all monotone functions from $P$ to natural numbers $\mathbb N$. Under what conditions on $P$ Is it possible to recover the order on $P$ from the knowledge of $\operatorname{Hom}(P, \mathbb N)$?

I should mention here that the only example I am interested in is the poset of prime ideals in a commutative Noetherian ring.

It would be great if you could include references.

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3 Answers 3

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Sitting inside $Hom(P,{\bf N})$ is the set $P^*$ of monotone functions from $P$ into $\{0,1\}$. This set carries the structure of a "Stone lattice", and the normal lattice homomorphisms from any Stone lattice into $\{0,1\}$ will be a poset, in the case of $P^*$ it recovers $P$. In fact I have just described a dual equivalence between the category of posets with order preserving maps and the category of Stone lattices with normal lattice homomorphisms. See Theorem 6.33 of my book Lipschitz Algebras (second edition).

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  • $\begingroup$ Presumably in the case when $P$ is the poset of prime ideals of a ring, $P^∗$ is isomorphic to the lattice of semiprime ideals? $\endgroup$ Jul 19, 2012 at 21:37
  • $\begingroup$ Douglas, maybe something like that is true, but it isn't literally true in general --- take the ring ${\bf Z}$, its spectrum corresponds to prime numbers (as a poset, unordered), the semiprime ideals correspond to natural numbers, and $P^*$ corresponds to the power set of the set of prime numbers. $\endgroup$
    – Nik Weaver
    Jul 19, 2012 at 22:12
  • $\begingroup$ @Nik Weaver: Thanks for the answer Nik! Actually this still not the kind of answer I want, probably because I haven't stated my question very well. I've just asked another question here: mathoverflow.net/questions/102704/… $\endgroup$ Jul 19, 2012 at 23:41
  • $\begingroup$ Excellent. Incisive. An exemplar of "definitive answer" (if only the question were what the questioner truly intended, etc.) $\endgroup$ Jul 19, 2012 at 23:44
  • $\begingroup$ @Nik. What I had in mind amounts to taking finite subsets of the set of prime numbers whereas you are allowing arbitary subsets. (Actually in $Z$ shouldn't you be including the zero ideal as a prime ideal, so that $P$ isn't an unordered set?) Your approach seems to jettison the topology that is available if $P$ is a poset of prime ideals. But I suppose that not every poset is a poset of prime ideals of a ring? $\endgroup$ Jul 20, 2012 at 23:03
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If you have given all linear extensions of $\mathcal{L}(P)$ of a poset $P$. This is the set of all linear orderings (permutations) of the vertex set of $P$ preserving the order in $P$. The order in $P$ can than be recovered as $v < w$ in P iff $v <_\sigma w$ for all permutations $\sigma \in \mathcal{L}(P)$.

Proof:

One direction is obvious: if $v < w$ in P, then, by definition, $v <_\sigma w$ for all permutations $\sigma \in \mathcal{L}(P)$.

To obtain the other direction, observe that posets are in 1-1 correspondence with directed acyclic graphs, see e.g. the last paragraph of http://en.wikipedia.org/wiki/Partially_ordered_set#Strict_and_non-strict_partial_orders.

Let us consider $P$ to be identified with its directed acyclic graph. Since adding an edge either from $v$ to $w$ or from $w$ to $v$ does not create a cycle in this graph, we have posets $P_{v < w}$ and $P_{w < v}$ with $v < w$ and $w < v$ respectively. Since $$ \emptyset \neq \mathcal{L}(P_{v < w}), \mathcal{L}(P_{w < v}) \subseteq \mathcal{L}(P),$$ we finally found two permutations $\sigma,\tau \in \mathcal{L}(P)$ with $v < _ \sigma w$ and $w < _ \tau v$. $\qquad\square$

For references see the wiki page on linear extensions:

http://en.wikipedia.org/wiki/Linear_extension

Or am I misunderstanding something in your question?

Best, Christian

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  • $\begingroup$ Does this work for infinite posets? $\endgroup$ Jul 19, 2012 at 10:42
  • $\begingroup$ good point! I add a proof above which doesn't distinguish between finite and infinite posets. $\endgroup$ Jul 19, 2012 at 13:12
  • $\begingroup$ @Christian: I think what you are describing is a way to recover $P$ from the group of poset automorphisms of $P$. What I am looking at is a result similar to Yoneda's lemma in spirit. Yoneda's lemma says that knowing object $X$ in a category is equivalent to knowledge of the functor $Hom(X, - )$. I would like to know if in case of posets one can get away with only konwing $Hom(P,\mathbb N)$. $\endgroup$ Jul 19, 2012 at 23:06
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If I'm not mistaken, Novotny's paper "Ueber gewisse Eigenschaften von Kardinaloperationen" deals with [and I think solves, if you use the partial ordering on $Hom(P,\mathbb N)$] this question.

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    $\begingroup$ Bibliographical details: Novotný, Miroslav, Über gewisse Eigenschaften von Kardinaloperationen, Spisy Přírod. Fak. Univ. Brno 1960 1960 465–484, MR0133241 (24 #A3075). $\endgroup$ Apr 14, 2015 at 23:44

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