Welcome to MathOverflow

Question

2

1

Let $P$ be a poset and denote by $Hom(P, \mathbb N)$ 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 $Hom(P, \mathbb N)$?

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.

Thanks!

edited Jul 26 at 0:17

François G. Dorais♦
22.6k●2●48●113

asked Jul 19 at 7:03

Mikhail Gudim
407●6

Nik Weaver · Accepted Answer · 2012-07-19 19:22:01Z UTC

4

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 5.1.3 of my book Lipschitz Algebras.

answered Jul 19 at 19:22

Nik Weaver
4,426●1●7●20

	Presumably in the case when $P$ is the poset of prime ideals of a ring, $P^∗$ is isomorphic to the lattice of semiprime ideals? – Douglas Somerset Jul 19 at 21:37
	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. – Nik Weaver Jul 19 at 22:12
	@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/… – Mikhail Gudim Jul 19 at 23:41
	Excellent. Incisive. An exemplar of "definitive answer" (if only the question were what the questioner truly intended, etc.) – paul garrett Jul 19 at 23:44
	Thank you, Paul! – Nik Weaver Jul 20 at 4:49

show 2 more comments

Christian Stump · Answer 2 · 2012-07-19 13:31:35Z UTC

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

Characterizing Posets by Functions Into Natural Numbers

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.