Commutative Algebra 12

Some Results on Posets

In this article we have two goals in mind:

  • to introduce the idea of noetherian posets, and
  • to state Zorn’s lemma and give some examples.

The latter is of utmost importance in diverse areas of mathematics.

Definition.

A partial ordering on a set S is a relation ≤ on S × S, satisfying the following.

  • (Reflexive) For any x\in S, we have x\le x.
  • (Transitive) For any x, y, z\in S, if x\le y and y\le z, then x\le z.
  • (Anti-symmetric) For any x, y\in S, if x\le y and y\le x, then x=y.

total ordering is a partial ordering such that for any x,y \in S, either x\le y or y\le x.

partially ordered set (or just poset) is a set together with an assigned partial ordering. Similarly, we have totally ordered sets.

Given a partially ordered set (S, \le), for x,y\in S, we write:

  • x < y if x\le y and x\ne y;
  • x\ge y if y \le x;
  • x > y if y < x.

Examples

  1. The set of real numbers (or rational numbers, or integers) forms a totally ordered set under the usual arithmetic ordering.
  2. If X is a set, let P(X) be the collection of all subsets of X. This forms a partially ordered set under inclusion. If X has more than one element, P(X) is not totally ordered.
  3. Any subset T of a partially ordered set S gives a partially ordered set. If S is totally ordered, so is T.

Lemma 1 (Duality).

If (S, \le) is a poset, then so is (S, \ge).

Proof.

Easy exercise. ♦

Duality allows us to cut our work by half in a lot of cases.

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Bounds

Let (S, \le) be a poset.

Definition.

  • The maximum of S is an m_1 \in S such that for any x\in S we have x\le m_1.
  • The minimum of S is an m_0 \in S such that for any x\in S we have x \ge m_0.
  • maximal element of S is an m\in S such that for any x\in S, if x \ge m then x=m.
  • minimal element of S is an m'\in S such that for any x\in S, if x \le m' then x=m'.

warningThe naming is a little confusing, but one must differentiate between the maximum of a set and the maximal elements of a set. In the poset below, S has three maximal elements but no maximum.

For example, let X = {abc} and let S be the following subsets of P(X) under inclusion.

sample_poset

The following properties are obvious.

Lemma 2.

  • The maximum (resp. minimum) of a poset is unique if it exists.
  • If the maximum (resp. minimum) of a poset exists, it is the unique maximal (resp. minimal) element.

Proof

Easy exercise. ♦

Exercise

Suppose S has a unique maximal element m_1. Must m_1 be the maximum of S?

Finally, for a subset T of an ordered set S, we define the following.

Definition.

An upper bound (resp. lower bound) of T in S is an x\in S satisfying: for all y\in T, we have y\le x (resp. y \ge x).

Clearly, upper and lower bounds are not unique in general. E.g. for S = \mathbb Z under the arithmetic ordering, the subset T of even integers has no upper or lower bound. The subset T’ of squares has lower bounds 0, -1, -2, … but no upper bound.

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Noetherian Sets

This will be used a few times in our study of commutative algebra.

Definition.

A poset S is said to be noetherian if every non-empty subset T of S has a minimal element (in T).

It is easy to see that every finite poset S is noetherian.

  • Let T\subseteq S be any non-empty subset. Pick x_0 \in T. If x_0 is a minimal element of T we are done; otherwise there exists x_1 \in T, x_1 < x_0. Again if x_1 is a minimal element of T we are done; otherwise we repeat. The process terminates since T is finite because we cannot have x_0 > x_1 > \ldots in T. Thus T has a minimal element.

Examples

  1. The set \mathbb N of positive integers under ≤ is noetherian.
  2. The set \mathbb N \times \mathbb N is noetherian, where (m, n) \le (m', n') if m\le m' and n \le n'.
  3. The set \mathbb N is noetherian, where we take m\le n to mean m|n.

Note that examples 2 and 3 are not totally ordered. The key property of noetherian sets is the following.

Theorem (Noetherian Induction).

Let T be a subset of a noetherian poset S satisfying the following.

  • If x\in S is such that (y \in S, y < x \implies y \in T), then x\in T.

Then T = S.

Note

To paraphrase the condition in words: if T contains all elements of S smaller than x, then it must contain x itself.

Proof

If T\ne S, U = S - T is non-empty so it has a minimal element x. By minimality, any y\in S with y < x cannot lie in U so y\in T. But by the given condition this means x\in T, which is a contradiction. ♦

Here is an equivalent way of expressing the noetherian property.

Proposition 1.

A poset S is noetherian if and only if the following hold.

  • For any sequence of elements x_1 \ge x_2 \ge \ldots in S, there is an n such that x_n = x_{n+1} = x_{n+2} = \ldots.

Proof

(⇒) Suppose S is noetherian and x_1 \ge x_2 \ge \ldots are elements of S. The set \{x_n : n=1,2,\ldots\} thus has a minimal element x_n. Since we have x_n \ge x_{n+1} \ge \ldots, equality must hold by minimality.

(⇐) Suppose S is not noetherian; let T\subseteq S be a non-empty subset with no minimal element. Pick x_1 \in T; it is not minimal, so we can find x_2 \in T with x_2 < x_1. Again since x_2\in T is not minimal we can find x_3 \in T with x_3 < x_2. Repeat to obtain an infinitely decreasing sequence. ♦

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Zorn’s Lemma

Finally we have the following critical result.

Theorem (Zorn’s Lemma).

Let S be a poset. A chain in S is a subset which is totally ordered. Suppose S is a poset such that every chain T\subseteq S has an upper bound in S.

Then S has a maximal element.

A typical application of Zorn’s lemma is the following.

Proposition 2.

Every vector space V over a field k has a basis.

Note

This looks like an intuitively obvious result, but try finding a basis for the ℝ-space of all real functions f:\mathbb R\to \mathbb R. Using Zorn’s lemma, one can show that a basis exists but describing it does not seem possible. Generally, results that require Zorn’s lemma are of this nature: they claim existence of certain objects or constructions without exhibiting them explicitly. Some of these are rather unnerving, like the Banach-Tarski paradox.

Proof

Let \Sigma be the set of all linearly independent subsets of V. We define a partial order on \Sigma by inclusion, i.e. for C, D \in \Sigma, C\le D if and only if C\subseteq D. Next, we claim that every chain \Sigma' \subseteq \Sigma has an upper bound.

Let D =\cup_{C \in \Sigma'} C; we need to show that D is a linearly independent subset of V, so that D\in \Sigma is an upper bound of \Sigma'. Suppose

\alpha_1 v_1 + \ldots + \alpha_n v_n = 0, \quad v_1, \ldots, v_n \in D, \alpha_1, \ldots, \alpha_n \in k.

For each i=1, \ldots, n, we have v_i \in C_i for some C_i \in \Sigma'. But \Sigma', being a chain, is totally ordered so without loss of generality, we assume C_1 \supseteq C_i for all 1 \le i \le n. This means v_1, \ldots, v_n \in C_1; since C_1 is linearly independent, we have \alpha_1 = \ldots = \alpha_n = 0.

Now apply Zorn’s lemma and we see that \Sigma has a maximal element C. We claim that a maximal linearly independent subset of V must span V. If not, we can find v\in V outside the span of C. This means C \cup \{v\} is linearly independent, thus contradicting the maximality of C. ♦

warningZorn’s lemma holds vacuously when S is empty. However, in many applications of the lemma, constructing an upper bound for a chain T\subseteq S requires T to be non-empty. In such instances, we will mentally replace Zorn’s lemma with the following equivalent version.

  • If S is a non-empty poset such that every non-empty chain in S has an upper bound in S, then S has a maximal element..

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