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Here is a proof from Atiyah-Macdonald:

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For i) $\implies$ ii) could one not write "If $(x_n)$ is such that $x_m = x_{m+1} = \dots$ then obviously $x_m$ is a maximal element"?

I am asking because the book has been getting terser with proofs and has reached super-terse by now but "i) $\implies$ ii)" seems much longer than necessary. So I must be missing something.

Note: for the equivalence of these two statements we need choice (which is not mentioned in the book).

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Rudy the Reindeer
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3 Answers3

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It seems to me that the countability issue is something of a red herring.

$\Sigma$ is a partially ordered set, and $T$ is a nonempty subset of $\Sigma$. Even if $T$ is countable, it need not be an increasing chain, i.e., of the form $\{x_n\}_{n=1}^{\infty}$ with $x_1 \leq x_2 \leq \ldots \leq x_n \leq \ldots$. So if an increasing chain $\{x_n\}_{n=1}^{\infty}$ in $T$ stabilizes then the eventual value of the sequence is a maximal element of the subset $\{ x_n\}_{n=1}^{\infty}$ of $T$, which need not be a maximal element of $T$ itself. Thus the "longer" proof given is necessary.

Added: For my take on the result in question, see Lemma 10 of these notes on factorization. (This seems to be currently missing from my commutative algebra notes, which only mostly include the material in this shorter set of notes.) To my eye what I say is the same as what Atiyah-Macdonald say, but someone learning the material for the first time might (possibly) think otherwise.

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Pete L. Clark
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  • This is a good point. I don't have a copy of A-M handy, but I assume you are correct that $\Sigma$ is a poset. – Alex Becker Jul 20 '12 at 07:02
  • @Alex: well, for that matter, what I'm saying holds even for countable linearly ordered sets. For instance $\mathbb{Q}$ is such a thing but cannot be written as an increasing chain. – Pete L. Clark Jul 20 '12 at 07:10
  • Thank you Pete, for this enlightening answer. Would you mind showing an example of $T$ with an increasing chain such that the upper bound of the chain is not a maximal element in $T$? I'm having trouble visualising this case. – Rudy the Reindeer Jul 20 '12 at 09:50
  • @Clark: For instance take $T = \mathbb{Z}$ with the usual ordering, and take $x_n = 0$ for all $n$. – Pete L. Clark Jul 20 '12 at 10:30
  • Thank you. I somehow assumed that we have $x_n \neq x_m$ for $n \neq m$. That breaks constant sequences. If one assumes that sequences have to be "proper" (in this sense), then we'd have that every upper bound of every chain in $T$ is maximal, no? – Rudy the Reindeer Jul 20 '12 at 12:22
  • @Clark: I thought we were working under the condition that every chain was eventually constant. If you don't want that hypothesis, take for instance $T = \mathbb{Q}$ and $x_n = 1 - \frac{1}{n}$. – Pete L. Clark Jul 20 '12 at 14:42
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If $\Sigma$ is not countable, then not every subset $T$ is the set of terms of some sequence $(x_n)$.

Alex Becker
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I think the confusion is coming from the fact that $\Sigma$ a priori is any set that we don't know is countable or not - how on earth does not get a "sequence" out of this to contradict the ACC say in a proof? Perhaps the proof of (1) $\implies$ (2) from Isaacs' Algebra may be more illuminating:

Suppose $\Sigma$ satisfies the ACC and that there exists $S \subseteq \Sigma$ not empty such that $S$ has no maximal element. Then for every $a \in S$, the set

$$\{b\in S : b > a\}$$

is not empty. Now we invoke the axiom of choice: given any $a \in S$ I can always choose $b$ such that $b > a$. Viz, there exists a "choice function" $f : S \rightarrow S$ such that $f(a) > a$ for all $a$. Now choose just some $a \in S$ and set $a_1 = a$, $a_2 = f(a_1)$, in general we have $a_{n+1} = f(a_n)$. Then

$$a_1 < a_2 < a_3 < \ldots $$

is an increasing sequence that does not stabilise, contradicting the ACC.