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I am studying out of Matsumura's Commutative Ring Theory, and in the first section on modules he proves (following Kaplansky) that every projective module over a local ring is free.

My questions have more to do with an application of transfinite induction than the actual algebra.

In the proof of the above result, Matsumura uses a lemma, and it is the proof of the lemma that I have a question on. The proof can be seen at Google Books: link. I am referencing Lemma 1 on page 10.

In the lemma, we define a family of submodules $\{F_\alpha\}$ using transfinite induction. I can follow the proof, i.e., I understand the (very) basic mechanics of transfinite induction and how each $F_\alpha$ is constructed and why the desired result follows.

Here are my questions:

(1) For every ordinal $\alpha$, we define a submodule $F_\alpha$. The ordinals do not form a set. Is it then true that I cannot speak of the set of all submodules $F_\alpha$?

Matsumura writes about half way down the page that "if $F_\beta = F$ then the construction stops at $F_\beta$." Must this eventually happen? Must this construction terminate?

I am having a hard time wrapping my mind around the fact that we have modules (which are sets) and submodules (which are sets), and then all of the sudden we have a family of submodules which isn't (if I have all this right) a set.

(2) Why can I not define a sequence of submodules over $\omega$? Why do I need the full strength of transfinite induction? Why do I need all ordinals?

As is obvious, this transfinite stuff is quite new to me; I have never used these techniques before. I would appreciate any insight.

Martin Sleziak
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John Myers
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  • Let $M$ be any set, pick $N\subseteq M$ a subset and for each ordinal $\alpha$ let $F_\alpha$ be $N$. Then $(F_\alpha)$ is a family of subsets of $M$ which is not a set... This is exactly the same situation as the one causing you trouble in (1). As for the stopping: I cannot access the link but he probably shows that the construction stops, no? – Mariano Suárez-Álvarez Jun 16 '12 at 19:21
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    My guess is the $F_\alpha$ form a strictly increasing sequence of submodules: in that case, the construction has to stop because otherwise you would produce mode submodules of $F$ that it can possible can (because there are ordinals whose cardinal is larger than that of the set of submodules of $F$) – Mariano Suárez-Álvarez Jun 16 '12 at 19:24
  • Dang, I was worried the link wouldn't work. I apologize. The $F_\alpha$ are constructed to be a well-ordered and increasing sequence such that their union is $F$. – John Myers Jun 16 '12 at 19:33

1 Answers1

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The idea is to generalize the idea of induction because some things require more than just a countably infinite sequence of steps in order to construct them.

For example, Borel sets are constructed by a transfinite induction and the construction is of length $\omega_1$, that is to say that we do not exhaust this construction until we exhausted all the countable ordinals.

Secondly, the ordinals are simply used to index the modules. $F_\alpha$ is merely a subset of $F$. We can talk on the set of subsets of $F_\alpha$ which are submodules. This is like saying $A_n=\text{some set of real numbers}$ and then talking about $A_\omega=\bigcup A_n$. Does that imply that $A_n$ is a set of natural numbers? Not at all.

Lastly, these sort of construction are usually of the form $F_\alpha\subsetneq F_{\alpha+1}\subsetneq\ldots$ and so if the construction does not stop we can define the function $\alpha\mapsto F_{\alpha+1}\setminus F_\alpha$ which is injective (do you see why?) and its range is a subset of $\mathcal P(F)$. This means that $F$ is a set whose power "set" is actually a proper class which is a contradiction to the power set axiom (which says that the power set of a set is indeed a set).

Asaf Karagila
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  • Can you confirm that $\omega_1$-many steps are in fact sufficient for the construction of generated $\sigma$-algebras (of a family of sets) in general? I have a proof, it's easy, but on Wikipedia for instance it is only mentioned for the Borel algebra of a metric space. – Martin Brandenburg Dec 14 '20 at 21:14
  • @Martin: Yes, see also https://math.stackexchange.com/questions/54172/the-sigma-algebra-of-subsets-of-x-generated-by-a-set-mathcala-is-the-s/54179#54179 (I use the notation from descriptive set theory, but there's no real mentioning of topology there). – Asaf Karagila Dec 14 '20 at 21:17
  • ____Thanks!____ – Martin Brandenburg Dec 14 '20 at 23:55