Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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-Alvarez Jun 16 '12 at 19:21
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-Alvarez 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
up vote 1 down vote accepted

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.