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I know that it seems very loose as a title but I hope this post will be beneficial to all the forum members.

One thing I like about free modules is that they help one define maps directly as we do in a vector space by just defining the images of the elements of a base (if it exists).

My questions are:

  1. I have read somewhere that two minimal generating sets for a free module do not necessarily have the same cardinality, except if the corresponding ring is local. Is that true? What is the intuition behind a ring being "local" then?

  2. "A map (module map of course) from our free module to itself is bijective iff it is injective." In which general setting is that statement true?

I hope that post end up containing many examples and counterexamples that are certainly beneficial to beginners like myself.


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The fragment "in which is general setting that is available" is ungrammatical -- do you mean something like "in which general setting is this result available"? – joriki May 15 '11 at 1:40
it was a typo just corrected it sorry :) – El Moro May 15 '11 at 1:42
Look up invariant basis number. – Arturo Magidin May 15 '11 at 1:43
Your first question may confuse "basis" with "minimal generating set". In a local ring they are the same thing, but in a general ring such as the integers Z, one can have large minimal generating sets such as { 1 }, { 2, 3 }, { 6, 10, 15 }, and { 30, 42, 70, 105 }. Your second question is definitely an IBN question. – Jack Schmidt May 15 '11 at 1:54
Thank you for clarifying this. I know that the two notions are different but I appreciate that you stressed the difference – El Moro May 15 '11 at 2:03

1 In $\mathbb Z$ seen as a free module of rank one over itself, the sets consisting of $\{1\}$ on one hand and $\{2,3\}$ on the other are minimal generating sets of different sizes.

2 is false for every domain $A$ , unless it is a field. Indeed if injective implied surjective for the free module $A^1$, then for every non zero $a\in A$, the map $A^1 \to A^1:x\mapsto ax$ , being injective, would be surjective and so $a$ would have as multiplicative inverse the $x\in A$ mapping to $1$.

However a theorem of Vasconcelos states that an endomorphism of a finitely generated module (not assumed free) over any commutative ring is bijective if and only if it is surjective.

Full Disclosure My main motivation for answering this question is that it gives me an opportunity to advertise Vasconcelos's theorem , proved on page 9, Theorem 2.4, of Matsumura's Commutative Ring Theory . Even in Atiyah-Macdonald's excellent Introduction to commutative Algebra this result is given, in Exercise 6.1, only with the superfluous hypothesis that the finitely generated module be noetherian.

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It's very possible I'm misunderstanding something, and if so, please forgive this question, but how is $\{1,2\}$ a minimal generating set for $\mathbb{Z}$ as a module over itself when it properly contains the generating set $\{1\}$? – Keenan Kidwell May 16 '11 at 14:59
Keenan, you are perfectly right: I made a stupid typo. I meant the sets $\{1\}$ and $\{2,3\}$ . So it is not I who have to forgive you, but the other way round! I'm going to edit my answer immediately. Thanks for pointing this out. – Georges Elencwajg May 16 '11 at 22:37
This theorem of Vasconselos (1969) was proved a few years earlier by Strooker ("Lifting Projectvies", Nagoya Math. J. 27 (1966), 747--751; see the proof on p. 750). – KCd Jun 4 '12 at 0:03
@KCd: very surprising ! Have you known this for a long time? – Georges Elencwajg Jun 4 '12 at 0:38
@Georges: Not for a long time, just a couple of years. The paper by Strooker is in the references of Morris Orzech's article "Onto Endomorphisms are Isomorphisms" in the Amer. Math. Monthly 78 (1971), 357--362. That might be where I came across it. – KCd Jun 4 '12 at 1:11

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