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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.

Regards

Arturo Magidin
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El Moro
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    Look up invariant basis number. – Arturo Magidin May 15 '11 at 01:43
  • 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 01:54
  • 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 02:03
  • No problem. Then the first question is definitely misstated. Over every commutative (associative, unital) ring, every basis of any particular free module has the same cardinality. Local is irrelevant. All commutative rings have IBN. – Jack Schmidt May 15 '11 at 02:11
  • @Jack Hi Jack well actually i just read the excerpt from my textbook again and you are right. It is about minimal generating sets. so what does "local" add here? What is the rational behind a module being local in general? Thanks in advance – El Moro May 15 '11 at 02:21

1 Answers1

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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.

Georges Elencwajg
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  • 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 04 '12 at 00:03
  • @KCd: very surprising ! Have you known this for a long time? – Georges Elencwajg Jun 04 '12 at 00:38
  • @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 04 '12 at 01:11