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:
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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?
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"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
Best Answer
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.