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Feb 18, 2012 at 0:53 comment added Todd Trimble And thank you Ralph! This is really pretty stuff. I found an online reference: math.columbia.edu/algebraic_geometry/stacks-git/… which was closer to what I was hoping would be possible (using determinant tricks); see 13.4.
Feb 17, 2012 at 23:25 comment added Ralph If $A$ is comm. this is Theorem 2.4 in Matsumura's "Commutative ring theory".
Feb 17, 2012 at 21:53 comment added Todd Trimble @Harry: I should have been more clear that my first question was rhetorical. Thanks for the reference!
Feb 17, 2012 at 21:51 history made wiki Post Made Community Wiki by François G. Dorais
Feb 17, 2012 at 21:16 comment added Harry Altman It's obviously false if IBN fails, but it is true over commutative rings. It's a consequence of (an appropriate form of) Nakayama's lemma. I think you can find it in Atiyah & MacDonald. I have no idea about what happens if commutativity fails but IBN still holds.
Feb 17, 2012 at 21:11 comment added Uday @Todd Trimble I doubt this may be true in general. In vector space setting, we will need to have a condition of dimension of the range and kernel to be equal. In generalizing to module, this condition may demand IBN.
Feb 17, 2012 at 20:53 comment added Todd Trimble Is this true if the ring doesn't satisfy the invariant basis number property: en.wikipedia.org/wiki/Invariant_basis_number#Examples Or did you mean to consider only (say) commutative rings? (I am interested in your example; do you have a reference?)
Feb 17, 2012 at 20:44 comment added Uday Thanks. This statement is a direct generalization of finite dimensional vector spaces. May be we can extend the results to infinite dimensional modules and compact operators defined on them.
Feb 17, 2012 at 20:32 history answered Matthieu Romagny CC BY-SA 3.0