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Hi there,

Let R be a local commutative ring. If M and N are two R-modules with the condition that their direct sum is equal to R^n. How do I use Nakayama to show that M and N are in fact free R-modules?

I don't know if it was useful but I first showed that M and N are finite R-modules by making two surjective homomorphisms from R^n to M and R^n to N so that R^n/N is isomorphic to M and R^n/M is isomorphic to N. We know that R^n/N and R^n/M are finite R-modules, hence M and N are finite R-modules.

Please correct me if I'm doing stupid things here... and feel absolutely free to give me hints on how to use Nakayama's lemma to prove that M and N are free...



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closed as too localized by Daniel Litt, S. Carnahan Oct 4 '11 at 3:44

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This sounds like a homework exercise, in which case your question should probably be asked on But I could be wrong. – M Turgeon Oct 3 '11 at 23:28
Moreover, there seems to be a problem. If your ring is infinite, and you have proven that $M,N$ are finite, then the only way they could be free as well is if they are trivial. But you would have a counterexample to your "lemma" by simply taking $R\oplus R$. – M Turgeon Oct 3 '11 at 23:36
Lemma 2.2 in Weibel's K-Book chapter I ( ). Given that part of the proof relegated to the exercise and the rest isn't too much detailed either, you can safely consider it a hint. – darij grinberg Oct 3 '11 at 23:37
@M Turgeon: He means "finitely generated" when saying "finite". – darij grinberg Oct 3 '11 at 23:37
This is a standard homework exercise (essentially equivalent to one of my favorites); voting to close. – Daniel Litt Oct 3 '11 at 23:39

1 Answer 1

$M$ and $N$ are projective module over a local ring. Then they are free. (See, Matsumura: Commutative ring theory, Theorem 2.5). Is it your exercise?

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