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Math Help - free modules

  1. #1
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    free modules

    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...

    Thnx

    Jools
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  2. #2
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    Re: free modules

    Quote Originally Posted by Joolz View Post
    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...

    Thnx
    Jools
    i'll assume that you're familiar with projective modules. the problem is basically saying that "every finitely generated projective module over a local ring is free". this is a spacial case of this result, due to Kaplansky, that "every projective module over a local ring is free"! anyway, let's solve your problem:

    since R^n is a finitely generated R-module, both M and N are finitely generated R-modules. let \mathfrak{m} be the unique maximal ideal of R and put k = R/\mathfrak{m}. then M/\mathfrak{m}M is a finite dimensional k-vector space. let

    \dim_k M/\mathfrak{m}M = r.  \ \ \ \ \ \ \ \ \ (1)

    you should know that, as a result of Nakayama's lemma, M is generated, as an R-module, by r elements. (in fact, r is the minimum number of generators of M as an R-module.) anyway, so there exists an onto R-homomorphism f : R^r \longrightarrow M. let \ker f = K. then we have a short exact sequence 0 \to K \to R^r \to M \to 0 and thus, since M is projective, this sequence splits, i.e. R^r \cong M \oplus K. so M \oplus K is a free R-module of rank r. let

    F = M \oplus K. \ \ \ \ \ \ \ (2)

    then \mathfrak{m}F = \mathfrak{m}M \oplus \mathfrak{m}K and thus

    F/\mathfrak{m}F = (M \oplus K)/(\mathfrak{m}M \oplus \mathfrak{m}K}) \cong M/\mathfrak{m}M \oplus K/\mathfrak{m}K. \ \ \ \ \ \ \ (3)

    now, since F is a free R-module of rank r, we have \dim_k F/\mathfrak{m}F = \text{rank}(F)=r. we also have \dim_k M/\mathfrak{m}M = r by (1). thus \dim_k K/\mathfrak{m}K = 0, by (3). so K= \mathfrak{m}K and hence K = 0, by Nakayama's lemma. it now follows from (2) that F=M and so M is free.
    Last edited by NonCommAlg; October 4th 2011 at 03:48 PM.
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  3. #3
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    Re: free modules

    Thank you very much!! I think I now understand how I can use Nakayama's lemma... :-) The strange thing is that my book doesn't explain anything about projective modules... So I was wondering if there is another way to go from the exact sequence to the fact: R^r is isomorphic to the direct sum of M and K? Because not only do they not mention projective modules in my book, splitting of an exact sequence stays undefined as well.. Of course I did find them on the internet but I'm very curious on how to solve this without...

    So, what I do know from the book is the following: If M is a finitely generated R-module, then M is isomorphic to a quotient of R^n for some integer n>0.

    If I use this in the problem above then the following holds:

    M is isomorphic to R^r/K

    My question now is: Is this the same as saying R^r is isomorphic to the direct sum of M and K??

    Again thank you a-lot for your given explaination!!
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