# Thread: free modules

1. ## 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

2. ## Re: free modules

Originally Posted by Joolz
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.

3. ## 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!!