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

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

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

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

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

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

now, since $\displaystyle F$ is a free $\displaystyle R$-module of rank $\displaystyle r$, we have $\displaystyle \dim_k F/\mathfrak{m}F = \text{rank}(F)=r$. we also have $\displaystyle \dim_k M/\mathfrak{m}M = r$ by $\displaystyle (1)$. thus $\displaystyle \dim_k K/\mathfrak{m}K = 0,$ by $\displaystyle (3)$. so $\displaystyle K= \mathfrak{m}K$ and hence $\displaystyle K = 0$, by Nakayama's lemma. it now follows from $\displaystyle (2)$ that $\displaystyle F=M$ and so $\displaystyle M$ is free.