Let $R$ be a local ring (commutative with identity) and suppose we have $M$ and $N$ submodules of $R^n$ such that $M\oplus N\cong R^n$. Then $M\cong R^s$ and $N\cong R^t$ for some $s$ and $t$ such that $s+t=n$.

I am completely lost here.

  • $\begingroup$ Note that finitely generated projective modules over a local ring are free and that you get the additivity of dimensions by vector space theory (you obtain vector spaces over $R/m$ by tensoring). $\endgroup$ – Daniel Valenzuela Sep 4 '15 at 10:28

Direct summands of free $R$-modules are called projective $R$-modules, so I'd be using that terminology here.

Let $(A, \mathfrak{m}, k)$ be a local ring with maximal ideal $\mathfrak{m}$ and residue field $k = A/\mathfrak{m}$. Let $M$ be a finitely generated projective $A$-module.

Pick a minimal generating set of cardinality $n$ for $M$ (you can do this by picking a minimal set of generators for the $k$-vector space $M/\mathfrak{m}M$ and lifting it to $M$). There is a short exact sequence $$0 \to N \to A^n \stackrel{f}{\to} M \to 0$$ where the map $f$ sends the standard generators of $A^n$ to the generators of $M$. $N$ is the kernel of $f$.

As $M$ is projective, the sequence splits so that we have $M \oplus N \cong A^n$. Tensoring with $k$, we have $M/\mathfrak{m}M \oplus N/\mathfrak{m}N \cong k^n$. $M/\mathfrak{m}M$ is a vector space of dimension $n$ over $k$, so using the fact that two $F$-vector spaces are isomorphic iff they have the same dimension, we get $N/\mathfrak{m}N = 0$

By Nakayama's lemma, $N = 0$, as $\mathfrak{m}$ is obviously in the Jacoboson radical of $R$ (which is itself $\mathfrak{m}$). Hence, $f$ must be an isomorphism. $\blacksquare$

For infinitely generated projective modules, this is a hard theorem of Kaplansky. See here for the original article.

As a side-note, a corollary of this nice fact is that if $M$ is a projective $A$-module, then given a prime $\wp \subset A$, the localization $M_\wp$ is a free $A_\wp$-module. Thus, projective modules are said to be "locally free". If you're familiar with vector bundles, then you can see that this looks suspiciously similar to trivializations of vector bundles. Indeed, projective modules are similar to vector bundles in many ways. The celebrated Serre-Swan theorem states that if $X$ is a compact Hausdorff topological space, and $C(X)$ is the ring of real-valued functions on $X$, then the category of vector bundles on $X$ is equivalent to the category of projective $C(X)$-modules. Inspired by this, Serre later conjectured that projective modules over polynomial rings are free, as an analogue of the fact that vector bundles over $\Bbb C^n$ are all trivial (which is easy to see, as $\Bbb C^n$ is contractible). However, this turned out to be a very hard problem, which was finally settled independently by homotopy theorist (!) Quillen and Suslin.


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