Finitely generated projective modules over local rings are free

Remarks

Choose $x_1, \ldots, x_n \in P$ whose classes form a basis of the $k$-vector space $P/\mathfrak{m}P$. By Nakayama's lemma, they generate $P$, so there is a surjection $\pi \colon R^n \to P$; it splits because $P$ is projective, so $R^n \cong P \oplus K$ with $K = \ker \pi$. Reducing modulo $\mathfrak{m}$ gives $k^n \cong k^n \oplus K/\mathfrak{m}K$, so $K/\mathfrak{m}K = 0$. As a direct summand of $R^n$, $K$ is finitely generated, and Nakayama's lemma gives $K = 0$; thus $P \cong R^n$.

The local hypothesis is necessary. Over $R = \Z/6\Z$, the module $\Z/2\Z$ is a direct summand of $R$, hence projective, but it has $2$ elements, while every finitely generated free $R$-module has $6^n$ elements for some $n \ge 0$. Kaplansky's theorem removes the finiteness hypothesis: every projective module over a local ring is free. More generally, a finitely presented flat module over a local ring is free.

Going deeper

For a student

This is why finitely generated projective modules over any commutative ring are called locally free: their localizations at prime ideals are free. The rank at each prime gives a locally constant function on the spectrum, and rank one projectives form the Picard group. Geometrically, finitely generated projective modules are vector bundles on $\operatorname{Spec} R$, and the theorem says that a vector bundle is trivial near each point. Whether such modules are globally free over polynomial rings is the content of the Quillen–Suslin theorem.