[proofplan] We choose lifts in $P$ of a basis of the residue [vector space](/page/Vector%20Space) $P/\mathfrak m P$ and use them to define a map $R^r \to P$. A finite-generation form of [Nakayama's lemma](/theorems/2935), proved inside the argument, shows that this map is surjective. Projectivity splits the surjection, and reducing the resulting direct-sum decomposition modulo $\mathfrak m$ shows that the complementary summand has zero residue quotient. A second application of the same Nakayama argument forces the complement to vanish, so $P$ is isomorphic to $R^r$. [/proofplan]

[step:Prove the finite-generation form of Nakayama's lemma] We will use the following elementary form of Nakayama's lemma. [claim:Nakayama for finitely generated modules over a local ring] Let $M$ be a finitely generated $R$-module. If $M=\mathfrak m M$, then $M=0$. [/claim] [proof] Assume, toward a contradiction, that $M \neq 0$. Since $M$ is finitely generated, among all finite generating sets of $M$ choose one with minimal cardinality, say $x_1,\dots,x_n \in M$, where $n \geq 1$. The equality $M=\mathfrak m M$ gives coefficients $a_1,\dots,a_n \in \mathfrak m$ such that \begin{align*} x_n=\sum_{i=1}^n a_i x_i. \end{align*} Rearranging, \begin{align*} (1-a_n)x_n=\sum_{i=1}^{n-1} a_i x_i. \end{align*} Because $R$ is local with maximal ideal $\mathfrak m$ and $a_n \in \mathfrak m$, the element $1-a_n$ is a unit in $R$. Hence \begin{align*} x_n=(1-a_n)^{-1}\sum_{i=1}^{n-1} a_i x_i. \end{align*} Thus $x_1,\dots,x_{n-1}$ still generate $M$, contradicting the minimality of $n$. Therefore $M=0$. [/proof] [/step]

[step:Choose residue classes and construct the comparison map] Let $k:=R/\mathfrak m$ be the residue field. Since $P$ is finitely generated over $R$, the quotient $P/\mathfrak m P$ is a finite-dimensional $k$-vector space. Let $r:=\dim_k(P/\mathfrak m P)$, and choose elements $p_1,\dots,p_r \in P$ whose images \begin{align*} \overline{p}_1,\dots,\overline{p}_r \in P/\mathfrak m P \end{align*} form a $k$-basis. If $r=0$, this means $P/\mathfrak m P=0$. Define the $R$-[linear map](/page/Linear%20Map) \begin{align*} f:R^r &\to P \\ (a_1,\dots,a_r) &\mapsto \sum_{i=1}^r a_i p_i. \end{align*} The induced $k$-linear map \begin{align*} \overline f:k^r &\to P/\mathfrak m P \\ (\overline a_1,\dots,\overline a_r) &\mapsto \sum_{i=1}^r \overline a_i\,\overline p_i \end{align*} is an isomorphism because $\overline p_1,\dots,\overline p_r$ is a $k$-basis of $P/\mathfrak m P$. [/step]

[step:Use Nakayama to show the comparison map is surjective] Let $C:=\operatorname{coker}(f)$ be the cokernel of $f:R^r \to P$. Since $P$ is finitely generated, $C$ is a finitely generated $R$-module. Reducing the exact sequence \begin{align*} R^r \xrightarrow{f} P \to C \to 0 \end{align*} modulo $\mathfrak m$ gives the exact sequence \begin{align*} k^r \xrightarrow{\overline f} P/\mathfrak m P \to C/\mathfrak m C \to 0. \end{align*} Since $\overline f$ is an isomorphism, it is surjective, so $C/\mathfrak m C=0$. Equivalently, $C=\mathfrak m C$. By the Nakayama lemma proved above, $C=0$. Therefore $f$ is surjective. [/step]

[step:Split the surjection using projectivity] Because $P$ is projective and $f:R^r \to P$ is surjective, the identity map $\operatorname{id}_P:P\to P$ lifts through $f$. Hence there exists an $R$-linear map \begin{align*} s:P &\to R^r \end{align*} such that $f\circ s=\operatorname{id}_P$. Let $K:=\ker(f)$. The splitting $s$ gives an $R$-module isomorphism \begin{align*} K\oplus P &\to R^r \\ (x,p) &\mapsto x+s(p). \end{align*} Indeed, every $y\in R^r$ decomposes as \begin{align*} y=(y-s(f(y)))+s(f(y)), \end{align*} where $y-s(f(y))\in K$, and the intersection $K\cap s(P)$ is zero because if $s(p)\in K$, then \begin{align*} p=f(s(p))=0. \end{align*} Thus $R^r\cong K\oplus P$. [/step]

[step:Reduce the splitting modulo $\mathfrak m$ and kill the kernel] Since $K$ is a direct summand of the finitely generated module $R^r$, the module $K$ is finitely generated. Reducing the direct-sum decomposition $R^r\cong K\oplus P$ modulo $\mathfrak m$ gives a $k$-vector space decomposition \begin{align*} k^r \cong K/\mathfrak m K \oplus P/\mathfrak m P. \end{align*} Under this reduction, the induced map $k^r\to P/\mathfrak m P$ is precisely $\overline f$, which is an isomorphism. Therefore the summand $K/\mathfrak m K$ must be zero. Equivalently, $K=\mathfrak m K$. Applying the Nakayama lemma proved above to the finitely generated module $K$, we obtain $K=0$. Hence $\ker(f)=0$. [/step]

[step:Conclude that the projective module is free] The map \begin{align*} f:R^r &\to P \end{align*} is both surjective and injective, so it is an $R$-module isomorphism. Therefore \begin{align*} P\cong R^r. \end{align*} Thus every finitely generated projective module over the commutative local ring $(R,\mathfrak m)$ is free. [/step]