[proofplan] We prove the equivalence by moving between three standard models for finite projectives: lifting splittings, direct summands of finite free modules, and images of idempotent endomorphisms. Finite generation gives a surjection from a finite free module onto $P$, and projectivity splits that surjection. A direct-sum decomposition gives a projection endomorphism, which is represented by an idempotent matrix. Conversely, an idempotent matrix gives complementary projections $p$ and $I_n-p$, so its image is a direct summand of the finite free module $R^n$, hence finitely generated and projective. [/proofplan]

[step:Split a finite free presentation using projectivity]Assume that $P$ is finitely generated and projective as a right $R$-module. Choose generators $x_1,\dots,x_n \in P$ with $n \ge 1$; if $P=0$, take $n=1$ and $x_1=0$. Define the right $R$-[linear map](/page/Linear%20Map) \begin{align*} \pi:R^n \to P, \qquad (r_1,\dots,r_n)^\top \mapsto \sum_{i=1}^n x_i r_i. \end{align*} The chosen elements generate $P$, so $\pi$ is surjective. Since $P$ is projective, the identity map $\operatorname{id}_P:P\to P$ lifts through the surjection $\pi$. Thus there exists a right $R$-linear map $s:P\to R^n$ such that $\pi\circ s=\operatorname{id}_P$. Let $Q=\ker \pi$, regarded as a right $R$-submodule of $R^n$. Define \begin{align*} \Phi:P\oplus Q \to R^n, \qquad (x,q)\mapsto s(x)+q. \end{align*} This map is right $R$-linear. If $\Phi(x,q)=0$, then applying $\pi$ gives $x=0$, and then $q=0$, so $\Phi$ is injective. For any $v\in R^n$, put $x=\pi(v)$ and $q=v-s(\pi(v))$. Then $q\in Q$ because \begin{align*} \pi(q)=\pi(v)-\pi(s(\pi(v)))=\pi(v)-\pi(v)=0. \end{align*} Hence $v=s(x)+q=\Phi(x,q)$, so $\Phi$ is surjective. Therefore $P\oplus Q\cong R^n$.[/step]

[guided]We start from the hypotheses that $P$ is finitely generated and projective, and we must manufacture a finite free module that contains $P$ as a direct summand. Since $P$ is finitely generated, choose elements $x_1,\dots,x_n\in P$ that generate $P$, with $n\ge 1$. In the special case $P=0$, we may take $n=1$ and $x_1=0$, so the convention that natural numbers start at $1$ causes no difficulty. Define a map from the finite free right module $R^n$ onto $P$ by \begin{align*} \pi:R^n \to P, \qquad (r_1,\dots,r_n)^\top \mapsto \sum_{i=1}^n x_i r_i. \end{align*} This is right $R$-linear: for $a,b\in R$, vectors $v=(r_1,\dots,r_n)^\top$ and $w=(s_1,\dots,s_n)^\top$ in $R^n$, one has \begin{align*} \pi(va+w b)=\sum_{i=1}^n x_i(r_i a+s_i b)=\left(\sum_{i=1}^n x_i r_i\right)a+\left(\sum_{i=1}^n x_i s_i\right)b=\pi(v)a+\pi(w)b. \end{align*} It is surjective precisely because the elements $x_1,\dots,x_n$ generate $P$. Now we use projectivity in its lifting form. The map $\pi:R^n\to P$ is surjective, and the identity map $\operatorname{id}_P:P\to P$ is a right $R$-linear map. Since $P$ is projective, there exists a right $R$-linear map $s:P\to R^n$ such that \begin{align*} \pi\circ s=\operatorname{id}_P. \end{align*} This equation says that $s$ is a section of $\pi$. Let $Q=\ker \pi$. We claim that $R^n$ is the internal direct sum of $s(P)$ and $Q$. Define \begin{align*} \Phi:P\oplus Q \to R^n, \qquad (x,q)\mapsto s(x)+q. \end{align*} The map $\Phi$ is right $R$-linear because both $s$ and the inclusion $Q\subset R^n$ are right $R$-linear. If $\Phi(x,q)=0$, then $s(x)+q=0$. Applying $\pi$ gives \begin{align*} 0=\pi(s(x)+q)=\pi(s(x))+\pi(q)=x+0=x. \end{align*} Thus $x=0$, and then $q=0$, so $\Phi$ is injective. To prove surjectivity, take an arbitrary vector $v\in R^n$. Define $x=\pi(v)\in P$ and define $q=v-s(\pi(v))\in R^n$. Then \begin{align*} \pi(q)=\pi(v)-\pi(s(\pi(v)))=\pi(v)-\pi(v)=0, \end{align*} so $q\in Q$. With this choice, \begin{align*} v=s(\pi(v))+q=\Phi(\pi(v),q). \end{align*} Hence $\Phi$ is surjective. Therefore $\Phi$ is an isomorphism of right $R$-modules, and $P\oplus Q\cong R^n$.[/guided]

[step:Convert a direct summand decomposition into an idempotent matrix] Assume that there exist an integer $n\ge 1$, a right $R$-module $Q$, and a right $R$-module isomorphism $\alpha:P\oplus Q\to R^n$. Let \begin{align*} \rho:P\oplus Q\to P\oplus Q, \qquad (x,q)\mapsto (x,0) \end{align*} be the projection onto the first summand. Define \begin{align*} e:R^n\to R^n, \qquad v\mapsto \alpha(\rho(\alpha^{-1}(v))). \end{align*} Then $e$ is a right $R$-linear endomorphism of $R^n$ and \begin{align*} e^2=\alpha\rho\alpha^{-1}\alpha\rho\alpha^{-1}=\alpha\rho^2\alpha^{-1}=\alpha\rho\alpha^{-1}=e. \end{align*} Thus $e$ is an idempotent projection. Its image is \begin{align*} \operatorname{im}(e)=\alpha(P\oplus 0), \end{align*} so $P\cong \operatorname{im}(e)$ by the restriction of $\alpha$ to $P\oplus 0$. Let $u_1,\dots,u_n\in R^n$ be the standard column vectors. For each $j$, write \begin{align*} e(u_j)=(p_{1j},\dots,p_{nj})^\top \end{align*} with entries $p_{ij}\in R$. Let $p=(p_{ij})\in M_n(R)$. For a vector $v=(r_1,\dots,r_n)^\top\in R^n$, right $R$-linearity gives \begin{align*} e(v)=\sum_{j=1}^n e(u_j)r_j=pv. \end{align*} Hence $e$ is left multiplication by the matrix $p$. Since $e^2=e$, left multiplication by $p^2-p$ is the zero endomorphism of $R^n$. Applying it to each standard vector $u_j$ shows every column of $p^2-p$ is zero, so $p^2=p$. Therefore $P\cong \operatorname{im}(p:R^n\to R^n)$ for an idempotent matrix $p\in M_n(R)$. [/step]

[step:Decompose the finite free module using an idempotent matrix] Assume that there exist an integer $n\ge 1$ and an idempotent matrix $p\in M_n(R)$ such that $P\cong \operatorname{im}(p:R^n\to R^n)$. Let $I_n\in M_n(R)$ denote the identity matrix, and define $q=I_n-p\in M_n(R)$. Since $p^2=p$, we have \begin{align*} q^2=(I_n-p)^2=I_n-2p+p^2=I_n-p=q. \end{align*} Also \begin{align*} pq=p(I_n-p)=p-p^2=0 \end{align*} and \begin{align*} qp=(I_n-p)p=p-p^2=0. \end{align*} We claim that \begin{align*} R^n=\operatorname{im}(p)\oplus \operatorname{im}(q). \end{align*} For every $v\in R^n$, \begin{align*} v=I_n v=(p+q)v=pv+qv, \end{align*} with $pv\in \operatorname{im}(p)$ and $qv\in \operatorname{im}(q)$. Thus the two images span $R^n$. If $w\in \operatorname{im}(p)\cap \operatorname{im}(q)$, then $w=pa=qb$ for some $a,b\in R^n$. Hence \begin{align*} w=pw=p(qb)=(pq)b=0. \end{align*} The intersection is zero, so the sum is direct. Therefore $\operatorname{im}(p)$ is a direct summand of $R^n$. [/step]

[step:Show that a direct summand of a finite free module is finitely generated and projective] Let $A$ be a right $R$-module for which there exist an integer $n\ge 1$ and a right $R$-module $B$ such that $A\oplus B\cong R^n$. We prove that $A$ is finitely generated and projective. Let $\beta:A\oplus B\to R^n$ be a right $R$-module isomorphism. Let \begin{align*} \iota:A\to A\oplus B, \qquad a\mapsto (a,0) \end{align*} be the inclusion, and let \begin{align*} \lambda:A\oplus B\to A, \qquad (a,b)\mapsto a \end{align*} be the projection. Define \begin{align*} j:A\to R^n, \qquad a\mapsto \beta(\iota(a)) \end{align*} and \begin{align*} r:R^n\to A, \qquad v\mapsto \lambda(\beta^{-1}(v)). \end{align*} Then $r\circ j=\operatorname{id}_A$. To prove finite generation, let $u_1,\dots,u_n\in R^n$ be the standard column vectors. For every $a\in A$, write \begin{align*} j(a)=\sum_{i=1}^n u_i r_i \end{align*} with uniquely determined entries $r_i\in R$. Applying $r$ gives \begin{align*} a=(r\circ j)(a)=\sum_{i=1}^n r(u_i)r_i. \end{align*} Thus the finite set $\{r(u_1),\dots,r(u_n)\}$ generates $A$. To prove projectivity, let $\gamma:M\to N$ be a surjective homomorphism of right $R$-modules, and let $f:A\to N$ be a right $R$-linear map. Since $R^n$ is free on $u_1,\dots,u_n$, choose elements $m_i\in M$ with \begin{align*} \gamma(m_i)=f(r(u_i)) \end{align*} for each $i\in\{1,\dots,n\}$. Define the right $R$-linear map \begin{align*} h:R^n\to M, \qquad \sum_{i=1}^n u_i r_i \mapsto \sum_{i=1}^n m_i r_i. \end{align*} Then $\gamma\circ h=f\circ r$ on the basis vectors $u_i$, hence on all of $R^n$ by right $R$-linearity. Define \begin{align*} \ell:A\to M, \qquad a\mapsto h(j(a)). \end{align*} Then \begin{align*} \gamma(\ell(a))=(\gamma\circ h)(j(a))=(f\circ r)(j(a))=f(a) \end{align*} for every $a\in A$. Thus every map from $A$ to a quotient lifts, so $A$ is projective. Applying this result to $A=\operatorname{im}(p)$ and $B=\operatorname{im}(I_n-p)$ from the previous step shows that $\operatorname{im}(p)$ is finitely generated and projective. Since $P\cong \operatorname{im}(p)$, the module $P$ is finitely generated and projective. This proves condition 3 implies condition 1, and the three conditions are equivalent. [/step]