[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 $P\oplus Q\cong R^n$.
[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]
[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]
