[proofplan]
We first verify that the operation stays inside $V(R)$ by showing that finite direct sums of finitely generated projective modules are again finitely generated and projective. We then check that the formula $[P]+[Q]=[P\oplus Q]$ is independent of the chosen representatives of the isomorphism classes. Finally, the standard associativity, symmetry, and zero-module isomorphisms for direct sums descend to associativity, commutativity, and the identity law on isomorphism classes.
[/proofplan]
[step:Show finite direct sums remain finitely generated projective modules]
Let $P$ and $Q$ be finitely generated projective left $R$-modules. Choose finite generating sets $\{p_1,\dots,p_m\}\subset P$ and $\{q_1,\dots,q_n\}\subset Q$. Then the finite set
\begin{align*}
\{(p_1,0),\dots,(p_m,0),(0,q_1),\dots,(0,q_n)\}\subset P\oplus Q
\end{align*}
generates $P\oplus Q$ as a left $R$-module, so $P\oplus Q$ is finitely generated.
To prove projectivity, let $A$, $B$, and $C$ be left $R$-modules, let $\pi:A\to B$ be a surjective $R$-[linear map](/page/Linear%20Map), and let $f:P\oplus Q\to B$ be an $R$-linear map. Define $R$-linear maps $f_P:P\to B$ and $f_Q:Q\to B$ by
\begin{align*}
f_P(p):=f(p,0)
\end{align*}
and
\begin{align*}
f_Q(q):=f(0,q).
\end{align*}
Since $P$ and $Q$ are projective, there exist $R$-linear maps $\widetilde f_P:P\to A$ and $\widetilde f_Q:Q\to A$ such that
\begin{align*}
\pi\circ \widetilde f_P=f_P
\end{align*}
and
\begin{align*}
\pi\circ \widetilde f_Q=f_Q.
\end{align*}
Define an $R$-linear map $\widetilde f:P\oplus Q\to A$ by
\begin{align*}
\widetilde f(p,q):=\widetilde f_P(p)+\widetilde f_Q(q).
\end{align*}
For every $(p,q)\in P\oplus Q$,
\begin{align*}
(\pi\circ \widetilde f)(p,q)=\pi(\widetilde f_P(p)+\widetilde f_Q(q)).
\end{align*}
By $R$-linearity of $\pi$,
\begin{align*}
(\pi\circ \widetilde f)(p,q)=f_P(p)+f_Q(q).
\end{align*}
By the definitions of $f_P$ and $f_Q$,
\begin{align*}
(\pi\circ \widetilde f)(p,q)=f(p,0)+f(0,q).
\end{align*}
Since $f$ is $R$-linear,
\begin{align*}
(\pi\circ \widetilde f)(p,q)=f(p,q).
\end{align*}
Thus every map from $P\oplus Q$ lifts across every surjective $R$-linear map, so $P\oplus Q$ is projective. Hence $[P\oplus Q]\in V(R)$.
[guided]
We need the formula $[P]+[Q]=[P\oplus Q]$ to define an operation on $V(R)$, so the first point is closure: if $[P]$ and $[Q]$ are elements of $V(R)$, then $[P\oplus Q]$ must also be an element of $V(R)$. By definition of $V(R)$, this means that $P\oplus Q$ must again be finitely generated and projective.
First consider finite generation. Since $P$ is finitely generated, choose elements $p_1,\dots,p_m\in P$ that generate $P$ as a left $R$-module. Since $Q$ is finitely generated, choose elements $q_1,\dots,q_n\in Q$ that generate $Q$. Every element of $P\oplus Q$ has the form $(p,q)$ with $p\in P$ and $q\in Q$. Write
\begin{align*}
p=\sum_{i=1}^{m} r_i p_i
\end{align*}
with coefficients $r_1,\dots,r_m\in R$, and write
\begin{align*}
q=\sum_{j=1}^{n} s_j q_j
\end{align*}
with coefficients $s_1,\dots,s_n\in R$. Then
\begin{align*}
(p,q)=\sum_{i=1}^{m} r_i(p_i,0)+\sum_{j=1}^{n} s_j(0,q_j).
\end{align*}
Thus the finite set
\begin{align*}
\{(p_1,0),\dots,(p_m,0),(0,q_1),\dots,(0,q_n)\}
\end{align*}
generates $P\oplus Q$, so $P\oplus Q$ is finitely generated.
Now we prove projectivity by using the lifting property. Let $\pi:A\to B$ be a surjective $R$-linear map of left $R$-modules, and let $f:P\oplus Q\to B$ be an $R$-linear map. To lift $f$, we split it into its two components. Define $R$-linear maps $f_P:P\to B$ and $f_Q:Q\to B$ by
\begin{align*}
f_P(p):=f(p,0)
\end{align*}
and
\begin{align*}
f_Q(q):=f(0,q).
\end{align*}
Because $P$ is projective and $\pi$ is surjective, there is an $R$-linear map $\widetilde f_P:P\to A$ satisfying $\pi\circ \widetilde f_P=f_P$. Because $Q$ is projective and the same map $\pi$ is surjective, there is an $R$-linear map $\widetilde f_Q:Q\to A$ satisfying $\pi\circ \widetilde f_Q=f_Q$.
We now recombine these two lifts. Define $\widetilde f:P\oplus Q\to A$ by
\begin{align*}
\widetilde f(p,q):=\widetilde f_P(p)+\widetilde f_Q(q).
\end{align*}
This map is $R$-linear because $\widetilde f_P$ and $\widetilde f_Q$ are $R$-linear and addition in $A$ is $R$-bilinear. For every $(p,q)\in P\oplus Q$, we compute
\begin{align*}
(\pi\circ \widetilde f)(p,q)=\pi(\widetilde f_P(p)+\widetilde f_Q(q)).
\end{align*}
Using $R$-linearity of $\pi$, this becomes
\begin{align*}
(\pi\circ \widetilde f)(p,q)=\pi(\widetilde f_P(p))+\pi(\widetilde f_Q(q)).
\end{align*}
Using the defining equations for the two component lifts, we get
\begin{align*}
(\pi\circ \widetilde f)(p,q)=f_P(p)+f_Q(q).
\end{align*}
By the definitions of $f_P$ and $f_Q$,
\begin{align*}
(\pi\circ \widetilde f)(p,q)=f(p,0)+f(0,q).
\end{align*}
Since $f$ is $R$-linear and $(p,q)=(p,0)+(0,q)$ in $P\oplus Q$, this equals
\begin{align*}
f(p,q).
\end{align*}
Therefore $\pi\circ \widetilde f=f$. This proves the lifting property for $P\oplus Q$, so $P\oplus Q$ is projective. Combining finite generation and projectivity, $P\oplus Q$ is a finitely generated projective left $R$-module, and therefore $[P\oplus Q]$ belongs to $V(R)$.
[/guided]
[/step]
[step:Prove the operation is independent of representatives]
Let $P$, $P'$, $Q$, and $Q'$ be finitely generated projective left $R$-modules with $P\cong P'$ and $Q\cong Q'$. Choose $R$-module isomorphisms $\alpha:P\to P'$ and $\beta:Q\to Q'$. Define
\begin{align*}
\alpha\oplus\beta:P\oplus Q\to P'\oplus Q'
\end{align*}
by
\begin{align*}
(\alpha\oplus\beta)(p,q):=(\alpha(p),\beta(q)).
\end{align*}
The inverse map is $\alpha^{-1}\oplus\beta^{-1}$, so $\alpha\oplus\beta$ is an $R$-module isomorphism. Hence $P\oplus Q\cong P'\oplus Q'$, and therefore
\begin{align*}
[P\oplus Q]=[P'\oplus Q'].
\end{align*}
Thus $[P]+[Q]:=[P\oplus Q]$ is well-defined on isomorphism classes.
[/step]
[step:Descend associativity of direct sum to isomorphism classes]
Let $P$, $Q$, and $S$ be finitely generated projective left $R$-modules. Define the $R$-linear map
\begin{align*}
a_{P,Q,S}:(P\oplus Q)\oplus S\to P\oplus(Q\oplus S)
\end{align*}
by
\begin{align*}
a_{P,Q,S}((p,q),s):=(p,(q,s)).
\end{align*}
Its inverse sends $(p,(q,s))$ to $((p,q),s)$, so $a_{P,Q,S}$ is an $R$-module isomorphism. Therefore
\begin{align*}
[(P\oplus Q)\oplus S]=[P\oplus(Q\oplus S)].
\end{align*}
Using the definition of the operation on $V(R)$, this gives
\begin{align*}
([P]+[Q])+[S]=[P]+([Q]+[S]).
\end{align*}
Thus the operation is associative.
[/step]
[step:Descend the symmetry of direct sum to commutativity]
Let $P$ and $Q$ be finitely generated projective left $R$-modules. Define the $R$-linear map
\begin{align*}
\tau_{P,Q}:P\oplus Q\to Q\oplus P
\end{align*}
by
\begin{align*}
\tau_{P,Q}(p,q):=(q,p).
\end{align*}
The inverse map is $\tau_{Q,P}:Q\oplus P\to P\oplus Q$, so $\tau_{P,Q}$ is an $R$-module isomorphism. Hence
\begin{align*}
[P\oplus Q]=[Q\oplus P].
\end{align*}
By the definition of addition in $V(R)$,
\begin{align*}
[P]+[Q]=[Q]+[P].
\end{align*}
Thus the operation is commutative.
[/step]
[step:Verify the zero module gives the identity element]
The zero left $R$-module $0$ is finitely generated, since it is generated by the empty set, and it is projective, since the unique map $0\to B$ lifts uniquely through every surjective $R$-linear map $\pi:A\to B$. Thus $[0]\in V(R)$.
For every finitely generated projective left $R$-module $P$, define
\begin{align*}
\lambda_P:0\oplus P\to P
\end{align*}
by
\begin{align*}
\lambda_P(0,p):=p,
\end{align*}
and define
\begin{align*}
\rho_P:P\oplus 0\to P
\end{align*}
by
\begin{align*}
\rho_P(p,0):=p.
\end{align*}
Both maps are $R$-module isomorphisms, with inverse maps $p\mapsto(0,p)$ and $p\mapsto(p,0)$ respectively. Therefore
\begin{align*}
[0]+[P]=[P]
\end{align*}
and
\begin{align*}
[P]+[0]=[P].
\end{align*}
Combining closure, well-definedness, associativity, commutativity, and the identity law, $V(R)$ is a commutative monoid under $[P]+[Q]=[P\oplus Q]$ with identity element $[0]$.
[/step]
