[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 $[P\oplus Q]\in V(R)$.[/step]

[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: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]