[proofplan] We use the standard Morita equivalence between left $R$-modules and left $M_n(R)$-modules. The equivalence is given by tensoring with the natural column module $R^n$, and its inverse is obtained by applying the corner idempotent $E_{11}$. We verify explicitly that these functors are inverse, that they preserve finitely generated projective modules, and therefore that they induce an isomorphism on the commutative monoids of finite projective modules. Finally, the universal property of group completion gives the asserted isomorphism on $K_0$. [/proofplan]

[step:Construct the standard Morita functors] Let $A=M_n(R)$, and for $1\le i,j\le n$ let $E_{ij}\in A$ denote the matrix with $1_R$ in the $(i,j)$ entry and $0_R$ elsewhere. Let $e=E_{11}$. Regard $R^n$ as the column module of length $n$. It is an $(A,R)$-bimodule: $A$ acts on the left by matrix multiplication and $R$ acts on the right componentwise. Define a functor \begin{align*} F:R\operatorname{-Mod} &\to A\operatorname{-Mod} \end{align*} \begin{align*} P &\mapsto R^n\otimes_R P \end{align*} where $A$ acts on the first tensor factor. For a left $A$-module $Q$, define \begin{align*} G(Q)=eQ=\{eq:q\in Q\}. \end{align*} This is a left $R$-module by \begin{align*} r\cdot x=(rE_{11})x \end{align*} for $r\in R$ and $x\in eQ$. For an $A$-[linear map](/page/Linear%20Map) $h:Q\to Q'$, define $G(h):eQ\to eQ'$ by restriction, $G(h)(x)=h(x)$. This gives a functor \begin{align*} G:A\operatorname{-Mod}\to R\operatorname{-Mod}. \end{align*} [/step]

[step:Verify that the two functors are naturally inverse]For a left $R$-module $P$, define \begin{align*} \alpha_P:e(R^n\otimes_R P)&\to P \end{align*} \begin{align*} e(v\otimes p)&\mapsto v_1p, \end{align*} where $v=(v_1,\dots,v_n)^\top\in R^n$ and $p\in P$. Since $eR^n$ consists exactly of columns whose only possibly nonzero entry is the first one, this map is well-defined. Its inverse is the map \begin{align*} \beta_P:P&\to e(R^n\otimes_R P) \end{align*} \begin{align*} p&\mapsto u_1\otimes p, \end{align*} where $u_1=(1_R,0_R,\dots,0_R)^\top\in R^n$. The identities $\alpha_P\beta_P=\operatorname{id}_P$ and $\beta_P\alpha_P=\operatorname{id}_{e(R^n\otimes_R P)}$ follow immediately from the description of $eR^n$. Thus $GF\cong\operatorname{id}_{R\operatorname{-Mod}}$ naturally in $P$. Now let $Q$ be a left $A$-module. For $v=(v_1,\dots,v_n)^\top\in R^n$, define $C(v)\in A$ to be the matrix whose first column is $v$ and whose other columns are zero. Define \begin{align*} \theta_Q:R^n\otimes_R eQ&\to Q \end{align*} \begin{align*} v\otimes x&\mapsto C(v)x. \end{align*} This is well-defined over $R$ because $C(vr)=C(v)(rE_{11})$ for every $r\in R$. It is $A$-linear because $C(av)=aC(v)$ for every $a\in A$ and $v\in R^n$. Define \begin{align*} \eta_Q:Q&\to R^n\otimes_R eQ \end{align*} \begin{align*} q&\mapsto \sum_{i=1}^n u_i\otimes E_{1i}q, \end{align*} where $u_i\in R^n$ is the $i$-th standard column vector. Since $eE_{1i}=E_{1i}$, each $E_{1i}q$ lies in $eQ$, so $\eta_Q$ is well-defined. We compute \begin{align*} \theta_Q(\eta_Q(q))=\sum_{i=1}^n C(u_i)E_{1i}q. \end{align*} Since $C(u_i)=E_{i1}$, we have $C(u_i)E_{1i}=E_{ii}$, and therefore \begin{align*} \theta_Q(\eta_Q(q))=\sum_{i=1}^n E_{ii}q=q, \end{align*} because $\sum_{i=1}^n E_{ii}=1_A$. Conversely, for $v=(v_1,\dots,v_n)^\top\in R^n$ and $x\in eQ$, \begin{align*} \eta_Q(\theta_Q(v\otimes x))=\sum_{i=1}^n u_i\otimes E_{1i}C(v)x. \end{align*} The matrix $E_{1i}C(v)$ equals $v_iE_{11}$, where $v_iE_{11}$ denotes the matrix with $v_i$ in the $(1,1)$ entry and zero elsewhere. Hence \begin{align*} \eta_Q(\theta_Q(v\otimes x))=\sum_{i=1}^n u_i\otimes (v_iE_{11})x. \end{align*} By the tensor relation over $R$ and the definition of the $R$-module structure on $eQ$, \begin{align*} u_i\otimes (v_iE_{11})x=(u_iv_i)\otimes x. \end{align*} Thus \begin{align*} \eta_Q(\theta_Q(v\otimes x))=\sum_{i=1}^n u_iv_i\otimes x=v\otimes x. \end{align*} Therefore $FG\cong\operatorname{id}_{A\operatorname{-Mod}}$ naturally in $Q$.[/step]

[guided]The point of the construction is that $e=E_{11}$ extracts the first coordinate of an $A$-module, and the column module $R^n$ rebuilds the remaining coordinates by multiplying with the matrix units. First take a left $R$-module $P$. The object $F(P)=R^n\otimes_R P$ is a left $A$-module because $A=M_n(R)$ acts on the column factor. Applying $G$ means taking the first corner: \begin{align*} GF(P)=e(R^n\otimes_R P). \end{align*} Since $eR^n$ consists of columns of the form $(r,0,\dots,0)^\top$, the map \begin{align*} \alpha_P:e(R^n\otimes_R P)&\to P \end{align*} \begin{align*} e(v\otimes p)&\mapsto v_1p \end{align*} recovers exactly the $R$-module element encoded in the first coordinate. Its inverse is \begin{align*} \beta_P:P&\to e(R^n\otimes_R P) \end{align*} \begin{align*} p&\mapsto u_1\otimes p. \end{align*} The composite $\alpha_P\beta_P$ sends $p$ to $p$, and the composite $\beta_P\alpha_P$ fixes every tensor in $e(R^n\otimes_R P)$ because such a tensor only has a first-coordinate contribution. Hence $GF\cong\operatorname{id}_{R\operatorname{-Mod}}$. Now take a left $A$-module $Q$. We must show that $R^n\otimes_R eQ$ reconstructs all of $Q$. For a column vector $v=(v_1,\dots,v_n)^\top\in R^n$, let $C(v)\in A$ be the matrix with first column $v$ and all other columns zero. Define \begin{align*} \theta_Q:R^n\otimes_R eQ&\to Q \end{align*} \begin{align*} v\otimes x&\mapsto C(v)x. \end{align*} This is balanced over $R$: if $r\in R$, then $C(vr)=C(v)(rE_{11})$, so \begin{align*} \theta_Q(vr\otimes x)=C(vr)x=C(v)(rE_{11})x=\theta_Q(v\otimes r\cdot x). \end{align*} It is also $A$-linear, since for $a\in A$ one has $C(av)=aC(v)$. The inverse map distributes an element of $Q$ into its matrix-unit components: \begin{align*} \eta_Q:Q&\to R^n\otimes_R eQ \end{align*} \begin{align*} q&\mapsto \sum_{i=1}^n u_i\otimes E_{1i}q. \end{align*} Each $E_{1i}q$ lies in $eQ$ because $eE_{1i}=E_{1i}$. Applying $\theta_Q$ gives \begin{align*} \theta_Q(\eta_Q(q))=\sum_{i=1}^n C(u_i)E_{1i}q. \end{align*} Since $C(u_i)=E_{i1}$, this becomes \begin{align*} \theta_Q(\eta_Q(q))=\sum_{i=1}^n E_{i1}E_{1i}q=\sum_{i=1}^n E_{ii}q=q. \end{align*} The last equality uses $\sum_{i=1}^n E_{ii}=1_A$. For the other composite, take $v=(v_1,\dots,v_n)^\top\in R^n$ and $x\in eQ$. Then \begin{align*} \eta_Q(\theta_Q(v\otimes x))=\sum_{i=1}^n u_i\otimes E_{1i}C(v)x. \end{align*} The product $E_{1i}C(v)$ has only one possibly nonzero entry, namely $v_i$ in the $(1,1)$ position, so $E_{1i}C(v)=v_iE_{11}$. Therefore \begin{align*} \eta_Q(\theta_Q(v\otimes x))=\sum_{i=1}^n u_i\otimes (v_iE_{11})x. \end{align*} By the tensor relation defining $R^n\otimes_R eQ$, this equals \begin{align*} \sum_{i=1}^n u_iv_i\otimes x=v\otimes x. \end{align*} Thus $\theta_Q$ and $\eta_Q$ are inverse $A$-linear maps, naturally in $Q$. Hence $F$ and $G$ are mutually inverse equivalences.[/guided]

[step:Restrict the equivalence to finitely generated projective modules] The functor $F$ preserves finite direct sums because [tensor product](/page/Tensor%20Product) over $R$ commutes with finite direct sums. It sends the finite free left $R$-module $R^m$ to \begin{align*} F(R^m)=R^n\otimes_R R^m\cong (R^n)^m, \end{align*} which is a finitely generated left $A$-module. Since $R^n\cong Ae$ as a left $A$-module and $Ae$ is a direct summand of the free module $A$, the module $R^n$ is finitely generated projective over $A$. Hence $F(R^m)$ is finitely generated projective over $A$. If $P$ is finitely generated projective over $R$, then there exists an integer $m\ge 1$ and a left $R$-module $P'$ such that \begin{align*} P\oplus P'\cong R^m. \end{align*} Applying $F$ gives \begin{align*} F(P)\oplus F(P')\cong F(R^m), \end{align*} so $F(P)$ is a direct summand of a finitely generated projective, hence is finitely generated projective. Similarly, since $G$ is an inverse equivalence to $F$, the same direct-summand argument applied through the natural isomorphisms $FG\cong\operatorname{id}_{A\operatorname{-Mod}}$ and $GF\cong\operatorname{id}_{R\operatorname{-Mod}}$ shows that $G$ sends finitely generated projective left $A$-modules to finitely generated projective left $R$-modules. [/step]

[step:Pass from projective modules to the monoid $V$] Let $V(R)$ denote the commutative monoid of isomorphism classes of finitely generated projective left $R$-modules under direct sum. Let $V(A)$ be defined analogously for $A=M_n(R)$. Because $F$ and $G$ preserve finitely generated projective modules and finite direct sums, they induce monoid homomorphisms \begin{align*} V(F):V(R)&\to V(A), & [P]&\mapsto [F(P)], \end{align*} and \begin{align*} V(G):V(A)&\to V(R), & [Q]&\mapsto [G(Q)]. \end{align*} The natural isomorphisms $GF\cong\operatorname{id}_{R\operatorname{-Mod}}$ and $FG\cong\operatorname{id}_{A\operatorname{-Mod}}$ imply \begin{align*} V(G)\circ V(F)&=\operatorname{id}_{V(R)}, & V(F)\circ V(G)&=\operatorname{id}_{V(A)}. \end{align*} Therefore $V(F)$ and $V(G)$ are inverse monoid isomorphisms. [/step]

[step:Apply group completion to obtain the $K_0$ isomorphism] By the definition of $K_0$ for a unital ring, $K_0(R)$ is the group completion of $V(R)$, and $K_0(A)$ is the group completion of $V(A)$. Since $V(G):V(A)\to V(R)$ is a monoid isomorphism, the universal property of group completion, as in [citetheorem:8633], induces a group isomorphism \begin{align*} K_0(A)\to K_0(R). \end{align*} Substituting $A=M_n(R)$ gives \begin{align*} K_0(M_n(R))\cong K_0(R). \end{align*} This is the isomorphism induced by the standard Morita equivalence, and its construction is natural for the standard matrix-ring Morita context. [/step]