[proofplan]
We use the standard Morita equivalence implemented by the column module $R^n$ and its dual row module. This equivalence is exact and restricts to finitely generated projective modules, so it identifies the monoids of projective-module classes and hence their Grothendieck groups, giving the $K_0$ isomorphism. For $K_1$, we identify matrices over $M_n(R)$ with larger matrices over $R$, check compatibility with stabilization and elementary subgroups, and then pass to the quotient definition of $K_1$. Naturality follows because all constructions are entrywise and commute with a unital ring homomorphism $R\to S$.
[/proofplan]
[step:Construct the standard Morita equivalence between $M_n(R)$ and $R$]
Let $A:=M_n(R)$. Let $e\in A$ denote the matrix idempotent whose $(1,1)$-entry is $1_R$ and whose other entries are $0_R$. Define
\begin{align*}
F: A\operatorname{-Mod} \to R\operatorname{-Mod}
\end{align*}
by $F(M):=eM$ for a left $A$-module $M$. This is a left $R$-module by identifying $eAe$ with $R$ through the ring isomorphism sending $eae$ to the $(1,1)$-entry of $a$.
Define the left $A$-module
\begin{align*}
P:=Ae.
\end{align*}
It is also a right $R$-module through the identification $eAe\cong R$. Define a functor
\begin{align*}
G:R\operatorname{-Mod}&\to A\operatorname{-Mod}
\end{align*}
by
\begin{align*}
G(N):=P\otimes_R N.
\end{align*}
Here $R\operatorname{-Mod}$ denotes the category of left $R$-modules and $A\operatorname{-Mod}$ denotes the category of left $A$-modules.
The standard Morita theorem for full matrix rings gives natural isomorphisms of functors $F\circ G \cong \operatorname{id}_{R\operatorname{-Mod}}$ and $G\circ F \cong \operatorname{id}_{A\operatorname{-Mod}}$. Equivalently, $F$ and $G$ are quasi-inverse equivalences of categories. Since equivalences preserve and reflect exact sequences, finite generation, direct summands, and finite direct sums, they restrict to an exact equivalence
\begin{align*}
\operatorname{Proj}(A)\simeq \operatorname{Proj}(R),
\end{align*}
where $\operatorname{Proj}(A)$ and $\operatorname{Proj}(R)$ denote the exact categories of finitely generated projective left modules over $A$ and $R$, respectively.
[guided]
We first make the Morita equivalence explicit enough to track what it does to projective modules. Put $A:=M_n(R)$, and let $e\in A$ be the matrix with $1_R$ in the $(1,1)$-entry and $0_R$ in every other entry. This element satisfies $e^2=e$, so it is an idempotent. The corner ring $eAe$ is naturally isomorphic to $R$: an element of $eAe$ is a matrix whose only possibly nonzero entry is the $(1,1)$-entry, and multiplication of such matrices is exactly multiplication in $R$.
For a left $A$-module $M$, define
\begin{align*}
F(M):=eM.
\end{align*}
Because $eM$ is a module over the corner ring $eAe$, and because $eAe\cong R$, this is a left $R$-module. To build the inverse direction, define
\begin{align*}
P:=Ae.
\end{align*}
This is a left $A$-module by left multiplication, and it is a right $R$-module through the identification $eAe\cong R$. Hence, for every left $R$-module $N$, the [tensor product](/page/Tensor%20Product) $P\otimes_R N$ is defined and is a left $A$-module through the left action on $P$. This gives a functor
\begin{align*}
G:R\operatorname{-Mod}&\to A\operatorname{-Mod}
\end{align*}
by
\begin{align*}
G(N):=P\otimes_R N.
\end{align*}
The standard Morita theorem for full matrix rings states that these two functors are quasi-inverse equivalences. In this case, the reason is concrete: $e$ is a full idempotent in $M_n(R)$, meaning $AeA=A$, and the column module $Ae$ is the usual progenerator implementing the equivalence between $M_n(R)$-modules and $R$-modules. Thus there are natural isomorphisms
\begin{align*}
F\circ G \cong \operatorname{id}_{R\operatorname{-Mod}}
\end{align*}
and
\begin{align*}
G\circ F \cong \operatorname{id}_{A\operatorname{-Mod}}.
\end{align*}
We also need this equivalence on finitely generated projective modules, not merely on all modules. An equivalence of additive categories preserves finite direct sums and direct summands. Therefore a module is finitely generated projective if and only if its image under the equivalence is finitely generated projective. Since equivalences of module categories are exact, this gives an exact equivalence
\begin{align*}
\operatorname{Proj}(A)\simeq \operatorname{Proj}(R).
\end{align*}
This is the categorical input for the $K_0$ part.
[/guided]
[/step]
[step:Pass the projective-module equivalence to $K_0$]
Let $V(A)$ denote the commutative monoid of isomorphism classes of finitely generated projective left $A$-modules under direct sum, and define $V(R)$ analogously. The equivalence $F:\operatorname{Proj}(A)\to \operatorname{Proj}(R)$ sends direct sums to direct sums up to natural isomorphism, so it induces a monoid isomorphism
\begin{align*}
V(A) \to V(R), \qquad [M] \mapsto [F(M)].
\end{align*}
Its inverse is induced by $G$.
By definition, $K_0(A)$ and $K_0(R)$ are the group completions of $V(A)$ and $V(R)$. A monoid isomorphism induces an isomorphism on group completions. Therefore
\begin{align*}
K_0(M_n(R))=K_0(A)\cong K_0(R).
\end{align*}
[/step]
[step:Identify the stable general linear groups]
For each $m\ge 1$, define the block-matrix ring isomorphism
\begin{align*}
\Theta_m: M_m(A)&\to M_{mn}(R)
\end{align*}
by replacing each entry of an $m\times m$ matrix over $A=M_n(R)$ by its corresponding $n\times n$ block over $R$. This restricts to a group isomorphism
\begin{align*}
\Theta_m:GL_m(A)&\to GL_{mn}(R).
\end{align*}
The stabilization map $GL_m(A)\to GL_{m+1}(A)$ sends $B$ to $\operatorname{diag}(B,1_A)$. Under $\Theta_m$ and $\Theta_{m+1}$, this corresponds to the stabilization map
\begin{align*}
GL_{mn}(R)&\to GL_{(m+1)n}(R),
\end{align*}
given by
\begin{align*}
C&\mapsto \operatorname{diag}(C,I_n).
\end{align*}
The subsequence of matrix sizes $n,2n,3n,\dots$ is cofinal among positive integers for stable general linear groups, because every $q\ge 1$ embeds into some $mn$ with $mn\ge q$. Hence the maps $\Theta_m$ induce a canonical group isomorphism
\begin{align*}
\Theta:GL(A)&\to GL(R),
\end{align*}
where both sides are stable general linear groups.
[guided]
The $K_1$ argument starts by comparing invertible matrices over $M_n(R)$ with ordinary invertible matrices over $R$. For a positive integer $m$, define
\begin{align*}
\Theta_m:M_m(M_n(R))&\to M_{mn}(R)
\end{align*}
as follows: an $m\times m$ matrix whose entries are $n\times n$ matrices over $R$ is regarded as one large $mn\times mn$ block matrix over $R$. This is a ring isomorphism because block addition and block multiplication are exactly the addition and multiplication rules in $M_m(M_n(R))$.
A ring isomorphism sends units to units, so $\Theta_m$ restricts to a group isomorphism
\begin{align*}
\Theta_m:GL_m(M_n(R))&\to GL_{mn}(R).
\end{align*}
We must also check that this comparison is compatible with stabilization, because $K_1$ is defined using stable matrices. The stabilization in $GL_m(M_n(R))$ sends an invertible matrix $B$ to $\operatorname{diag}(B,1_{M_n(R)})$. Since $1_{M_n(R)}=I_n$ after viewing it as an $n\times n$ matrix over $R$, the block identification sends this stabilized matrix to
\begin{align*}
\operatorname{diag}(\Theta_m(B),I_n)\in GL_{(m+1)n}(R).
\end{align*}
Thus stabilization over $M_n(R)$ corresponds exactly to stabilization over $R$ by an identity block of size $n$.
Finally, stable $GL(R)$ does not depend on whether we pass through all sizes $1,2,3,\dots$ or only through the cofinal sizes $n,2n,3n,\dots$. Indeed, every integer $q\ge 1$ is at most $mn$ for some $m\ge 1$, so every finite invertible matrix over $R$ appears after stabilization inside one of the groups $GL_{mn}(R)$. Therefore the compatible isomorphisms $\Theta_m$ induce a group isomorphism
\begin{align*}
\Theta:GL(M_n(R))&\to GL(R).
\end{align*}
[/guided]
[/step]
[step:Check that elementary subgroups match under the stable identification]
Let $E(A)\le GL(A)$ and $E(R)\le GL(R)$ denote the stable elementary subgroups. We show that
\begin{align*}
\Theta(E(A))=E(R).
\end{align*}
First let $X=(x_{ij})\in M_n(R)=A$, and let $e_{ab}(X)\in GL_m(A)$ be an elementary transvection with $a\ne b$. Under $\Theta_m$, it becomes the block elementary matrix whose $(a,b)$ block is $X$. This block matrix is the product, over $1\le i,j\le n$, of the elementary matrices over $R$ with entry $x_{ij}$ in the row indexed by $(a,i)$ and the column indexed by $(b,j)$. Since $a\ne b$, these row and column indices are distinct, so each factor is elementary over $R$. Thus $\Theta(E(A))\subseteq E(R)$.
Conversely, take an elementary matrix $e_{\alpha\beta}(r)\in GL_q(R)$ with $\alpha\ne\beta$ and $r\in R$. Stabilize it into $GL_{mn}(R)$ for some $m\ge 2$ large enough that the two indices $\alpha,\beta$ occur among the first $mn$ basis vectors. Write the index set as block indices $(a,i)$ with $1\le a\le m$ and $1\le i\le n$. If the two indices lie in different blocks, then $e_{\alpha\beta}(r)$ is directly one of the elementary factors appearing in the image of a block elementary matrix over $A$.
If the two indices lie in the same block, choose a block index $c$ different from that block index. Let $k=1$. The elementary commutator relation in stable matrix groups gives
\begin{align*}
[e_{(a,i),(c,k)}(r),e_{(c,k),(a,j)}(1_R)]=e_{(a,i),(a,j)}(r).
\end{align*}
Both elementary matrices on the left lie between different blocks and hence belong to $\Theta(E(A))$. Since $\Theta(E(A))$ is a subgroup, the commutator also belongs to $\Theta(E(A))$. Therefore every generator of $E(R)$ belongs to $\Theta(E(A))$, and so $E(R)\subseteq \Theta(E(A))$.
[/step]
[step:Pass to the quotient defining $K_1$]
By [citetheorem:8655], the [stable elementary subgroup is normal](/theorems/8655) in the stable general linear group, so the quotient groups below are well-defined. Using the stable elementary definition of algebraic $K_1$, we have
\begin{align*}
K_1(A)=GL(A)/E(A)
\end{align*}
and
\begin{align*}
K_1(R)=GL(R)/E(R).
\end{align*}
The group isomorphism
\begin{align*}
\Theta:GL(A)\to GL(R)
\end{align*}
satisfies $\Theta(E(A))=E(R)$, so it descends to a group isomorphism
\begin{align*}
GL(A)/E(A)&\to GL(R)/E(R).
\end{align*}
Therefore
\begin{align*}
K_1(M_n(R))=K_1(A)\cong K_1(R).
\end{align*}
[/step]
[step:Verify naturality with respect to ring homomorphisms]
Let $\varphi:R\to S$ be a unital ring homomorphism, and let
\begin{align*}
M_n(\varphi):M_n(R)&\to M_n(S)
\end{align*}
be the entrywise ring homomorphism. For $K_0$, extension of scalars along $M_n(\varphi)$ commutes with the standard Morita functors because the idempotent $e\in M_n(R)$ maps to the corresponding idempotent in $M_n(S)$ and because tensor product is associative up to its canonical natural isomorphism. Hence the square
\begin{align*}
K_0(M_n(R))\to K_0(R)\to K_0(S)
\end{align*}
agrees with
\begin{align*}
K_0(M_n(R))\to K_0(M_n(S))\to K_0(S)
\end{align*}
under the constructed Morita isomorphisms.
For $T\in\{R,S\}$, define
\begin{align*}
\Theta_{m,T}: M_m(M_n(T)) \to M_{mn}(T)
\end{align*}
to be the block-matrix identification. Then
\begin{align*}
\Theta_{m,S}\circ M_m(M_n(\varphi))=M_{mn}(\varphi)\circ \Theta_{m,R}.
\end{align*}
This holds because both sides apply $\varphi$ entrywise to the same $mn\times mn$ block matrix. Passing to stable general linear groups and then to quotients by elementary subgroups gives the corresponding commutative square on $K_1$. Thus the isomorphisms $K_0(M_n(R))\cong K_0(R)$ and $K_1(M_n(R))\cong K_1(R)$ are natural in the unital ring $R$. This proves the theorem.
[/step]
