## Formalized Name Morita Invariance of $K_1$ for Matrix Rings ## Formalized Statement Let $R$ be a unital, not necessarily commutative, ring, let $n \geq 1$, and put $A:=M_n(R)$. For every unital ring $C$, define $GL(C):=\varinjlim_q GL_q(C)$ under the stabilization maps $X\mapsto \operatorname{diag}(X,1_C)$, let $E(C)\leq GL(C)$ be the stable elementary subgroup, and define $K_1(C):=GL(C)/E(C)$. Let $e=e_{11}(1_R)\in A$, and identify $eAe$ with $R$ by $r\mapsto e_{11}(r)$. Then block expansion $M_q(M_n(R))\cong M_{qn}(R)$ induces an isomorphism of abelian groups $K_1(M_n(R))\cong K_1(R)$. Its inverse is induced by the corner-stabilization map which sends a matrix $X=(x_{ij})\in GL_q(R)$ to the matrix over $A$ acting as $X$ on the $e$-summand and as the identity on the complementary summands. This is the $K_1$ isomorphism associated to the standard Morita idempotent $e$. ## Proof [proofplan] We prove the matrix-ring case directly in the stable Whitehead model, without appealing to a general Morita-invariance theorem. The block expansion is compatible with stabilization and identifies the stable group $GL(M_n(R))$ with the cofinal stable group $GL(R)$ on matrix sizes divisible by $n$. We then check that the stable elementary subgroup is carried onto the stable elementary subgroup, including the elementary transvections lying inside a single $n$-block by an explicit commutator computation. Finally, we identify the inverse map with the corner-stabilization construction and verify that its composite with block expansion is ordinary stabilization in $GL(R)$. [/proofplan] [step:Fix the stable matrix conventions] Let $A:=M_n(R)$. For $1\leq a,b\leq n$ and $r\in R$, let $e_{ab}(r)\in A$ denote the matrix with $r$ in the $(a,b)$-entry and $0_R$ elsewhere. Put \begin{align*} e:=e_{11}(1_R). \end{align*} Then $e$ is an idempotent, and the map $R\to eAe$ given by $r\mapsto e_{11}(r)$ is a unital ring isomorphism. For a unital ring $C$ and an integer $q\geq 1$, let $I_q(C)$ denote the identity matrix in $M_q(C)$. The stable elementary subgroup $E(C)$ is the subgroup of $GL(C)$ generated by all stable classes of elementary transvections $I_q(C)+cE_{ij}$ with $c\in C$ and $i\neq j$. The elementary commutator identities imply that $E(C)$ is normal in $GL(C)$, so the quotient group $K_1(C)=GL(C)/E(C)$ is defined. [/step] [step:Expand matrices over $M_n(R)$ into matrices over $R$] For each integer $q\geq 1$, define \begin{align*} \beta_q:GL_q(A)\to GL_{qn}(R) \end{align*} by block expansion. Namely, if $B=(B_{ij})$ with $B_{ij}\in A$ and $B_{ij}=(b_{ij,ab})_{1\leq a,b\leq n}$, then $\beta_q(B)$ is the $qn\times qn$ matrix over $R$ whose rows and columns are indexed by pairs $(i,a)$ and whose $((i,a),(j,b))$-entry is $b_{ij,ab}$. This construction is the standard ring isomorphism \begin{align*} M_q(M_n(R))\cong M_{qn}(R). \end{align*} It sends invertible matrices to invertible matrices and satisfies \begin{align*} \beta_{q+1}(\operatorname{diag}(B,1_A))=\operatorname{diag}(\beta_q(B),I_n(R)). \end{align*} Thus the maps $\beta_q$ induce a group homomorphism \begin{align*} \beta:GL(A)\to GL(R). \end{align*} The homomorphism $\beta$ is an isomorphism. Indeed, the positive integers divisible by $n$ form a cofinal subsystem of the directed system defining $GL(R)$. The maps $\beta_q$ identify the directed system $GL_q(A)$ with this cofinal subsystem $GL_{qn}(R)$, and passage to a direct limit over a cofinal subsystem gives the same direct limit. [/step] [step:Show that block expansion identifies the elementary subgroups] First let $B=I_q(A)+CE_{ij}$ be an elementary matrix in $GL_q(A)$, where $i\neq j$ and $C=(c_{ab})\in A$. Since all nonidentity entries of the expansion lie in the block from the $j$th summand to the $i$th summand, and since $i\neq j$, the corresponding elementary transvections over $R$ have pairwise zero cross-products. Hence \begin{align*} \beta_q(B)=\prod_{a=1}^{n}\prod_{b=1}^{n}\left(I_{qn}(R)+c_{ab}E_{(i,a),(j,b)}\right). \end{align*} Each factor is elementary over $R$, so $\beta(E(A))\subseteq E(R)$. Conversely, let $I_N(R)+rE_{uv}$ be an elementary transvection over $R$, with $u\neq v$. Stabilize it, if necessary, so that $N=qn$ for some $q\geq 2$ and so that there is an $n$-block different from the block containing both $u$ and $v$. Write $u=(i,a)$ and $v=(j,b)$ with $1\leq i,j\leq q$ and $1\leq a,b\leq n$. If $i\neq j$, then $I_{qn}(R)+rE_{(i,a),(j,b)}$ is the block expansion of the elementary matrix $I_q(A)+e_{ab}(r)E_{ij}$ over $A$. If $i=j$, then $a\neq b$. Choose $k\in\{1,\dots,q\}$ with $k\neq i$, and set $w:=(k,1)$. The elementary commutator identity in $GL_{qn}(R)$ gives \begin{align*} [I_{qn}(R)+rE_{u,w},I_{qn}(R)+E_{w,v}]=I_{qn}(R)+rE_{u,v}. \end{align*} The two elementary matrices on the left have indices in different $n$-blocks, so each lies in $\beta(E(A))$ by the previous paragraph. Therefore $I_{qn}(R)+rE_{u,v}$ also lies in $\beta(E(A))$. Thus every stable elementary transvection over $R$ lies in $\beta(E(A))$, and hence \begin{align*} \beta(E(A))=E(R). \end{align*} Since $\beta:GL(A)\to GL(R)$ is a group isomorphism carrying $E(A)$ onto $E(R)$, it descends to an isomorphism \begin{align*} \overline{\beta}:K_1(A)\to K_1(R). \end{align*} [guided] The only point requiring care is surjectivity on elementary subgroups. Block expansion visibly sends an elementary matrix over $M_n(R)$ to a product of elementary matrices over $R$: an off-diagonal block $C=(c_{ab})$ is decomposed into its individual entries $c_{ab}$, and each entry gives one elementary transvection in the expanded matrix. For the reverse inclusion, take an elementary transvection $I_N(R)+rE_{uv}$ over $R$. Because $K_1$ is stable, we may add identity rows and columns. We therefore choose a size $N=qn$ with $q\geq 2$ and write the two coordinates as $u=(i,a)$ and $v=(j,b)$, where $i,j$ are block indices and $a,b$ are positions inside a block. If $i\neq j$, the transvection already comes from a single elementary matrix over $A=M_n(R)$: it is the expansion of $I_q(A)+e_{ab}(r)E_{ij}$. The only missing case is $i=j$. Then $u$ and $v$ lie inside the same $n$-block, so the transvection is not itself the expansion of an off-diagonal elementary matrix over $A$. We use one extra block to express it as a commutator of two transvections that do cross between different blocks. Choose $k\neq i$ and set $w=(k,1)$. The standard elementary commutator identity gives \begin{align*} [I_{qn}(R)+rE_{u,w},I_{qn}(R)+E_{w,v}]=I_{qn}(R)+rE_{u,v}. \end{align*} Here $u$ and $w$ are in different blocks, and $w$ and $v$ are in different blocks. Hence both factors on the left are block expansions of elementary matrices over $A$. Their commutator therefore lies in $\beta(E(A))$, and it is exactly the desired elementary transvection. This proves $\beta(E(A))=E(R)$. [/guided] [/step] [step:Define the corner-stabilization map] For $X=(x_{ij})\in GL_q(R)$, define $\alpha_q(X)\in GL_q(A)$ by \begin{align*} (\alpha_q(X))_{ij}=e_{11}(x_{ij}) \end{align*} when $i\neq j$, and by \begin{align*} (\alpha_q(X))_{ii}=e_{11}(x_{ii})+\sum_{a=2}^{n}e_{aa}(1_R) \end{align*} when $i=j$. Equivalently, $\alpha_q(X)$ acts as $X$ on the direct summands selected by $e$ and as the identity on the complementary summands selected by $e_{22}(1_R),\dots,e_{nn}(1_R)$. The orthogonality of these idempotents gives \begin{align*} \alpha_q(XY)=\alpha_q(X)\alpha_q(Y) \end{align*} for all $X,Y\in GL_q(R)$, and the same formula applied to $X^{-1}$ gives the inverse of $\alpha_q(X)$. Hence $\alpha_q(X)\in GL_q(A)$. The construction is compatible with stabilization because \begin{align*} \alpha_{q+1}(\operatorname{diag}(X,1_R))=\operatorname{diag}(\alpha_q(X),1_A). \end{align*} Thus the maps $\alpha_q$ induce a group homomorphism \begin{align*} \alpha:GL(R)\to GL(A). \end{align*} If $X=I_q(R)+rE_{ij}$ is elementary over $R$, with $i\neq j$, then \begin{align*} \alpha_q(X)=I_q(A)+e_{11}(r)E_{ij}, \end{align*} which is elementary over $A$. Therefore $\alpha(E(R))\subseteq E(A)$, and $\alpha$ descends to a homomorphism \begin{align*} \overline{\alpha}:K_1(R)\to K_1(A). \end{align*} [/step] [step:Compute the composite on $K_1(R)$] Let $X\in GL_q(R)$. The expanded matrix $\beta_q(\alpha_q(X))\in GL_{qn}(R)$ acts as $X$ on the coordinates $(1,1),\dots,(q,1)$ and as the identity on all coordinates $(i,a)$ with $2\leq a\leq n$. Let $P\in GL_{qn}(R)$ be the permutation matrix that moves the coordinates $(1,1),\dots,(q,1)$ to the first $q$ positions, preserving their order, and moves the remaining coordinates after them. Then \begin{align*} P\beta_q(\alpha_q(X))P^{-1}=\operatorname{diag}(X,I_{q(n-1)}(R)). \end{align*} The right-hand side represents the same element of the stable group $GL(R)$ as $X$. Since equality in $GL(R)$ is already obtained after applying the stabilization and permutation that define the cofinal direct-limit identification, we have \begin{align*} \beta\alpha([X]_{GL(R)})=[X]_{GL(R)}. \end{align*} Passing to the quotient by elementary subgroups gives \begin{align*} \overline{\beta}\overline{\alpha}=\operatorname{id}_{K_1(R)}. \end{align*} [/step] [step:Conclude that the corner map is the inverse Morita map] The map $\overline{\beta}:K_1(A)\to K_1(R)$ is an isomorphism because $\beta:GL(A)\to GL(R)$ is an isomorphism carrying $E(A)$ onto $E(R)$. The previous step proves that $\overline{\beta}\overline{\alpha}=\operatorname{id}_{K_1(R)}$. Therefore $\overline{\alpha}$ is the inverse of $\overline{\beta}$, so also \begin{align*} \overline{\alpha}\overline{\beta}=\operatorname{id}_{K_1(A)}. \end{align*} Under the identification $eAe\cong R$ by $r\mapsto e_{11}(r)$, the map $\overline{\alpha}$ is exactly the corner-stabilization map associated to the full idempotent $e=e_{11}(1_R)$. The idempotent is full because \begin{align*} 1_A=\sum_{a=1}^{n}e_{a1}(1_R)e e_{1a}(1_R). \end{align*} Thus this is the $K_1$ isomorphism associated to the standard Morita idempotent for $M_n(R)$, and we obtain \begin{align*} K_1(R)\cong K_1(M_n(R)). \end{align*} [/step]