[proofplan] We first assemble the ordinary determinants $\det_n:GL_n(R)\to R^\times$ into a single stable determinant on $GL(R)$ by checking compatibility with the stabilization maps $A\mapsto \operatorname{diag}(A,1)$. We then use multiplicativity of determinant to see that this stable determinant is a [group homomorphism](/page/Group%20Homomorphism). Finally, every elementary transvection has determinant $1$, so the stable elementary subgroup $E(R)$ lies in the kernel, and the determinant descends uniquely through the quotient $K_1(R)=GL(R)/E(R)$. [/proofplan] [step:Assemble the ordinary determinants into a stable determinant on $GL(R)$] For each integer $n\geq 1$, let $GL_n(R)$ denote the group of invertible $n\times n$ matrices over $R$, and let \begin{align*} s_n:GL_n(R)\to GL_{n+1}(R) \end{align*} be the stabilization homomorphism defined by \begin{align*} s_n(A)=\operatorname{diag}(A,1_R). \end{align*} Let $GL(R)=\varinjlim_n GL_n(R)$ be the stable general linear group with respect to these maps. Since $R$ is commutative, the ordinary determinant is defined on matrices over $R$. For each $n\geq 1$, write \begin{align*} \det_n:GL_n(R)\to R^\times \end{align*} for the determinant homomorphism. If $A\in GL_n(R)$, then expansion of the determinant of the block diagonal matrix gives \begin{align*} \det_{n+1}(s_n(A))=\det_{n+1}(\operatorname{diag}(A,1_R))=\det_n(A)\cdot 1_R=\det_n(A). \end{align*} Thus the maps $\det_n$ are compatible with stabilization. Therefore they define a unique map \begin{align*} \det_{\mathrm{st}}:GL(R)\to R^\times \end{align*} such that, for every $n\geq 1$ and every $A\in GL_n(R)$, the image of the class of $A$ in $GL(R)$ is $\det_n(A)$. [/step] [step:Verify that the stable determinant is a group homomorphism] Let $g,h\in GL(R)$. Choose an integer $n\geq 1$ and matrices $A,B\in GL_n(R)$ representing $g$ and $h$ after applying enough stabilization. This is possible by the definition of the filtered colimit $GL(R)=\varinjlim_n GL_n(R)$. The product $gh$ is represented by $AB\in GL_n(R)$. Since $R$ is commutative, the determinant is multiplicative on $GL_n(R)$, so \begin{align*} \det_{\mathrm{st}}(gh)=\det_n(AB)=\det_n(A)\det_n(B)=\det_{\mathrm{st}}(g)\det_{\mathrm{st}}(h). \end{align*} Hence \begin{align*} \det_{\mathrm{st}}:GL(R)\to R^\times \end{align*} is a group homomorphism. [guided] We must check that passing to the stable group has not destroyed multiplicativity. The point is that two elements of $GL(R)$ can always be represented in one common matrix size: if one initially comes from $GL_m(R)$ and the other from $GL_n(R)$, stabilize both into $GL_N(R)$ for some integer $N\geq \max\{m,n\}$. So let $g,h\in GL(R)$, and choose an integer $N\geq 1$ and matrices $A,B\in GL_N(R)$ representing them. In the stable group, the product $gh$ is represented by the ordinary matrix product $AB\in GL_N(R)$. The determinant is available here because $R$ is commutative; over a noncommutative ring, the ordinary determinant is not a multiplicative map with values in $R^\times$ in this form. Therefore multiplicativity in $GL_N(R)$ gives \begin{align*} \det_{\mathrm{st}}(gh)=\det_N(AB)=\det_N(A)\det_N(B). \end{align*} By the definition of $\det_{\mathrm{st}}$, the right-hand side is \begin{align*} \det_{\mathrm{st}}(g)\det_{\mathrm{st}}(h). \end{align*} Thus $\det_{\mathrm{st}}$ preserves multiplication, and so it is a group homomorphism from $GL(R)$ to $R^\times$. [/guided] [/step] [step:Show that the stable elementary subgroup lies in the kernel] For $n\geq 2$, for distinct indices $1\leq i\neq j\leq n$, and for an element $r\in R$, let \begin{align*} e_{ij}(r)=I_n+rE_{ij}\in GL_n(R) \end{align*} denote the elementary transvection, where $E_{ij}$ is the matrix with $1_R$ in the $(i,j)$ entry and $0_R$ elsewhere. Since $e_{ij}(r)$ differs from $I_n$ by adding $r$ times column $i$ to column $j$, its determinant is unchanged from that of $I_n$. Hence \begin{align*} \det_n(e_{ij}(r))=1_R. \end{align*} The stable elementary subgroup $E(R)\leq GL(R)$ is generated by the images of all such elementary transvections under stabilization. Since $\det_{\mathrm{st}}$ is a group homomorphism and sends every generator of $E(R)$ to $1_R$, it sends every element of $E(R)$ to $1_R$. Therefore \begin{align*} E(R)\subseteq \ker(\det_{\mathrm{st}}). \end{align*} [/step] [step:Descend the determinant through the quotient defining $K_1(R)$] By the convention in the statement, \begin{align*} K_1(R)=GL(R)/E(R). \end{align*} Let \begin{align*} q:GL(R)\to K_1(R) \end{align*} be the quotient homomorphism. Since $E(R)\subseteq \ker(\det_{\mathrm{st}})$, the universal property of the [quotient group](/theorems/790) gives a unique group homomorphism \begin{align*} \det:K_1(R)\to R^\times \end{align*} such that \begin{align*} \det\circ q=\det_{\mathrm{st}}. \end{align*} Equivalently, if $A\in GL_n(R)$ represents a class in $K_1(R)$, then \begin{align*} \det([A])=\det_n(A). \end{align*} This is well-defined because stabilization preserves determinant and multiplication by elementary matrices does not change the determinant. Thus the stable determinant defines the claimed group homomorphism \begin{align*} \det:K_1(R)\to R^\times. \end{align*} [/step]