[proofplan] The determinant is defined on $K_1(R)$ because $R$ is commutative, and $SK_1(R)$ is, by definition, its kernel. The inclusion $GL_1(R)=R^\times\subset GL(R)$ gives a section of the determinant on $K_1(R)$, so the determinant is surjective and the resulting short exact sequence splits. Finally, a split short exact sequence of abelian groups gives an explicit direct-sum decomposition. [/proofplan] [step:Use commutativity to define the stable determinant on $K_1(R)$] Since $R$ is commutative and unital, the stable determinant is a well-defined [group homomorphism](/page/Group%20Homomorphism) \begin{align*} \det:K_1(R)\to R^\times. \end{align*} This is precisely the determinant homomorphism supplied by [citetheorem:8658]. We define \begin{align*} SK_1(R):=\ker(\det). \end{align*} Let \begin{align*} \kappa:SK_1(R)\to K_1(R) \end{align*} denote the inclusion homomorphism. [guided] The commutativity hypothesis is the point that allows determinants of matrices over $R$ to behave multiplicatively and hence to pass to stable $K_1$. By [citetheorem:8658], the stable determinant defines a group homomorphism \begin{align*} \det:K_1(R)\to R^\times. \end{align*} Here $R^\times$ is the abelian group of units of $R$ under multiplication. The special Whitehead group in this theorem is defined by the kernel convention \begin{align*} SK_1(R):=\ker(\det). \end{align*} Thus an element $x\in K_1(R)$ lies in $SK_1(R)$ exactly when \begin{align*} \det(x)=1_R. \end{align*} We write \begin{align*} \kappa:SK_1(R)\to K_1(R) \end{align*} for the inclusion map. This map is injective because it is the inclusion of a subgroup. [/guided] [/step] [step:Split the determinant by the $GL_1$ inclusion] Let $GL_1(R)=R^\times$ be the group of invertible $1\times 1$ matrices over $R$, and let $GL(R)=\varinjlim_n GL_n(R)$ be the stable general linear group under the stabilization maps $A\mapsto \operatorname{diag}(A,1_R)$. By [citetheorem:8659], the homomorphism \begin{align*} s:R^\times\to K_1(R),\qquad u\mapsto [u] \end{align*} induced by the inclusion $GL_1(R)\subset GL(R)$ is a section of the determinant. At the matrix level, the representative of $s(u)$ is the stabilized diagonal matrix $\operatorname{diag}(u,1_R,\dots,1_R)$, whose determinant is $u$. Hence \begin{align*} \det\circ s=\operatorname{id}_{R^\times}. \end{align*} In particular, $\det$ is surjective. [/step] [step:Identify the exact sequence] We prove exactness of \begin{align*} 0\to SK_1(R)\xrightarrow{\kappa}K_1(R)\xrightarrow{\det}R^\times\to 0. \end{align*} The map $\kappa$ is injective because it is an inclusion. Exactness at $K_1(R)$ follows from the definition \begin{align*} \operatorname{im}(\kappa)=SK_1(R)=\ker(\det). \end{align*} Exactness at $R^\times$ follows from the surjectivity of $\det$, proved by the existence of the section $s$. [/step] [step:Construct the direct-sum isomorphism from the splitting] Since $K_1(R)$ is an abelian group and the sequence above is split by $s$, define \begin{align*} \Phi:R^\times\oplus SK_1(R)\to K_1(R) \end{align*} by \begin{align*} \Phi(u,a):=s(u)+\kappa(a). \end{align*} Define \begin{align*} \Psi:K_1(R)\to R^\times\oplus SK_1(R) \end{align*} by \begin{align*} \Psi(x):=(\det(x),x-s(\det(x))). \end{align*} The second component of $\Psi(x)$ lies in $SK_1(R)$ because \begin{align*} \det(x-s(\det(x)))=\det(x)\det(s(\det(x)))^{-1}=\det(x)\det(x)^{-1}=1_R. \end{align*} Using $\det\circ s=\operatorname{id}_{R^\times}$ and the injectivity of $\kappa$, one checks directly that \begin{align*} \Psi(\Phi(u,a))=(u,a) \end{align*} for all $(u,a)\in R^\times\oplus SK_1(R)$, and \begin{align*} \Phi(\Psi(x))=x \end{align*} for all $x\in K_1(R)$. Hence $\Phi$ is an isomorphism of abelian groups, so \begin{align*} K_1(R)\cong R^\times\oplus SK_1(R). \end{align*} [/step]