[proofplan] We verify directly that the class of a unit $u\in R^\times$, viewed as the $1\times 1$ matrix $(u)$, has determinant $u$. This proves that the inclusion-induced homomorphism $s:R^\times\to K_1(R)$ is a section of the stable determinant. The direct-summand statement then follows from the standard splitting argument: every class in $K_1(R)$ differs from an element of $s(R^\times)$ by an element of $\ker(\det)$, and the intersection of these two subgroups is zero. [/proofplan]

[step:Define the unit inclusion and verify that it is a group homomorphism] For each $u\in R^\times$, let $(u)\in GL_1(R)$ denote the $1\times 1$ invertible matrix whose only entry is $u$. Let $q:GL(R)\to K_1(R)$ denote the quotient map from the stable general linear group to $K_1(R)$. The inclusion $GL_1(R)\subset GL(R)$ sends $(u)$ to an element of $GL(R)$, and hence defines \begin{align*} s:R^\times\to K_1(R), \qquad u\mapsto q((u))=[(u)]. \end{align*} For $u,v\in R^\times$, multiplication in $GL_1(R)$ gives $(u)(v)=(uv)$. By the definition of $K_1(R)$ used in the statement, the group law on $K_1(R)$ is induced by multiplication in $GL(R)$ and is written additively. Therefore \begin{align*} s(uv)=[(uv)]=[(u)(v)]=[(u)]+[(v)]=s(u)+s(v). \end{align*} Also $s(1_R)=[(1_R)]$, the identity element of $K_1(R)$. Thus $s$ is a [group homomorphism](/page/Group%20Homomorphism). [/step]

[step:Compute the determinant of the embedded unit]Because $R$ is commutative, the stable determinant descends to a homomorphism \begin{align*} \det:K_1(R)\to R^\times \end{align*} by [citetheorem:8658], and this is the homomorphism fixed in the statement. For $u\in R^\times$, the determinant of the $1\times 1$ matrix $(u)$ is exactly $u$. Hence \begin{align*} (\det\circ s)(u)=\det([(u)])=\det((u))=u. \end{align*} Since this holds for every $u\in R^\times$, \begin{align*} \det\circ s=\operatorname{id}_{R^\times}. \end{align*}[/step]

[guided]The point of this step is to check the splitting identity on an arbitrary element of the unit group. Since $R$ is commutative, the determinant of an invertible matrix over $R$ is multiplicative and, by [citetheorem:8658], it induces a well-defined group homomorphism \begin{align*} \det:K_1(R)\to R^\times. \end{align*} Now fix $u\in R^\times$. The map $s$ sends $u$ to the class in $K_1(R)$ represented by the $1\times 1$ invertible matrix $(u)$. Applying the determinant to this representative gives \begin{align*} \det((u))=u, \end{align*} because the determinant of a $1\times 1$ matrix is its unique entry. Therefore \begin{align*} (\det\circ s)(u)=\det(s(u))=\det([(u)])=u. \end{align*} Since the calculation used only that $u$ was an arbitrary unit of $R$, it proves the identity of homomorphisms \begin{align*} \det\circ s=\operatorname{id}_{R^\times}. \end{align*} This is precisely the section property: $s$ chooses, for each unit $u$, a class in $K_1(R)$ whose determinant is $u$.[/guided]

[step:Split $K_1(R)$ into determinant and determinant-zero parts] In this step we write the abelian group laws on both $K_1(R)$ and $R^\times$ additively; thus the zero element of $R^\times$ in this notation is the multiplicative unit $1_R$. First, $s$ is injective. Indeed, if $s(u)=s(v)$ for $u,v\in R^\times$, then applying $\det$ gives \begin{align*} u=(\det\circ s)(u)=(\det\circ s)(v)=v. \end{align*} We now prove the internal direct-sum decomposition. Let $x\in K_1(R)$, and define \begin{align*} u:=\det(x)\in R^\times. \end{align*} Then \begin{align*} \det(x-s(u))=\det(x)-\det(s(u))=u-u=0 \end{align*} in the additive notation for the abelian group $R^\times$. Hence $x-s(u)\in\ker(\det)$, and therefore \begin{align*} x=s(u)+(x-s(u))\in s(R^\times)+\ker(\det). \end{align*} Thus \begin{align*} K_1(R)=s(R^\times)+\ker(\det). \end{align*} It remains to check that the sum is direct. If $y\in s(R^\times)\cap\ker(\det)$, then $y=s(u)$ for some $u\in R^\times$, and \begin{align*} 0=\det(y)=\det(s(u))=u \end{align*} in additive notation on $R^\times$. Thus $u=1_R$ multiplicatively, and hence \begin{align*} y=s(1_R)=0 \end{align*} in $K_1(R)$. Therefore \begin{align*} s(R^\times)\cap\ker(\det)=\{0\}. \end{align*} Consequently \begin{align*} K_1(R)=s(R^\times)\oplus\ker(\det). \end{align*} Since $s:R^\times\to s(R^\times)$ is an isomorphism, this gives \begin{align*} K_1(R)\cong R^\times\oplus\ker(\det) \end{align*} as abelian groups. In particular, $R^\times$ is a direct summand of $K_1(R)$. [/step]