[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*}
[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]
[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]
