[proofplan]
The determinant gives a homomorphism from $K_1(A)$ to $A^\times$ because elementary matrices have determinant $1$. The inclusion of units as $1 \times 1$ invertible matrices gives a section, so the determinant is split surjective. The kernel is the special Whitehead group, and the deep arithmetic input is the Bass-Milnor-Serre theorem, which says that this kernel vanishes for rings of integers and rings of $S$-integers in number fields. Therefore the determinant is both surjective and injective.
[/proofplan]
[step:Define the stable determinant on $K_1(A)$]
For each integer $n\geq 1$, define $GL_n(A)$ to be the group of invertible $n\times n$ matrices with entries in $A$. Let $I_n\in GL_n(A)$ denote the $n\times n$ identity matrix, and for $1\leq i,j\leq n$ let $E_{ij}\in M_n(A)$ be the matrix whose $(i,j)$ entry is $1_A$ and whose other entries are $0_A$. Define the stable general linear group
\begin{align*}
GL(A):=\varinjlim_n GL_n(A)
\end{align*}
under the stabilization maps $B\mapsto \operatorname{diag}(B,1)$. For $n\geq 2$, $1\leq i\neq j\leq n$, and $a\in A$, define the elementary transvection $e_{ij}(a)\in GL_n(A)$ by $e_{ij}(a)=I_n+aE_{ij}$, and let $E(A)\leq GL(A)$ be the stable elementary subgroup generated by all such transvections. We use the standard Whitehead quotient model
\begin{align*}
K_1(A)=GL(A)/E(A),
\end{align*}
which is abelian in this model because $K_1(A)$ is the abelianization of $GL(A)$.
Since $A$ is a commutative unital ring, the stable determinant induces a [group homomorphism](/page/Group%20Homomorphism)
\begin{align*}
\det:K_1(A)\longrightarrow A^\times
\end{align*}
by [Determinant Descends To K One][citetheorem:8658]. Concretely, if $B\in GL_n(A)$ represents a class in $GL(A)$, then the image of its class in $K_1(A)$ is $\det(B)\in A^\times$.
This is well-defined on $K_1(A)$ because every elementary transvection has determinant $1$. Thus the determinant is constant on cosets modulo $E(A)$.
[/step]
[step:Split the determinant by including units as one by one matrices]
Define a map
\begin{align*}
s:A^\times&\longrightarrow K_1(A)
\end{align*}
as follows: for $u\in A^\times$, let $s(u)$ be the class in $K_1(A)$ of the matrix $(u)\in GL_1(A)\subset GL(A)$. This is a group homomorphism because multiplication in $GL_1(A)$ is multiplication in $A^\times$.
For every $u\in A^\times$, the stable determinant satisfies
\begin{align*}
(\det\circ s)(u)=\det((u))=u.
\end{align*}
Hence $s$ is a section of $\det$, so $\det:K_1(A)\to A^\times$ is surjective. Equivalently, this is the determinant splitting for commutative rings from [Determinant Splitting][citetheorem:8659].
[guided]
We first build a right inverse to the determinant. The group $A^\times$ is the group of invertible elements of the ring $A$, while $K_1(A)$ is built from stable invertible matrices over $A$. The most direct way to view a unit as a stable invertible matrix is to put it in size $1$.
Define
\begin{align*}
s:A^\times\longrightarrow K_1(A)
\end{align*}
by sending a unit $u\in A^\times$ to the class of the $1\times 1$ matrix $(u)$ in $K_1(A)$. This map is a group homomorphism: if $u,v\in A^\times$, then in $GL_1(A)$ one has $(u)(v)=(uv)$, so the corresponding classes in $K_1(A)$ satisfy $s(u)s(v)=s(uv)$.
Now compute the composite with the determinant. For each $u\in A^\times$,
\begin{align*}
(\det\circ s)(u)=\det((u))=u
\end{align*}
Thus $\det\circ s=\operatorname{id}_{A^\times}$. A homomorphism with a right inverse is surjective, so the determinant map
\begin{align*}
\det:K_1(A)\longrightarrow A^\times
\end{align*}
is surjective. This is precisely the splitting statement in [Determinant Splitting][citetheorem:8659].
[/guided]
[/step]
[step:Identify the kernel with $SK_1(A)$]
By definition, the special Whitehead group of the commutative ring $A$ is
\begin{align*}
SK_1(A):=\ker\left(\det:K_1(A)\longrightarrow A^\times\right).
\end{align*}
Therefore proving that the determinant is injective is exactly the same as proving
\begin{align*}
SK_1(A)=0.
\end{align*}
[/step]
[step:Apply Bass-Milnor-Serre vanishing for arithmetic Dedekind domains]
The repaired statement defines $\mathcal O_{F,S}$ as the usual ring of $S$-integers:
\begin{align*}
\mathcal O_{F,S}=\{x\in F: v_{\mathfrak p}(x)\geq 0\text{ for every nonzero prime ideal }\mathfrak p\notin S\}.
\end{align*}
Thus the stated hypotheses place $A$ in exactly one of the arithmetic cases covered by [Localization For S Integers][citetheorem:8693]: either $A=\mathcal O_F$, or $A=\mathcal O_{F,S}$ for a finite set $S$ of nonzero prime ideals of $\mathcal O_F$. The $K_1$ assertion in that theorem is equivalent to the Bass-Milnor-Serre vanishing statement
\begin{align*}
SK_1(A)=0.
\end{align*}
Equivalently, every stable determinant-one class over this ring $A$ is represented, after stabilization, by a product of elementary transvections. This is the deep arithmetic input in the proof.
Since $SK_1(A)$ is the kernel of the determinant, the equality $SK_1(A)=0$ implies
\begin{align*}
\ker(\det)=0.
\end{align*}
Thus the determinant is injective.
[/step]
[step:Conclude that the determinant is an isomorphism]
The determinant homomorphism
\begin{align*}
\det:K_1(A)\longrightarrow A^\times
\end{align*}
is surjective by the unit-section construction and injective by Bass-Milnor-Serre vanishing of $SK_1(A)$. Therefore it is an isomorphism of abelian groups:
\begin{align*}
K_1(A)\cong A^\times.
\end{align*}
This proves the theorem.
[/step]
