[proofplan]
The point is that the determinant on the matrix algebra is the determinant on $R$ after passing through the standard Morita invariance isomorphism. Thus $SK_1(M_n(R))$ is exactly the kernel of the composite $\det_R \circ \mu_n$. Since $\mu_n$ is an isomorphism, it carries this kernel isomorphically onto the kernel of $\det_R$, which is $SK_1(R)$ by definition.
[/proofplan]
[step:Choose the Morita isomorphism and the compatible determinant map]
Let
\begin{align*}
\mu_n:K_1(M_n(R)) \to K_1(R)
\end{align*}
denote the inverse of the standard Morita invariance isomorphism supplied by [citetheorem:8663]. Since $R$ is commutative, the stable determinant is a [group homomorphism](/page/Group%20Homomorphism)
\begin{align*}
\det_R:K_1(R) \to R^\times
\end{align*}
by [citetheorem:8658]. By the definition of the reduced norm, or determinant, for the matrix algebra $M_n(R)$ under Morita invariance, the determinant map on $K_1(M_n(R))$ is the composite
\begin{align*}
\det_{M_n(R)/R}:K_1(M_n(R)) \to R^\times
\end{align*}
given by
\begin{align*}
\det_{M_n(R)/R}=\det_R \circ \mu_n.
\end{align*}
Therefore
\begin{align*}
SK_1(M_n(R))=\ker(\det_R \circ \mu_n).
\end{align*}
[/step]
[step:Restrict the Morita isomorphism to the determinant kernels]
We prove that $\mu_n$ maps $\ker(\det_R \circ \mu_n)$ isomorphically onto $\ker(\det_R)$. If $x \in \ker(\det_R \circ \mu_n)$, then
\begin{align*}
\det_R(\mu_n(x))=1_R,
\end{align*}
so $\mu_n(x)\in \ker(\det_R)$. Hence $\mu_n$ restricts to a homomorphism
\begin{align*}
\mu_n\big|_{\ker(\det_R \circ \mu_n)}:\ker(\det_R \circ \mu_n) \to \ker(\det_R).
\end{align*}
Because $\mu_n$ is an isomorphism, this restricted homomorphism is injective. To prove surjectivity, let $y \in \ker(\det_R)$. Since $\mu_n$ is surjective, there exists $x \in K_1(M_n(R))$ such that $\mu_n(x)=y$. Then
\begin{align*}
(\det_R \circ \mu_n)(x)=\det_R(y)=1_R,
\end{align*}
so $x\in \ker(\det_R \circ \mu_n)$. Thus $y$ lies in the image of the restricted map.
[guided]
The only algebraic point is the following general fact: an isomorphism carries the kernel of a composite onto the kernel of the second map. Here the isomorphism is
\begin{align*}
\mu_n:K_1(M_n(R)) \to K_1(R),
\end{align*}
and the second map is
\begin{align*}
\det_R:K_1(R) \to R^\times.
\end{align*}
The composite is
\begin{align*}
\det_R \circ \mu_n:K_1(M_n(R)) \to R^\times.
\end{align*}
First take an element $x \in \ker(\det_R \circ \mu_n)$. By the definition of kernel, this means
\begin{align*}
(\det_R \circ \mu_n)(x)=1_R.
\end{align*}
Equivalently,
\begin{align*}
\det_R(\mu_n(x))=1_R.
\end{align*}
Hence $\mu_n(x)\in \ker(\det_R)$. This proves that $\mu_n$ sends $\ker(\det_R \circ \mu_n)$ into $\ker(\det_R)$, so the restricted map
\begin{align*}
\mu_n\big|_{\ker(\det_R \circ \mu_n)}:\ker(\det_R \circ \mu_n) \to \ker(\det_R)
\end{align*}
is well-defined.
The restriction is injective because $\mu_n$ itself is injective. For surjectivity, choose an arbitrary element $y \in \ker(\det_R)$. Since $\mu_n$ is surjective, there exists $x \in K_1(M_n(R))$ with $\mu_n(x)=y$. Now compute the composite on this chosen preimage:
\begin{align*}
(\det_R \circ \mu_n)(x)=\det_R(y).
\end{align*}
Because $y\in \ker(\det_R)$, we have $\det_R(y)=1_R$. Therefore
\begin{align*}
(\det_R \circ \mu_n)(x)=1_R,
\end{align*}
so $x\in \ker(\det_R \circ \mu_n)$. Thus every element of $\ker(\det_R)$ is hit by the restricted map, proving surjectivity.
[/guided]
[/step]
[step:Identify the two kernels with the two special Whitehead groups]
By definition,
\begin{align*}
SK_1(M_n(R))=\ker(\det_{M_n(R)/R}).
\end{align*}
Since $\det_{M_n(R)/R}=\det_R \circ \mu_n$, this gives
\begin{align*}
SK_1(M_n(R))=\ker(\det_R \circ \mu_n).
\end{align*}
Also, by definition for the commutative ring $R$,
\begin{align*}
SK_1(R)=\ker(\det_R).
\end{align*}
The restricted isomorphism from the previous step is therefore an isomorphism
\begin{align*}
SK_1(M_n(R)) \to SK_1(R).
\end{align*}
This is precisely the claimed identification under the standard Morita invariance isomorphism.
[/step]
