[proofplan]
We define the proposed section $s:\mathbb Z\to K_0(R)$ by sending $1$ to the class $[R]$ of the free rank-one module. Since $R$ has constant rank $1$, the composite $\operatorname{rank}\circ s$ is the identity on $\mathbb Z$. We then prove the elementary splitting directly: every class in $K_0(R)$ decomposes uniquely as its rank part plus an element of the kernel of the rank map.
[/proofplan]
[step:Define the section generated by the class of the rank-one free module]
Let $[R]\in K_0(R)$ denote the class of $R$ as a finitely generated projective $R$-module. Define $s:\mathbb Z\to K_0(R)$ by $s(n)=n[R]$. Since $K_0(R)$ is an abelian group under addition of classes and $\mathbb Z$ is the free abelian group on $1$, this formula defines a [group homomorphism](/page/Group%20Homomorphism). Explicitly, for $m,n\in\mathbb Z$,
\begin{align*}
s(m+n)=(m+n)[R]=m[R]+n[R]=s(m)+s(n).
\end{align*}
[/step]
[step:Compute the rank of the image of each integer]
Because $R$ is a nonzero commutative unital ring, every localization $R_{\mathfrak p}$ at a prime ideal $\mathfrak p\in\operatorname{Spec}(R)$ is a free $R_{\mathfrak p}$-module of rank $1$ over itself. Hence the finitely generated projective module $R$ has constant rank $1$, so
\begin{align*}
\operatorname{rank}([R])=1.
\end{align*}
Since $\operatorname{rank}:K_0(R)\to\mathbb Z$ is a group homomorphism, for every $n\in\mathbb Z$,
\begin{align*}
\operatorname{rank}(s(n))=\operatorname{rank}(n[R])=n\operatorname{rank}([R])=n.
\end{align*}
Therefore
\begin{align*}
(\operatorname{rank}\circ s)(n)=n
\end{align*}
for every $n\in\mathbb Z$, so
\begin{align*}
\operatorname{rank}\circ s=\operatorname{id}_{\mathbb Z}.
\end{align*}
Thus $s$ is a section of the rank homomorphism.
[guided]
The point of the proof is to show that the rank map has a distinguished right inverse. The natural candidate sends the integer $1$ to the class $[R]$, because $R$ is the free module of rank $1$ over itself. Thus we define $s:\mathbb Z\to K_0(R)$ by $s(n)=n[R]$. This is a group homomorphism because addition in $K_0(R)$ is written additively and $\mathbb Z$ is generated by $1$. For $m,n\in\mathbb Z$,
\begin{align*}
s(m+n)=(m+n)[R]=m[R]+n[R]=s(m)+s(n).
\end{align*}
We now compute the composite with the rank map. For every prime ideal $\mathfrak p\in\operatorname{Spec}(R)$, the localization $R_{\mathfrak p}$ is a free module of rank $1$ over itself. Therefore the module $R$ has constant rank $1$, and its class satisfies
\begin{align*}
\operatorname{rank}([R])=1.
\end{align*}
Since $\operatorname{rank}:K_0(R)\to\mathbb Z$ is a homomorphism of abelian groups, it respects integer multiples. Hence, for every $n\in\mathbb Z$,
\begin{align*}
\operatorname{rank}(s(n))=\operatorname{rank}(n[R])=n\operatorname{rank}([R])=n.
\end{align*}
This proves
\begin{align*}
\operatorname{rank}\circ s=\operatorname{id}_{\mathbb Z}.
\end{align*}
So $s$ is exactly a section of the rank map: applying $s$ and then taking rank recovers the original integer.
[/guided]
[/step]
[step:Split $K_0(R)$ into rank and reduced rank-zero parts]
Define
\begin{align*}
\widetilde K_0(R):=\ker(\operatorname{rank}:K_0(R)\to\mathbb Z).
\end{align*}
We construct an isomorphism $\Phi:\mathbb Z\oplus \widetilde K_0(R)\to K_0(R)$ by $\Phi(n,x)=s(n)+x$. This is a homomorphism because both $s$ and the inclusion $\widetilde K_0(R)\hookrightarrow K_0(R)$ are homomorphisms.
Define also $\Psi:K_0(R)\to \mathbb Z\oplus \widetilde K_0(R)$ by $\Psi(y)=(\operatorname{rank}(y),y-s(\operatorname{rank}(y)))$. For $y\in K_0(R)$, the second component lies in $\widetilde K_0(R)$ because
\begin{align*}
\operatorname{rank}(y-s(\operatorname{rank}(y)))=\operatorname{rank}(y)-(\operatorname{rank}\circ s)(\operatorname{rank}(y))=\operatorname{rank}(y)-\operatorname{rank}(y)=0.
\end{align*}
Thus $\Psi$ is well-defined. It is also a homomorphism: for $y,z\in K_0(R)$, using additivity of $\operatorname{rank}$ and of $s$ gives
\begin{align*}
\Psi(y+z)=(\operatorname{rank}(y)+\operatorname{rank}(z),y+z-s(\operatorname{rank}(y))-s(\operatorname{rank}(z)))=\Psi(y)+\Psi(z).
\end{align*}
For $n\in\mathbb Z$ and $x\in\widetilde K_0(R)$, using $\operatorname{rank}(x)=0$ gives
\begin{align*}
\Psi(\Phi(n,x))=\Psi(s(n)+x)=(\operatorname{rank}(s(n)+x),s(n)+x-s(\operatorname{rank}(s(n)+x)))=(n,x).
\end{align*}
For $y\in K_0(R)$,
\begin{align*}
\Phi(\Psi(y))=s(\operatorname{rank}(y))+y-s(\operatorname{rank}(y))=y.
\end{align*}
Hence $\Phi$ and $\Psi$ are inverse group homomorphisms. Therefore
\begin{align*}
K_0(R)\cong \mathbb Z\oplus \widetilde K_0(R).
\end{align*}
This proves the asserted rank splitting.
[/step]
