[proofplan]
The nil groups are defined as kernels of the maps on $K_0$ and $K_1$ induced by evaluation $R[t]\to R$ at $t=0$. The inclusion of constant polynomials $R\to R[t]$ is a section of evaluation, so functoriality gives a corresponding split pair on low $K$-groups. Homotopy invariance for regular Noetherian rings says that the inclusion-induced map is an isomorphism in degrees $0$ and $1$. Therefore the evaluation-induced map is its inverse and has zero kernel in both degrees.
[/proofplan]
[step:Identify the evaluation map and its section by constant polynomials]
Let
\begin{align*}
\iota:R\to R[t]
\end{align*}
be the unital ring homomorphism sending each element $r\in R$ to the constant polynomial $r$. Let
\begin{align*}
\operatorname{ev}_0:R[t]\to R
\end{align*}
be the unital ring homomorphism sending a polynomial $f(t)\in R[t]$ to its value $f(0)\in R$. For every $r\in R$,
\begin{align*}
(\operatorname{ev}_0\circ \iota)(r)=r.
\end{align*}
Hence
\begin{align*}
\operatorname{ev}_0\circ \iota=\operatorname{id}_R.
\end{align*}
[guided]
The two ring maps in the argument are the standard maps relating $R$ and the [polynomial ring](/page/Polynomial%20Ring) $R[t]$. First, define
\begin{align*}
\iota:R\to R[t]
\end{align*}
by sending $r\in R$ to the constant polynomial with value $r$. This is a unital ring homomorphism because constant polynomials preserve addition, multiplication, and the unit.
Second, define
\begin{align*}
\operatorname{ev}_0:R[t]\to R
\end{align*}
by evaluation at $t=0$. Thus, if
\begin{align*}
f(t)=a_0+a_1t+\cdots+a_mt^m
\end{align*}
with coefficients $a_0,\dots,a_m\in R$, then
\begin{align*}
\operatorname{ev}_0(f(t))=a_0.
\end{align*}
This map is also a unital ring homomorphism because evaluating a polynomial at $0$ preserves sums, products, and the constant polynomial $1_R$.
Now compute the composite on an arbitrary element $r\in R$. The map $\iota$ sends $r$ to the constant polynomial $r$, and evaluating that constant polynomial at $0$ gives back $r$. Therefore
\begin{align*}
(\operatorname{ev}_0\circ \iota)(r)=r.
\end{align*}
Since this holds for every $r\in R$, the composite is the identity homomorphism:
\begin{align*}
\operatorname{ev}_0\circ \iota=\operatorname{id}_R.
\end{align*}
This is the algebraic splitting that will become a splitting on $K_0$ and $K_1$ after applying functoriality.
[/guided]
[/step]
[step:Pass the splitting to low algebraic K groups]
Fix $i\in\{0,1\}$. By [citetheorem:8627], the unital ring homomorphisms $\iota$ and $\operatorname{ev}_0$ induce group homomorphisms
\begin{align*}
K_i(\iota):K_i(R)\to K_i(R[t])
\end{align*}
and
\begin{align*}
K_i(\operatorname{ev}_0):K_i(R[t])\to K_i(R).
\end{align*}
The same functoriality theorem gives compatibility with composition and identity homomorphisms. Applying it to
\begin{align*}
\operatorname{ev}_0\circ \iota=\operatorname{id}_R
\end{align*}
yields
\begin{align*}
K_i(\operatorname{ev}_0)\circ K_i(\iota)=K_i(\operatorname{id}_R)=\operatorname{id}_{K_i(R)}.
\end{align*}
Thus $K_i(\iota)$ is a section of $K_i(\operatorname{ev}_0)$.
[/step]
[step:Use homotopy invariance to make the section an isomorphism]
Since $R$ is a commutative unital regular Noetherian ring, [citetheorem:8673] applies to the inclusion
\begin{align*}
\iota:R\to R[t].
\end{align*}
It gives that, for $i=0$ and $i=1$, the induced homomorphism
\begin{align*}
K_i(\iota):K_i(R)\to K_i(R[t])
\end{align*}
is an isomorphism.
[/step]
[step:Conclude that the nil kernels vanish]
Fix $i\in\{0,1\}$. We have shown that
\begin{align*}
K_i(\operatorname{ev}_0)\circ K_i(\iota)=\operatorname{id}_{K_i(R)}
\end{align*}
and that $K_i(\iota)$ is an isomorphism. Therefore $K_i(\operatorname{ev}_0)$ is the inverse homomorphism to $K_i(\iota)$, so $K_i(\operatorname{ev}_0)$ is injective. By the defining convention for the Bass nil group,
\begin{align*}
NK_i(R)=\ker\bigl(K_i(\operatorname{ev}_0):K_i(R[t])\to K_i(R)\bigr).
\end{align*}
Since $K_i(\operatorname{ev}_0)$ is injective, its kernel is the zero subgroup. Hence
\begin{align*}
NK_i(R)=0.
\end{align*}
Taking $i=0$ and $i=1$ gives
\begin{align*}
NK_0(R)=0
\end{align*}
and
\begin{align*}
NK_1(R)=0.
\end{align*}
[/step]
