[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*}[/step]

[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: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]