NK groups

Remarks

The inclusion $R \to R[t]$ followed by $\varepsilon$ is the identity of $R$. So $K_n(R)$ is a direct summand of $K_n(R[t])$, and $$K_n(R[t]) \cong K_n(R) \oplus NK_n(R).$$ The construction is a functor of $R$.

Example: let $k$ be a field and $R = k[\varepsilon]/(\varepsilon^2)$. The element $u = 1 + \varepsilon t$ is a unit of the commutative ring $R[t]$, with inverse $1 - \varepsilon t$. The determinant $K_1(R[t]) \to R[t]^\times$ sends the class of $u$ to $u \ne 1$, so the class is not zero. Since $\varepsilon_*(u) = 1$, the class of $u$ is a nonzero element of $NK_1(R)$.

Non-example: for a regular Noetherian ring $R$, such as a field or $\Z$, the map $\varepsilon_*$ is an isomorphism, so $NK_n(R) = 0$ for all $n$. Then $K_n(R[t])$ has no part outside $K_n(R)$.

Some authors write $N K_n(R)$ as the cokernel of $K_n(R) \to K_n(R[t])$; it is isomorphic to the kernel above by the splitting.