When $R $ is a cusp then $K_0(R) \ncong K_0(R[s])$

Question

Quillen's classical result shows that if $R$ is a regular ring then $K_0(R) \cong K_0(R[t_1,...,t_m])$ for all $m \in \mathbb{N}$. So I wanted to construct some elementary examples where $K_0(R)$ fails to be isomorphic to $K_0(R[s])$. Thus my interest in computing $K_0(k[t^2,t^3])$ and $K_0(k[t^2,t^3,s])$ to see if they are isomorphic or not.

For $K_0(k[t^2,t^3])$ I have used the fact that since it is Noetheiran of dimension $1$, it will be isomorphic to $\mathbb{Z} \oplus Pic(k[t^2,t^3])$. To compute the Picard group I took the help of the conductor ideals to show that it is $ k.$ But for $K_0(k[t^2,t^3,s])$ I can understand that it will be isomorphic to a quotient of $\mathbb{Z} \oplus Pic(k[t^2,t^3,s])$ but I am unable to compute it explicitly. I would be grateful for any help or suggestions in this direction. Thank you.

Answer 1

Presumably you are looking at the conductor square on the left below, where $\epsilon^2=0$:

$$\matrix{k[t^2,t^3]&\rightarrow& k[t]\cr \downarrow&&\downarrow\cr k&\rightarrow &k[\epsilon]\cr}\qquad \matrix{k[t^2,t^3,S]&\rightarrow& k[t,S]\cr \downarrow&&\downarrow\cr k[S]&\rightarrow &k[\epsilon,S]\cr}$$

The associated Mayer-Vietoris sequence gives an isomorphism $k^+\cong k[\epsilon]^*/k^*\cong Pic(k[t^2,t^3])$ where the upper-star denotes units.

The same calculation using the square on the right will show that $Pic(k[t^2,t^3,S])=k[\epsilon,S]^*/k^*$.

Now look at the unit $1+S\epsilon\in k[\epsilon,S]^*$. This does not come from $k[\epsilon]^*$, and it gives an element of $Pic(k[t^2,t^3,S])$ that does not come from $Pic(k[t^2,t^3])$.

Chasing around the conductor square, you can verify that the projective module you've constructed is generated over $k[t^2,t^3,S]$ by $1+tS$ and $(tS)^2$.