Give an example of a commutative ring with unity $R$, such that $K_0(R) \ncong K_0(R[t])$

Question

In Weibel's $K$-book I have read that it is Quillen's classical result that if $R$ is a regular Noetherian ring then $K_0(R) \cong K_0(R[t])$.

So out of curiosity I have tried and failed quite a lot of times to see what happens for non-regular rings. But I am unable to find such a ring for which $K_0(R) \ncong K_0(R[t]$. I have tried with the coordinate rings of node and cusp with the help of Milnor's patching data but have failed to compute the explicit $K_0$ groups.

Any hints or suggestions are highly appreciated. Thank you.

I assume $K_0$ means $K$- group of projective modules. Then, the cusp should you such an example. If $R=k[[x^2,x^3]]$, $K_0(R)=\mathbb{Z}$, but $K_0(R[t])$ is $\mathbb{Z}\oplus k$. — Mohan, Sep 10 '23 at 20:33
I think $K_0(R) = \mathbb{Z} \oplus k $, since it is Noetherian of dimension 1 — Divya, Sep 11 '23 at 01:59
Where is the $k$ summand coming from in $K_0(R)$? The ring $R$ is local, so every projective module over it is free — Brendan Murphy, Sep 11 '23 at 02:17
I understand that since $R= k[[x^2,x^3]]$is local $K_0(R) = \mathbb{Z}$, but $\mathbb{Z} \oplus k$ for $K_0(R[t])$ I am unable to compute. — Divya, Sep 11 '23 at 05:09
Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see Guidelines for good use of $\rm\LaTeX$ in question titles. — soupless, Sep 12 '23 at 14:54
Thank you, I have edited as per the guidance, and will keep it in mind. — Divya, Sep 12 '23 at 15:05

Divya · Accepted Answer · 2023-09-12T13:38:42.797

I will try to write an answer to my own question here. Please correct me if I have made a mistake anywhere.

Let $R = k[[x^2,x^3]]$ and $S = k[[x^2,x^3]][y]$.

Since $R$ is a local ring $K_0(R) \cong \mathbb{Z}$. Now for a commutative ring with unity $R'$ there is always a natural surjection $$\operatorname{rank} \oplus \det: K_0(R') \rightarrow [\operatorname{Spec}(R'),\mathbb{Z}] \oplus \operatorname{Pic}(R'),$$ where $[\operatorname{Spec}(R),\mathbb{Z}]$ is the ring of all continuous functions from $\operatorname{Spec}(R)$ to $\mathbb{Z}$.

I will first compute $\operatorname{Pic}(S)$ by constructing the following Milnor square $$\matrix{k[[x^2,x^3]][s]&\rightarrow& k[[x]][s]\cr \downarrow&&\downarrow\cr k[s]&\rightarrow &k[[x]][s]/(x^2)[s]\cr}$$ Now because of the Unit Pic sequence, and the fact that Picard group over a UFD is trivial, we conclude that $\operatorname{Pic}(S) = (k[[x]][y]/(x^2)[y])^{\times}/k^{\times}$ (unit group). Which is nontrivial. Thus $K_0(S)/I \cong \mathbb{Z} \oplus\operatorname{ Pic}(S)$. Now if $K_0(S) \cong \mathbb{Z}$ then we have a contradiction.

Give an example of a commutative ring with unity $R$, such that $K_0(R) \ncong K_0(R[t])$

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