New answers tagged picard-group
4
votes
Accepted
$K_0((k[x]/(x^2))[y])$
Let $P$ be a finitely generated projective module over your ring $S=R[y]=k[x,y]/y^2$. Then $P/yP$ is a finitely generated projective module over $k[x]$, and so is free by the Quillen-Suslin theorem. ...
- 53.9k
9
votes
Accepted
When $R $ is a cusp then $K_0(R) \ncong K_0(R[s])$
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[...
- 22.2k
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