For a simple, really concrete example you can also look at:
$A=k[x,y,u,v]/(xy+ux^2+vy^2)$, $X =Spec(A)$, $I=(x,y)$, $U = D(I)$.
Then the functions $f=frac{-v}{x}=frac{y+ux}{y^2}$ and $g=frac{-u}{y}=frac{x+vy}{x^2}$ are defined on $U$. But $yf+xg=1$, so $U$ is affine!
Cheers,
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