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 $f=-V/X=(Y+UX)/Y^2$ and $g=-U/Y=(X+VY)/X^2$ are defined on $U$. But $Yf+Xg=1$, so $U$ is affine!

Cheers,

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,

For a simple, really concrete example you can also looks at:

$A=k[X,Y,U,V]/(XY+UX^2+VY^2)$, $X =Spec(A)$, $I=(X,Y)$, $U = D(I)$.

Then the functions $f=-V/X=(Y+UX)/Y^2$ and $g=-U/Y=(X+VY)/X^2$ are defined on $U$. But $Yf+Xg=1$, so $U$ is affine!

Cheers,

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 $f=-V/X=(Y+UX)/Y^2$ and $g=-U/Y=(X+VY)/X^2$ are defined on $U$. But $Yf+Xg=1$, so $U$ is affine!

Cheers,