I'm self-teaching myself some basic algebraic geometry, and I wanted to double check something that seems too easy.
An exercise sheet I found asks to compute the domain of definition of the rational function $f(x,y)=\frac{1-y}{x}$ on the circle $x^2+y^2=1$ in $\mathbb{A}^2$, the affine $2$-space over an algebraically closed field $k$.
I think the domain of definition is everywhere on the curve except when $x=0$, so the domain of definition would be $$ \{(x,y)\in\mathbb{A}^2:x^2+y^2=1\}\setminus\{(0,1),(0,-1)\}. $$
Does something more nuanced happen in $\mathbb{A}^2$, or is that all there is to it here? Thanks.
When restricted to $x^2 + y^2 = 1$, we have $f(x,y) = (1 - y)/x = ( 1 - y^2)/ x(1+y) = x^2/ x(1+y) = x/(1+y)$, and so this function is also defined at the point $(0,1)$.