# Domain of definition of $(1-y)/x$ on $x^2+y^2=1$?

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.

• Your idea is exactly correct. – KReiser Nov 27 '14 at 2:39
• @KReiser: Did you consider what happens when you multiply the top and bottom by $1+y$? – tracing Nov 27 '14 at 2:45
• Hmm, good eye! I did not see that. – KReiser Nov 27 '14 at 6:06

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 $(0,1)$.
• I would have said that $f$ can be continuously extended at $(0,1)$ by this fomula. Your last simplification works only when $x\neq 0$. – Taladris Nov 27 '14 at 2:50