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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.

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  • $\begingroup$ Your idea is exactly correct. $\endgroup$ – KReiser Nov 27 '14 at 2:39
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    $\begingroup$ @KReiser: Did you consider what happens when you multiply the top and bottom by $1+y$? $\endgroup$ – tracing Nov 27 '14 at 2:45
  • $\begingroup$ Hmm, good eye! I did not see that. $\endgroup$ – KReiser Nov 27 '14 at 6:06
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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 $(0,1)$.

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  • $\begingroup$ I would have said that $f$ can be continuously extended at $(0,1)$ by this fomula. Your last simplification works only when $x\neq 0$. $\endgroup$ – Taladris Nov 27 '14 at 2:50
  • $\begingroup$ @Taladris: The question is about the domain of definition of a rational function on a smooth curve, which (in the absence of any indication to the contrary) usually means the maximal domain of definition. $\endgroup$ – tracing Nov 27 '14 at 4:11
  • $\begingroup$ Thanks, I knew I probably made an oversight. $\endgroup$ – Ankita Desari Nov 27 '14 at 6:33

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