# when is rational function regular?

In general, how does one determine if a rational function is regular? I have the particular problem of determining in which points of the circle $V(x^2+y^2-1) \subseteq A^2$is the rational function $\alpha= \frac{y-1}{x}$ regular?

• For your example: no, because $(0, -1)$ is in the curve. Mar 26 '14 at 13:21

To say that $$\phi= \frac{y-1}{x}$$ is regular on the circle means that there exist polynomials $$p(X,Y), q(X,Y)\in k[X,Y]$$ such that: $$Y-1=X\cdot p(X,Y)+q(X,Y)\cdot(X^2+Y^2-1)\in k[X,Y]$$ But substituting $$X=0$$ in that equality yields $$Y-1=q(0,Y)\cdot(Y^2-1) \in k[Y]$$ which is impossible since the left hand side has degree $$1$$ whereas the right hand side is zero or has degree $$\geq2$$.
Hence the rational function $$\phi$$ is not regular.
Since $$\phi$$ is not regular but is clearly regular at all points of the circle different from $$P=(0,-1)$$, it follows that $$P$$ is the only point where $$\phi$$ is not regular, i.e. the only pole of $$\phi$$.
• This is incorrect, since $\frac{1-y}{x}=\frac{x}{1+y}$ which is defined for almost all but $(0,-1)$. May 15 '19 at 21:48
• The question is determine in which points is regular, not only if it is regular or not. In your proof you are missing $(0,1)$. May 15 '19 at 23:47
• @José Alejandro Aburto Araneda: I am missing nothing in my proof, which shows that $\phi$ is not regular, and does not state explicitly where $\phi$ is regular. I had not addressed this question in my proof, since I thought it was perfectly clear once we know that $\phi$ is not rational. I have written an edit making this explicit. May 16 '19 at 7:55