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?

  • $\begingroup$ For your example: no, because $(0, -1)$ is in the curve. $\endgroup$
    – Zhen Lin
    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 $\phi$.

  • 5
    $\begingroup$ ...and this also shows that algebraic geometry is often far less frightful than supposed :-) $\endgroup$ Mar 26 '14 at 21:21
  • $\begingroup$ This is incorrect, since $\frac{1-y}{x}=\frac{x}{1+y}$ which is defined for almost all but $(0,-1)$. $\endgroup$ May 15 '19 at 21:48
  • $\begingroup$ @José Alejandro Aburto Araneda. What I wrote is perfectly correct, despite your irrelevant comment: a regular function must be defined at all points, and that a rational function is defined "for almost all" points (as you write) certainly does not make it regular. Actually all rational functions are defined at almost all points, but most of these rational functions are not regular! $\endgroup$ May 15 '19 at 22:20
  • $\begingroup$ The question is determine in which points is regular, not only if it is regular or not. In your proof you are missing $(0,1)$. $\endgroup$ May 15 '19 at 23:47
  • $\begingroup$ @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. $\endgroup$ May 16 '19 at 7:55

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