I'm doing the exercises from Eisenbud's Commutative Algebra with a view toward Algebraic Geometry, and I don't understand part of one of them, ex. 1.11 a):

Exercise 1.11 a: Over $\mathbb{C}$, consider a circle and a parabola represented by $\mathbb{C}[x,y]/(x^2+y^2-1) \cap (y-2-x^2)$ [...]. Show that the "projection from the north pole" gives a bijection (given by rational functions) between the circle minus one point and the line minus two points.

The only line appearing in the exercise is the line $x=0$, but since we can't project the circle from the north pole onto the line, let's assume that the line is $y=0$. We know that the stereographic projection gives a bijection between the circle minus one point and the whole line, so this is not the projection that we are looking for.

Taking a look over the Hints and solutions, we have that the projection is precisely $t \mapsto (4t/(t^2+4), (t^2-4)/(t^2+4))$, which satisfies that it is a bijection between the circle minus $(0,1)$ and $\mathbb{C} \setminus \{2i, -2i\}$.

My questions are the following: where does this map come from? Why is it a stereographic projection?


  • $\begingroup$ Oddly enough, its image is not even the zero set of $x^2+y^2-1$. $\endgroup$ – Karl Kronenfeld Feb 26 '14 at 11:32
  • $\begingroup$ The image of the map is on the zero set of $x^2+y^2-1$: $(4t)^2 + (t^2-4)^2= 16t^2 + t^4 - 8t^2 + 16 = (t^2+4)^2 \Rightarrow $ $(4t/(t^2+4))^2 + ((t^2-4)/(t^2+4))^2 = 1$. $\endgroup$ – Pedro A. Castillejo Feb 26 '14 at 13:07
  • $\begingroup$ Well, that's embarrassing. lol. I guess I need to practice multiplication, or something. $\endgroup$ – Karl Kronenfeld Feb 26 '14 at 13:19

Let $Z=\{(z,w)\in\mathbb C^2\colon z^2+w^2=1\}\setminus\{(0,1)\}$. The idea is that you take the stereographic projection in $\mathbb R^2$ and simply allow the parameter to be complex. Recall--or compute--the stereographic projection in $\mathbb R^2$ can be stated as a bijection of the line $y=-1$ onto $Z\cap \mathbb R^2$, given by $$f:(t,-1)\mapsto\left(\frac{4t}{t^2+4},\frac{t^2-4}{t^2+4}\right).$$ If we allow $t$ to vary in $\mathbb C\setminus\{\pm 2i\}$ the hope is that it is a bijection onto $Z$.

To construct an inverse, use the other direction of the stereographic projection map in $\mathbb R^2$, which one can calculate as being given by $$g:(z,w)\mapsto 2z/(1-w),$$ and extend to the rest of $Z$.

The maps $f:\mathbb C\setminus\{\pm 2i\}\to \mathbb C^2$ and $g:Z\to\mathbb C$ are inverses where ever it makes sense to say that. We just have to show that the image of $f$ is in the domain of $g$ and vice versa. It is easy to show directly that $f$ maps into $Z$. Also, if $z/(1-w)=i$, where $(z,w)\in Z$, then $z^2=-(1-w)^2$, so $2w=2$, a contradiction. Similarly, $-2i\not\in g(Z)$. Therefore, $g=f^{-1}$ as desired.

  • $\begingroup$ Thanks! As a remark, we have to consider $Z \setminus (0,1)$, not $Z$, and the mapping of Eisenbud projects the circle (of radius 1) minus the north pole to the tangent line passing through the south pole. Your projection maps to the line passing through the center of the circle, that's why the numbers are different, but the idea is exactly the same. $\endgroup$ – Pedro A. Castillejo Feb 26 '14 at 13:14
  • $\begingroup$ @PedroA.Castillejo Thanks, I am a little error-prone right now, I guess. The reason I even used to the notation Z was to not have to write "besides (0,1)" over and over again, but gah I forgot to remove the point in the definition. :P Oh, and thanks for the explanation behind Eisenbud's choice. I didn't figure that one out. $\endgroup$ – Karl Kronenfeld Feb 26 '14 at 13:28
  • $\begingroup$ Having gotten some sleep, I edited the answer to make it agree with Eisenbud's choice. For those who are interested in the effects of sleep-deprivation on the brain, the first revision and comments may be of interest. :D $\endgroup$ – Karl Kronenfeld Feb 28 '14 at 8:13

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