Intersection points of curves

Question

In my lecture notes there is the following example for intersection points of curves:

$$F(x, y, z)=xz^3-y^4 \\ G(x, y, z)=xz^2-y^3$$ in $\mathbb{P}^2(\mathbb{C})$, where $\mathbb{P}^2(\mathbb{C})=U_2 \cup H$

where $U_2=\{[x, y, 1] | x, y \in \mathbb{C}\}, H=V(z)=\{[x, y, 0] | x, y \in \mathbb{C}\}$

The intersection points in $U_2$: We set $z=1$

$$x-y^4=0 \\ x-y^3=0$$

$$\Rightarrow y^4-y^3=0 \Rightarrow y^3(y-1)=0 \\ \Rightarrow y=0 \text{ or } y=1$$

For $y=0 \Rightarrow x=0$

For $y=1 \Rightarrow x=1$

So the intersection points are, $$P_0=[0, 0, 1] , P_1=[1, 1, 1]$$

$$P_0 \to P_0'=(0, 0) \\ P_1 \to P_1'=(1, 1)$$

The possible intersection points at infitity.

For $z=0$

$$\Rightarrow y=0$$

$[x, 0, 0], x \neq 0 \\ =[1, 0, 0]$

We homogenize with respect to the variable $x$.

We set $x=1$

the system $$z^3-y^4=0 \\ z^2-y^3=0$$

for $y=0, z=0$ $$\to P_3=(0,0)$$

If $y \neq 0 \Rightarrow z \neq 0$

$z=y$

$[1, y, y]$

$$\to P_4=(1, 1)$$

$$$$

Can you explain to me this example?

1. What does this mean:

   $$P_0 \to P_0'=(0, 0) \\ P_1 \to P_1'=(1, 1)$$

   The possible intersection points at infitity.

   ?

2. What does "We homogenize with respect to the variable $x$." mean? Why do we have to do that?

Edit:

• So one way to find the intersection points is:

  The intersection points in $U_2$: We set $z=1$

  $$x-y^4=0 \\ x-y^3=0$$

  $$\Rightarrow y^4-y^3=0 \Rightarrow y^3(y-1)=0 \\ \Rightarrow y=0 \text{ or } y=1$$

  For $y=0 \Rightarrow x=0$

  For $y=1 \Rightarrow x=1$

  So the intersection points are, $$P_0=[0, 0, 1] , P_1=[1, 1, 1]$$

  $$P_0 \to P_0'=(0, 0) \\ P_1 \to P_1'=(1, 1)$$

  The possible intersection points at infitity.

  For $z=0$

  $$\Rightarrow y=0$$

  $[x, 0, 0], x \neq 0 \\ =[1, 0, 0]$

  $$$$

  and the other one is the dehomogenization??? So these are two different ways???

$$$$

• To find the intersection points do we set the two functions to zero because the curve consists of the point $P$ such that $h(P)=0$ ???

$$$$

• Can we write $$P_0 \to P_0'=(0, 0) \\ P_1 \to P_1'=(1, 1)$$ because in $U_2$ $z$ is always equal to $1$ ???

$$$$

• $P_0=[0, 0, 0]$ is at the projective plane, right? Why do we write $P_0'=(0, 0)$ ? Isn't this at the projective plane? Or why do we use at $P_0'$ parenthesis $( )$ and not $[ ]$ ?

$$$$

• The points at infinity have always $z=0$, correct???

Answer 1

1. The process of setting one of the homogeneous coordinates equal to 1 is called dehomogenization and not homogenization.

2. One could stop arguing after finding out x=[1,0,0]. Just plug in the coordinates and check that x is indeed a solution and you are done. The remaining part is superfluos: the point $P_3$ in homogenous coordinates is [1,0,0] -- we already found that one before. The point $P_4$ is [1,1,1], again found before.

3. In general one can try to find the solutions of a homogenous system of polynomial equations without dehomogenization. This can work but often it is easier to dehomogenize, because that way one gets rid of some monomials -- preferably some of high degree.