Lines in the projective plane

Question

In my lecture notes we have the following:

The set $$\mathbb{P}^2(K)=\{[x, y, z] | (x, y, z) \in (K^3)^{\star}\}$$ is called projective plane over $K$.

There are the following cases:

• $z \neq 0$ $$\left [x, y, z\right ]=\left [\frac{x}{z}, \frac{y}{z}, 1\right ]$$ $$\mathbb{P}^2(K) \ni \left [x, y, z\right ] \to \left (\frac{x}{z}, \frac{y}{z}\right ) \in A^2(K)$$ These points are the finite points of $\mathbb{P}^2(K)$.

• $z=0$

  The points $\left [x, y, 0\right ]$ are called points at infinity of $\mathbb{P}^2(K)$.

The finite points $\left [x, y, 1\right ] \in \mathbb{P}^2(K)$ geometrically correspond to the intersection points of the lines from $(0,0,0)$ with the plane $z=1$.

The points at infinity $\left [x, y, 0\right ]$ geometrically correspond to the lines of the plane $x0y$ that pass through $(0, 0, 0)$.

We define as line in $\mathbb{P}^2(K)$ each equation of the form $$ax+by+cz=0, (a,b,c) \neq (0,0,0)$$ that means the set $$E=\{\left [x, y, z\right ] \in \mathbb{P}^2(K) | ax+by+cz=0, (a, b, c) \neq (0,0,0) \}$$

If $z \neq 0$ : $\left [x, y, z\right ]=\left [x, y, 1\right ]$, that means that the equation is written $$ax+by+c=0$$ (and if $ab \neq 0$ then it is the known affine line)

$$$$

Can you explain to me what $$\mathbb{P}^2(K) \ni \left [x, y, z\right ] \to \left (\frac{x}{z}, \frac{y}{z}\right ) \in A^2(K)$$ means?

Also, at the end, why does it stand that $\left [x, y, z\right ]=\left [x, y, 1\right ]$ ? At the beginning didn't we have that $\left [x, y, z\right ]=\left [\frac{x}{z}, \frac{y}{z}, 1\right ]$ ?

Answer 1

Hint: Consider the set $U_2:=\{[x,y,z] \in \mathbb P^2(K):z\neq 0 \}.$ Now define a map $\phi:U_2 \to \mathbb A^2(K), [x,y,z] \mapsto \left(\frac{x}{z}, \frac{y}{z}\right).$ Prove that this is a bijection.

Answer 2

One way to form projective space $\mathbb{P}^2(K)$ is to start with $\mathbb{A}^3(K)\setminus\{0\}$ (i.e. this is just $K^3\setminus \{0\}$, so think of $\mathbb{R}^3\setminus \{0\}$ or $\mathbb{C}^3\setminus \{0\}$ if you like), and define the following equivalence relation: $x\sim y$ iff there is a $c\in K$ where $x=cy$, that is, two points are equivalent if they are on the same line through the origin. This means all points on the same line through the origin are identified as one point in projective space. This is what you mean when you say "geometrically corresponds" above. If you want a line in projective space, this corresponds to a plane in affine space before taking the quotient by the equivalence relation. A point in projective space corresponds to a line in affine space. The general equation for a plane in 3-space passing through the origin is $ax+by+cz=0$ as you wrote.

Now as nice as projective space is, we would like to work back in affine space sometimes, so we observe that we can cover $\mathbb{P}^2(K)$ with three (standard) affine charts. Since $x,y,z$ cannot all be 0 (remember we removed the origin from our computations above), then we can consider three cases: $x\neq 0$, $y\neq 0$ and $z\neq 0$. In the case that $z\neq 0$ as you have above, we can observe that $[x,y,z] = [\frac{x}{z},\frac{y}{z},1]$ since $(x,y,z) = z(\frac{x}{z},\frac{y}{z},1)$ in $\mathbb{A}^3(K)$ and so are equivalent in projective space. Now no matter what $z\neq 0 $ is, I can still make $\frac{x}{z}$ and $\frac{y}{z}$ into any element of $K$ by choosing appropriate $x$ and $y$. I may as well write $u=\frac{x}{z}$ and $v= \frac{y}{z}$. Then I have the map $A_z= \{[x,y,z]\in \mathbb{P}^2(K)|z\neq 0\} \rightarrow \mathbb{A}^2(K)$ via $[x,y,z] = [\frac{x}{z},\frac{y}{z},1] = [u,v,1] \rightarrow (u,v)$ which is a bijection.

You can do the same for $x\neq 0$ and $y\neq 0$ and get two other charts $A_x$ and $A_y$ and together these three charts cover $\mathbb{P}^2(K)$. The point of the map is to show each chart is really a copy of affine space. If you are in the $x$-chart, then the points $\{[0,y,z]\in \mathbb{P}^2(K)$ are your "points at infinity". In other words your points at infinity and your finite points as you put it are a matter of perspective.

Of course this process generalizes naturally to allow us to construct $\mathbb{P}^n(K)$ and to cover it with $n+1$ standard affine charts.