I'm reading Garrity's Algebraic Geometry: A Problem Solving Approach, and am confused about a couple of things regarding polynomials in $\mathbb{P}^2$.
Definition 1.1.4. Let $P(x,y)$ be a degree $n$ polynomial defined over $\mathbb{C}^2$. The corresponding homogeneous polynomial defined over $\mathbb{P}^2$ is $P(x,y,z)=z^nP\left(\frac{x}{z},\frac{y}{z}\right)$.
I don't understand what the domain of these type of polynomial is. Points in $\mathbb{P}^2$ have so far been written $(x:y:z)$. Do polynomials defined over $\mathbb{P}^2$ actually have their domains in $\mathbb{C}^3$? Are their codomains $\mathbb{C}$? This confusion bleeds over into the following exercise.
Exercise 1.4.20. Show that in $\mathbb{P}^2$, any two distinct lines will intersect in a point. Notice this implies that parallel lines in $\mathbb{C}^2$, when embedded in $\mathbb{P}^2$, will intersect at the line at infinity.
I guess two distinct lines in $\mathbb{P}^2$ would be zero sets of two distinct polynomials $L_1 = a_1x+b_1y+c_1z+d_1$ and $L_2 = a_2x+b_2y+c_2z+d_2$ in $\mathbb{C}^3$? If that's correct, it seems that each line in $\mathbb{P}^2$ corresponds to a plane in $\mathbb{C}^3$, so if the corresponding planes in $\mathbb{C}^3$ intersect in a line, then the lines in $\mathbb{P}^2$ should intersect in the point corresponding to that line. Otherwise, the two planes are parallel so there exists some constant $\lambda\in\mathbb{C}$ such that $\lambda=\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ (and $d_1\neq\lambda d_2$). Then I suppose I want to show that in this case $z=0$, but I have no idea how.
The polynomial $x$ takes multiple different values on the point $[1:0:0]=[2:0:0]$, so you can't really call it a function with domain $\Bbb P^2$. What's true instead is that given a homogeneous polynomial $f$, you can distinguish the points in $\Bbb P^n$ where it's zero from where it's nonzero: $f(\lambda x_0,\cdots,\lambda x_n)=\lambda^d f(x_0,\cdots,x_n)$, so any two representatives of a point in $\Bbb P^n$ both either evaluate to $0$ or don't.
For the exercise, lines in $\Bbb P^2$ are the zero sets of homogeneous polynomials $ax+by+cz=0$, and if $a$ or $b$ is nonzero, then they're the homogenization of the equation $ax+by+c=0$ for a line in $\Bbb A^2$. Two lines $V(ax+by+c)$ and $V(a'x+b'y+c')$ in $\Bbb A^2$ are parallel iff $(a,b)$ is a scalar multiple of $(a',b')$ - do you see how to find a common solution of the homogenizations?