("Algebraic Geometry: A Problem Solving Approach" by Thomas Garrity)
I am struggling with Exercise 1.4.12.1 in the above, which I quote with some context:
Here is my intuitive thinking:
(a) lines in $\mathbb{C}^3$ with some elevation from the x-y plane (ie where $z \neq 0$) are reduced to a single point, where that point is the intersection of the line with the plane $z =1$. This point is unique (which is why $\phi$ in 1.4.9 is a bijection), and it is labelled $(x:y:1)$
(b) the lines in the x-y plane in 1.4.11/1.4.12 do not hit $z=1$. Those lines are of the form $( x,\frac{-ax-c}{b},0 )$ , or alternatively $( bx,-ax-c,0 )$
But then :
I do not understand why under $\phi$ these lines are assigned the point stated in 1.4.12.1
even so, why are such points necessarily distinct from those assigned to lines for the prior case, where $z \neq 0$ ?
EDIT #1: I add the rest of the exercise (ie 1.14.12.3) to show the conclusion reached by the author. Personally, I feel able to reach that conclusion directly from my (b) above:
The map $\phi: \mathbb{C}^2 \rightarrow \mathbb{P}^2 \setminus \{z = 0\}$ sends $(x,y)$ to $[x:y:1]$.
Your line $\ell$ is defined as $y = (-ax-c)/b$, and therefore
$$ \phi((x, (-ax-c)/b)) = [x: (-ax-c)/b: 1] = [bx : -ax-c: b].$$
In your intuitive thinking, you should think on the affine plane as having $z$ coordinate equal to 1. In other words, in $\mathbb{C}^3$, you are in the plane $z=1$!
An affine line in $\mathbb{C}^2$ therefore corresponds to a plane in $\mathbb{C}^3$: each point $(x_0, y_0)$ of the line in $\mathbb{C}^2$ corresponds in $\mathbb{C}^3$ to the line passing through the origin and the point $(x_0, y_0, 1)$, and putting together all the points in $\mathbb{C}^2$ (resp. lines in $\mathbb{C}^3$) you get a line (resp. a plane). So the affine line in $\mathbb{C}^2$ not only intersects the plane $z=1$, but is cointaned in it!