Help with Exercise 3(c) in Cox et al's Projective Algebraic Geometry

[Math Amateur]

I am reading the undergraduate introduction to algebraic geometry entitled "Ideals, Varieties and Algorithms: An introduction to Computational Algebraic Geometry and Commutative Algebra (Third Edition) by David Cox, John Little and Donal O'Shea ... ...

I am currently focused on Chapter 8, Section 1: The Projective Plane ... ... and need help getting started with Exercise 3(c) ... Exercise 3 in Section 8.1 reads as follows:
View attachment 5726
I would very much appreciate someone helping me to start Exercise 3(c) ... ... My thoughts so far are as follows ... I would be grateful if someone could critique my analysis ...
I am assuming that a point $$(x, y)$$ in $$\mathbb{R}^2$$ can be written as $$(x, y, 1)$$ in $$\mathbb{P}^2 ( \mathbb{R} )$$ ... as homogeneous coordinates ...So $$( \frac{1+t^2}{1 - t^2} , \frac{2t}{1 - t^2} )$$ in $$\mathbb{R}^2$$ becomes $$( \frac{1+t^2}{1 - t^2} , \frac{2t}{1 - t^2} , 1 )$$ in $$\mathbb{P}^2 ( \mathbb{R} )$$Now we can multiply or divide through homogeneous coordinates by a constant without altering the coordinate ... BUT ... what gives us the right to multiply by the variable $$(1-t^2 )$$ ... ... in order to get $$( \frac{1+t^2}{1 - t^2} , \frac{2t}{1 - t^2} , 1 ) = (1 + t^2, 2t, 1 - t^2 )$$Then ... I assume that $$t = \pm 1$$ ... since it leads to $$z = 0$$ in the homogeneous coordinates ... gives points at infinity ... BUT ... is that correct? ... AND ... is there any more that can be deduced ... ... ?hope that someone can help ...

Peter======================================================================To give readers of the above post some idea of the context of the exercise and also the notation I am providing some relevant text from Cox et al ... ... as follows:
View attachment 5727
View attachment 5728
View attachment 5729
View attachment 5730
View attachment 5731
View attachment 5732

 

[GJA]

Hi Peter,

This is correct

My thoughts so far are as follows ... I would be grateful if someone could critique my analysis ...
I am assuming that a point $$(x, y)$$ in $$\mathbb{R}^2$$ can be written as $$(x, y, 1)$$ in $$\mathbb{P}^2 ( \mathbb{R} )$$ ... as homogeneous coordinates ...So $$( \frac{1+t^2}{1 - t^2} , \frac{2t}{1 - t^2} )$$ in $$\mathbb{R}^2$$ becomes $$( \frac{1+t^2}{1 - t^2} , \frac{2t}{1 - t^2} , 1 )$$ in $$\mathbb{P}^2 ( \mathbb{R} )$$

The next question is a good one and highlights what can certainly be a confusing element of dealing with projective space

Now we can multiply or divide through homogeneous coordinates by a constant without altering the coordinate ... BUT ... what gives us the right to multiply by the variable $$(1-t^2 )$$ ... ... in order to get $$( \frac{1+t^2}{1 - t^2} , \frac{2t}{1 - t^2} , 1 ) = (1 + t^2, 2t, 1 - t^2 )$$

The rule is that you're allowed to multiply by any non-zero number. As the authors say, two points $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ are equivalent provided there is some non-zero $\lambda$ such that $(x_{1},y_{1},z_{1})=\lambda (x_{2},y_{2},z_{2})$; i.e. $\lambda$ is a variable that can take on any non-zero value and, in doing so, completely assembles all of the elements in the equivalence class of the point $(x_{1},y_{1},z_{1})$. So, as long as $t\neq\pm 1,$ $(1-t^2)\neq 0$ and so can be used as a "lambda" value.

 

[Math Amateur]

Hi Peter,

This is correct

The next question is a good one and highlights what can certainly be a confusing element of dealing with projective space
The rule is that you're allowed to multiply by any non-zero number. As the authors say, two points $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ are equivalent provided there is some non-zero $\lambda$ such that $(x_{1},y_{1},z_{1})=\lambda (x_{2},y_{2},z_{2})$; i.e. $\lambda$ is a variable that can take on any non-zero value and, in doing so, completely assembles all of the elements in the equivalence class of the point $(x_{1},y_{1},z_{1})$. So, as long as $t\neq\pm 1,$ $(1-t^2)\neq 0$ and so can be used as a "lambda" value.

THanks GJA ... appreciate the help ...

Peter