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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

Further I would like to understand the general approach for taking an algebraic curve in \(\displaystyle \mathbb{R}^2\) and finding the corresponding curve in \(\displaystyle \mathbb{P}^2 ( \mathbb{R} )\) ... ... BUT ... how is this done ...Hope 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:https://www.physicsforums.com/attachments/5739

https://www.physicsforums.com/attachments/5740

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I am currently focused on Chapter 8, Section 1: The Projective Plane ... ... and need help getting started with

*... Exercise 4 in Section 8.1 reads as follows:https://www.physicsforums.com/attachments/5738Can someone please help me with Exercise 4(a) ... ... indeed, what is actually involved in (rigorously) showing that the equation \(\displaystyle x^2 - y^2 = z^2\) is a well-defined curve in \(\displaystyle \mathbb{P}^2 ( \mathbb{R} )\) ... but I am very unsure of exactly how this works ... ... Presumably, what is involved is not only (rigorously) showing that the equation \(\displaystyle x^2 - y^2 = z^2\) is a well-defined curve in \(\displaystyle \mathbb{P}^2 ( \mathbb{R} )\) but showing that \(\displaystyle x^2 - y^2 = z^2\) is the representation in \(\displaystyle \mathbb{P}^2 ( \mathbb{R} )\) of the curve \(\displaystyle x^2 - y^2 = 1\) in \(\displaystyle \mathbb{R}^2\) ... ... ?Indeed whatever the meaning of the question, I would like to be able to show that the curve \(\displaystyle x^2 - y^2 = 1\) in \(\displaystyle \mathbb{R}^2\) becomes the curve \(\displaystyle x^2 - y^2 = z^2\) in \(\displaystyle \mathbb{P}^2 ( \mathbb{R} )\) ... indeed I think this is true ... BUT ... how do you rigorously show this ...***Exercise 4(a)**Further I would like to understand the general approach for taking an algebraic curve in \(\displaystyle \mathbb{R}^2\) and finding the corresponding curve in \(\displaystyle \mathbb{P}^2 ( \mathbb{R} )\) ... ... BUT ... how is this done ...Hope 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:https://www.physicsforums.com/attachments/5739

https://www.physicsforums.com/attachments/5740

View attachment 5741

View attachment 5742

View attachment 5743

View attachment 5744

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