Projective Plane ... Cox et al - Section 8.1, Exs 5(a) & 5(b)

[Math Amateur]

Homework Statement

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 Exercises 5(a) and 5(b) ... ...Exercise 5 in Section 8.1 reads as follows:

Can someone please help me to get started on Exercises 5(a) and 5(b) shown above ...

2. Homework Equations

The definitions 1, 2 and 3 in Cox et al Section 8.1 may be relevant ... see the text provided below ...
...

The Attempt at a Solution

[/B]
I am very unsure how to start on this exercise ... but i suspect that the following transformation or map as given in Cox et al Section 8.1 directly after Definition 3 is involved:

##\mathbb{R}^2 \longrightarrow \mathbb{P}^2 ( \mathbb{R} ) ##

where ##(x, y) \in \mathbb{R}^2## is sent to the point ##p \in \mathbb{P}^2 ( \mathbb{R} )## whose homogeneous coordinates are ##(x, y, 1)##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:

 

[andrewkirk]

I can have a go at (a).
We want the equation to be compatible with the equation ##y=x^2## and we also want it to give a well-defined curve, which means it must be homogeneous in x,y and z.

A simple equation that satisfies both those is ##yz=x^2##. Then for ##z=1## this gives the original equation. Any point in ##\mathbb R^2## with nonzero ##z## is the same as a point with ##z=1##. The only other points are those with ##z=0##, which are at infinity. For such points we will also have, courtesy of the equation, ##x=0##. So the set of points on the curve at infinity are those on the y-axis in ##\mathbb R^2##. This comprises two equivalence classes: [(0,0,0)] and [(0,1,0)]. So there are two points at infinity, which sounds like what we would want for a parabola (which answers part (b)).

So far so good. I haven't thought about c or d yet.

 

[Math Amateur]

I can have a go at (a).
We want the equation to be compatible with the equation ##y=x^2## and we also want it to give a well-defined curve, which means it must be homogeneous in x,y and z.

A simple equation that satisfies both those is ##yz=x^2##. Then for ##z=1## this gives the original equation. Any point in ##\mathbb R^2## with nonzero ##z## is the same as a point with ##z=1##. The only other points are those with ##z=0##, which are at infinity. For such points we will also have, courtesy of the equation, ##x=0##. So the set of points on the curve at infinity are those on the y-axis in ##\mathbb R^2##. This comprises two equivalence classes: [(0,0,0)] and [(0,1,0)]. So there are two points at infinity, which sounds like what we would want for a parabola (which answers part (b)).

So far so good. I haven't thought about c or d yet.

Thanks for the help Andrew ... but ... your solution seems to have been achieved with some good insight ...

Is there a method or process for transforming curves in ##\mathbb{R}^2## to curves in ##\mathbb{P}^2 ( \mathbb{R} )##?

Thanks again ...

Peter

P.S. if you have any ideas about parts (c) and (d) please let me know ...

 

[andrewkirk]

I don't know of any general method but, for curves whose equations are polynomial in x and y, the method is easy enough. You just write out the equation and find the degree of each term, which is the exponent for x plus the exponent for y. Let M be the highest such degree in the equation. Then, to make the equation homogeneous, we multiply each term by ##z^{M-d}## where ##d## is the degree of the term. We then have an equation that is homogeneous - hence making a well-defined curve in the projective plane - and which also involves ##z##.

There are a couple of examples of this technique in the author's discussion following Proposition 4.

(d) is easy. Just set ##x## equal to any nonzero constant, such as 1. The equation then becomes ##yz=1##, which is the classic hyperbola equation.

To do (c) we'd need to understand what the author means by 'tangent to' in the context of a projective plane. Do you know what he means by that?

 

[Math Amateur]

I don't know of any general method but, for curves whose equations are polynomial in x and y, the method is easy enough. You just write out the equation and find the degree of each term, which is the exponent for x plus the exponent for y. Let M be the highest such degree in the equation. Then, to make the equation homogeneous, we multiply each term by ##z^{M-d}## where ##d## is the degree of the term. We then have an equation that is homogeneous - hence making a well-defined curve in the projective plane - and which also involves ##z##.

There are a couple of examples of this technique in the author's discussion following Proposition 4.

(d) is easy. Just set ##x## equal to any nonzero constant, such as 1. The equation then becomes ##yz=1##, which is the classic hyperbola equation.

To do (c) we'd need to understand what the author means by 'tangent to' in the context of a projective plane. Do you know what he means by that?

Thanks Andrew ...

By the way, your approach seems to be a general method to extend algebraic curves in the Euclidean plane to the Projective plane ... the book: "Conics and Cubics: A Concrete Introduction to Algebraic Curves"by Robert Bix outlines the method as follows: