Questions about algebraic curves and homogeneous polynomial equations

[Bobby Lee]

TL;DR

Here, I present a few questions on the concept of homogeneous polynomial equations and their connection with algebraic curves. Since the latter is a topic of great interest, I think these questions as well as their answers might be of interest to those who subscribe to PhysicsForums.

It is generally well-known that a plane algebraic curve is a curve in $\mathcal{CP}^{2}$ given by a homogeneous polynomial equation $f(x,y)= \sum^{N}_{i+j=0}a_{i\,j}x^{i}y^{j}=0$, where $i$ and $j$ are nonnegative integers and not all coefficients $a_{ij}$ are zero~[1].
In addition, if $f(x,y)$ is a homogeneous polynomial of degree $d$, then [2]
$$f(tx,ty)=t^{d}f(x,y)$$
With those definitions before us, may we say that the following equations,
$$320 x^4-512 x^3 y+176 x^3+272 x^2 y^2-184 x^2 y+34 x^2$$
$$-96 x y^3+160 x y^2-84 x y+14 x+160 y^4-272 y^3+172 y^2-48 y+5=0 $$
and
$$ 25 x^2-120 x y-10 x+244 y^2+124 y+26=0$$
are algebraic curves? if so, how may I prove it using the definitions presented above?

References

1. Jennings, G. A. (2012). Modern Geometry with Applications. Estados Unidos: Springer New York.
2. Prasolov, V. V., Lando, S. K., Kazaryan, M. E. (2019). Algebraic Curves: Towards Moduli Spaces. Alemanha: Springer International Publishing.

 

[Office_Shredder]

What makes you worried that these do not immediately follow the definition you give for an algebraic curve? If it makes you feel better you can write out for the first one if meets the definition with $a_{4,0}=320$ etc, but no one would be this pedantic usually.

 

[Infrared]

It is generally well-known that a plane algebraic curve is a curve in $\mathcal{CP}^{2}$ given by a homogeneous polynomial equation $f(x,y)= \sum^{N}_{i+j=0}a_{i\,j}x^{i}y^{j}=0$, where $i$ and $j$ are nonnegative integers and not all coefficients $a_{ij}$ are zero~[1].

This defines a (finite) subset of $\mathbb{C}P^1$, not $\mathbb{C}P^2.$ An algebraic curve in $\mathbb{C}P^2$ could be described as the zero set of a homogenous polynomial in $3$ variables.

When you have a possibly nonhomoenous polynomial in $n$ variables, you can "homogenize" it by multiplying each term by an appropriate power of a new variable. For example, the equation $x^2+y+1=0$ defines a curve in the affine plane, but $x^2+yz+z^2=0$ defines a curve in the projective plane. These curves are related because if you take the points $[x:y:z]$ where $z\neq 0$ and identify it with the affine plane by scaling $z$ to be $1$, then the two curves coincide.

 

[mathwonk]

your definition in #1 seems to be flawed. you have the value of i+j varying, whereas the definition of homogeneous is that i+j should be constant. However as infrared has pointed out, your example of a non homogeneous polynomial in 2 variables, can be homogenized to give a homogeneous polynomial in 3 variables, and thus in this way does define a projective plane curve in the projective plane. Namely, the polynomial you give defines an affine plane curve, but the projective plane is a compactification of the affine plane, and the homogenization defines the closure in the projective plane of this affine plane curve.

e.g. your last equation, after homogenization, becomes
x^2 -120xy -10xz +244y^2 +124yz +26z^2 = 0

 

[mathwonk]

by the way, you may be interested in that specific curve, but geometers usually consider plane curves only up to isomorphism, and that conic I gave is linearly isomorphic to the much simpler one given by x^2 + y^2 + z^2 = 0, i.e. they are the same, after a linear change of coordinates. Up to linear isomorphism, there are only 3 different conics, given by x^2 (double line), x^2+y^2 (two distinct lines), and x^2+y^2+z^2 (smooth irreducible conic). this transformation of an arbitrary conic into such simpler ones is called "diagonalizing a quadratic form". (I assume here you are working over the complex numbers, as implied by your notation CP^2.)