Math 8203 (Algebraic Geometry)

Fall 1997

Central Projections and Rational Parametrizations
of Plane Curves

Copyright (©) 1997 by Joel Roberts.

Let's begin by saying what we mean by a rational parametrization of a plane curve. Thus, let X be an irreducible algebraic curve in A2. A rational parametrization of X is a map \varphi: U - > X, where U is a subset of A1 whose complement (in A1) is finite, and the mapping has the following properties:

  1. for t in U, the coordinates of \varphi(t ) are rational functions of t ;
  2. the complement (in X ) of the image of \varphi is a finite set.

OUR FIRST EXAMPLE comes from the text. Thus, let X be a plane curve of degree 3 with a node at the origin. Central projection from the origin maps X - {(0,0)} to the line x = 1 in an essentially bijective fashion, and the inverse of the projection provides the desired rational parametrization. Explicitly, we take X to be given by the following equation:

y2 = x2 + x3 .
If x is nonzero, then the line joining (0,0) to (x,y) intersects the line x = 1 at (1,t), where t = y/x. Identifying A1 with the line x, we have the central projection \pi: X - {(0,0)} -> A1 given by the following formula:
\pi((x,y)) = (1,t) ,
where t = y/x. The following sketch illustrates the situation:

sketch of nodal cubic, showing projection

To find the parametrization, we set t = y/x, so that y = tx. We substitute this into the defining equation of our curve, and then solve for x in terms of tto the extent that this is possible. Thus:

f (x,tx) = 0
(tx)2 = x2 + x3
When x is nonzero, we can cancel x from both sides of the equation, thereby obtaining:
x = t - 1
Since t = y/x , we have y = tx, so that we obtain the following parametrization:
\varphi(t) = (t - 1, t (t - 1)) = (t - 1, t 2 - t )

OUR SECOND EXAMPLE is the curve which is given in polar coordinates by the following equation:

r = cos(3\theta).
To convert the equation into rectangular coordinates we use the following trigonometric identity, which follows from the usual addition formulas:
cos(3\theta) = 4cos3\theta - 3cos\theta .
If we multiply this by r 3 and then apply the usual formulas relating polar coordinates and rectangular coordinates, we obtain the following:
r 4 = 4r 3cos 3\theta - 3r 3cos \theta
(x2 +y2)2 = 4x3 - 3x(x2 +y2)
(x2 + y2)2 = x3 - 3xy2
or finally:
(x2 + y2)2 = x(x2 - 3y2)

As in the previous example, we identify A1 with the line x = 1, and then construct the projection from the origin exactly as in the previous example. This is shown in the following figure:

sketch of a plane quartic with a triple point

To obtain a parametrization, we proceed as in the first example. Thus, we substitute y = tx into the defining equation of our curve and then solve for x in terms of t, as follows:

(x2 + (tx)2)2 = x(x2 - 3(tx)2)
x4(1 + t 2)2 = x3(1 - 3t 2)
When x is nonzero, we can cancel x3 from both sides of the equation, thus obtaining the following parametrization:

\varphi(t) =

SOME COMMENTS The parametrization \varphi(t ) is undefined for t = i and for t = -i. This fact, of course, is invisible when we are looking at the set of real points. The values t = \sqrt(3) and t = - \sqrt(3) are both mapped to (0,0). It also is intuitively valid, at least when K = C, to claim that t = \infty is mapped to (0,0), because \varphi(t ) gets close to (0,0) when |t | becomes very large.

BOTH OF THESE EXAMPLES are instances of a more general result. In order to state this result, we need some terminology. So, let K be an algebraically closed field. If f (x,y) is an irreducible element of the polynomial ring K[x,y], then we say that the curve X = V(f ) is an irreducible plane curve. More generally, a hypersurface in An whose defining polynomial is irreducible is called an irreducible hypersurface. By definition, the degree of an irreducible plane curve, or more generally an irreducible hypersurface, is the degree of its defining polynomial.

If X = V(f ) is a plane curve, we will say that X has a point of multiplicity = m at (0,0) if (0,0) is a point of X and

  1. every monomial of degree < m occurs in f with coefficient = 0, and:
  2. some monomial of degree m occurs in f with nonzero coefficient.
Now, we can state our result.

PROPOSITION. Let X be an irreducible plane curve of degree d . If X has a point of multiplicity d - 1, then X has a rational parametrization.

PROOF. The hypothesis implies that the defining defining equation of X is of the following form:

f(x,y) = Fd -1(x,y) + Fd (x,y) ,
where Fd -1(x,y) and Fd (x,y) are homogeneous of degrees d - 1 and d respectively. Substituting y = tx into the equation f(x,y) = 0, we obtain:
Fd -1(x,tx) + Fd (x,tx) = 0
xd - 1Fd -1(1,t) + xdFd (1,t) = 0
If x is nonzero, then we can cancel xd - 1. Thus, we obtain the following parametrization:

\varphi(t) = (- F_{d-1}(1,t)/F_d (1,t), -t F_{d-1}(1,t)/F_d (1,t))

Q.E.D.

The parametrization \varphi is undefined at values of t where Fd (1,t) = 0, while the values of t such that Fd - 1(1,t)= 0 are mapped to the origin. Since our field is algebraically closed, both of these things will definitely occur, unless either Fd - 1(1,t) or Fd (1,t) is a nonzero constant. (That situation happens when Fd - 1 or Fd respectively is a power of x.

More generally, let X be an irreducible curve of degree d with a point of multiplicity m at (0,0). Then the irreducible defining polynomial of X is of the following form:

f (x,y)= Fm (x,y) + ... + Fd (x,y),
where Fj (x,y) is homogeneous of degree j for j = m, ...,d, while Fm and Fd are nonzero. We call Fm the initial form of f(x,y), and we call Fd the highest degree form of f (x,y).

Since K is algebraically closed, the homogeneous polynomials Fm (x,y) and Fd (x,y) factor as products of homogeneous linear polynomials. Indeed, the linear factors (except, possibly, for powers of x) correspond bijectively to the roots of Fm (1,t) and Fd (1,t) respectively. BY DEFINITION, the zero set of Fm (1,t) is called the tangent cone of X at the origin. Since Fm (1,t) is a product of homogeneous linear polynomials, it follows that the tangent cone is a union of lines through the origin. For instance, in the case of our fourth degree curve with a triple point the initial form factors as follows:

f3(x,y) = x(x2 - 3y2) = x(x + \sqrt(3)"y)(x - \sqrt(3)"y) .

AN ILLUSTRATION showing this curve and its tangent cone at the origin is linked here.

THE FACTORS OF THE HIGHESTDEGREE FORM were seen to correspond to asymptotic directions at infinity when K = C. In general, we can understand this in another way by "homogenizing" the equation f(x,y) = 0 to obtain the equation F(X,Y,Z) = 0 of the projective closure \Xbar, specifically:

F(X,Y,Z) = Zd-mFm(X,Y) + ... + Fd (X,Y) .
We can then consider the intersection of \Xbar with the line at infinity, i.e. the line Z = 0. We conclude that this intersection is the solution set of the equations:
Fm (X,Y) = Z = 0.
In other words, the points where \Xbar intersects the line at infinity correspond bijectively to the distinct linear factors of the highest degree form of the highest degree form Fd (X,Y) . (In the present context, we understand two linear factors to be distinct if neither one is a constant multiple of the other.)

AS A FINAL EXAMPLE to illustrate both of these concepts, we consider the plane curve of degree 5 whose equation is:

x5 + y5 + xy(x - y) = 0.
This curve has a triple point at the origin; the initial form is:
F3 = xy(x - y) .
The highest degree form is:
F5 = x5 + y5 .
This form has one real linear factor, namely x + y, so that the intersection of the curve with the line at infinity includes the point [X,Y,Z] = [1,-1,0], which visibly corresponds to the asymptotic direction y = - x in the sketch below. The other points at infinity on the curve are not real: for each point, the ratio of the two nonzero homogeneous coordinates is a primitive fifth root of unity. Investigation of other properties of this curve will be relegated to one of the exercises.

sketch of a plane quintic with a triple point

RELATED EXERCISES ARE LINKED BELOW.

Last updated October 10, 1997.

Please send comments and/or corrections to:

Joel Roberts
351 Vincent Hall
625-1076
e-mail: roberts@math.umn.edu
http://www.math.umn.edu/~roberts