Math 8203
Fall 1997

Solutions to exercises 1 and 2

Exercise 1.

The affine parametrization was given by the following formula:

\varphi(t) = ((1 - 3t2)/ (1 + t2)2 , t(1 - 3t2)/ (1 + t2)2).
Since we can clear denominators, the extension to \psi: P1 - > P2 is given for points at finite distance by the formula:
\psi[1,t] = [x,y,z] = [1 - 3t2, t(1 - 3t2, (1 + t2)2],
and therefore in general as follows:
\psi[ s,t] = [x,y,z] = [s2(s2 - 3t2), st(s2 - 3t2), (s2 + t™2)2].
The points [1,ˆ3], [1,-ˆ3] and [0,1] are mapped to the origin in A2 (we identify A2 with the set of points where z is nonzero). The third point is considered as the point at infinity on P1. The points [1,i] and [1,-i] are mapped to points at infinity (z = 0) on the projective closure, specifically [1,i,0] and [1,-i,0].

Exercise 2.

(a) The projection is defined by the formula \pi(x,y) = (2x/ (x+ y), 2y/ (x + y) ). It is (at least) 2 to 1 because \pi(x,y) = \pi(-x,-y). To see that each point on L has at most two preimages, we study equations which are similar to those encountered in the construction of the parametrization below. Basically, there are just two nonzero values for z.

The following points of L are omitted from the image of \pi:

(2,0), (1,1), and (0,2)
This is because the x-axis, the y-axis and the line y = x intersect the curve only at the origin. Furthermore, if \omega5 = -1 but \omega \neq -1, then the point (a,b) on L with b = \omegaa is omitted from the origin because there is no point (x,y) at finite distance on the curve with y = \omegax. Explicitly, these omitted points are of the form (2/ (1+\omega), 2\omega/ (1+ \omega)), so that they are certainly not real points.

(b) If we substitute (x,y) = ((1-t)z, (1+t)z) into the equation of the curve, then we obtain:

((1-t)z)5 + ((1+t)z) 5 - 2t z (1-t)z(1+t)z = 0,
so that:
((1-t)5 + (1+t)5) z5 - 2t(1- t2)z3 = 0.
(NOTE: Since we are not considering a projective plane curve, the auxiliary variable z is not the coordinate whose vanishing defines the line at infinity ...) In solving the equation, we disregard the trivial root z = 0, which occurs with multiplicity 3. Thus, we obtain:
(2 + 20t2 + 10t 4)z2 = 2t(1 - t2).
This implies that z must be one of the square roots of t(1 - t2)/ (1 + 10t2 + 5t 4). If t is distinct from 0, 1, and -1, then locally on C there are two branches which express z as an analytic function of t.. We multiply by 1 - t or 1 + t respectively to obtain expressions for x and y.

(c) To get a real point in the image of the projection (equivalent to projecting a real point of the curve), the quotient t(1 - t2)/ (1 + 10t2 + 5t 4) must be positive. We cannot allow the quotient to be zero, since that would amount to claiming that the projection of the origin is well defined. Because 1 + 10t2 + 5t 4 > 0 for all real t, this amounts to requiring that t(1-t2) > 0. So the conclusion is that we can have t < -1 or 0 < t < 1. Therefore the real image consists of the set of points of L which are below the x-axis (or equivalently, to the right of x = 2), together with the set of points on the line segment between (0,2) and (1,1).

(d) The formula for one of our real parametrizations is thus:

\varphi+(t) = ( (1-t){t(1 - t2)/ (1 + 10t2 + 5t 4) }1/2, (1+t) {t(1 - t2)/ (1 + 10t2 + 5t 4) }1/2 ).
We take the positive square root, which is meaningful because we are studying the real locus. It is defined for t \leq -1 and also for 0 \leq t \leq 1. (Note that the denominator has no real zeros!) The parametrization is at least continuous at the (finite) endpoints of the two intervals; all of the endpoints are mapped the origin.

NOTE. If we look at our picture of this curve again, we see that the "bow tie" is parametrized (in two pieces) by \varphi+ and \varphi-, defined on closed interval [1,2], while the infinite part of the curve is parametrized (again in two pieces) by \varphi+ and \varphi- defined on (-\infty,-1]. We can join the two halves of the "bow tie" if we make a suitable shift of one copy of the interval, thereby obtaining a differentiable parametrization. With sligtly more work, one probably join the two halves of the infinite part of the curve in a single differentiable parametrization. We cannot join the two large pieces into a single differentiable parametrization because there would have to be a "corner" at the origin. On the other hand, the "bow tie" and the infinite piece are not part of an algebraic decomposition. Indeed, we can prove that the equation of the curve is irreducible in K[x,y], provided that we assume char(K) is different from 2. (It is easy to see what happens when char(K) = 2. In that case x+y = x-y, so that x+y becomes a factor of the defining equation of the curve.)

Last updated November 2, 1997.

Comments (and ¡corrections! please)to:
Joel Roberts
351 Vincent Hall
625-1076
e-mail: roberts@math.umn.edu
http://www.math.umn.edu/~roberts