1.1 A brief introduction to computational algebraic geometry  (Page 5/20)

Rational curves

A rational function in one variable is a function given as a quotient of polynomials

where $p\left(t\right),q\left(t\right)\in \mathbf{R}\left[t\right]$ are polynomials. Note that, despite the name, a rational function isn't a well-defined function at the points where $q\left(t\right)=0$ .

Given a curve $X=V\left(f\right)$ , for $f\in \mathbf{R}[x,y]$ an irreducible Irreducible means it doesn't factor as a product of non-constant polynomials. We need this condition here because if $f(x,y)=g(x,y)h(x,y)$ where $g(x,y)$ defines a rational curve and $h(x,y)$ defines an irrational curve, we wouldn't want to call the whole curve $V\left(f\right)$ rational when it consists of both the zero set of $g(x,y)$ (which we can parametrize rationally) and the zero set of $h(x,y)$ which we can't. Even if $V\left(g\right)$ and $V\left(h\right)$ were both rational, we still wouldn't be able to parametrize all of $V\left(f\right)$ by a single rational function. polynomial, we say that a rational parametrization of $X$ is a pair of rational functions $\left(x,\left(,t,\right),,,y,\left(,t,\right)\right)$ so that

for all values of $t$ where it is defined. If $X$ admits a rational parametrization, we say that $X$ is a rational curve.

Proposition The circle $X=V(x^2+y^2-1)$ is a rational curve.

To find a rational parametrization of the circle, we use the fact that a non-vertical line through a point $(-1,0)$ of the circle meets the circle in exactly one other point.

We can thus try to use the slope of the line through $(-1,0)$ as a rational parameter for the other point of the circle that the line intersects. The line with slope $t$ through $(-1,0)$ is defined by $y=tx+t$ . Combining this with the equation $x^2+y^2=1$ for the circle, we get

Expanding and grouping terms, we get

and we see that, as expected, $x=-1$ is a solution for all $t$ . This allows us to factor the equation as

and we find that the $x$ -coordinate of the point of intersection other than $(-1,0)$ is

and that

We've thus found that $\left(\frac{1-t^2}{1+t^2},,,\frac{2t}{1+t^2}\right)$ is a rational parametrization of the unit circle (which we could check by substituting back into the implicit equation).

Of course, in the case of the circle this may not immediately seem so exciting, since we already have a very convenient (but non-rational) parametrization by $(cost,sint)$ , but it turns out the rationality (together with the nice integer coefficients) of our parametrization in this example has a fairly interesting corollary:

A Pythagorean triple is a triple of three positive integers $(a,b,c)$ such that

Examples include $(3,4,5)$ , $(5,12,13)$ , and $(15,8,17)$ . Of course, if $(a,b,c)$ is a Pythagorean triple, then so is $(da,db,dc)$ , so we may as well restrict our attention to the case where $a$ , $b$ , and $c$ have no common factor. A computation modulo 4 also shows that in this case $c$ must be odd and exactly one of $a$ and $b$ must be even, so we may as well assume $b$ is even.

The connection to our parametrization of the circle is that if $a^2+b^2=c^2$ , then

so that $\left(\frac{a}{c},,,\frac{b}{c}\right)$ is a point with rational coordinates on the curve $x^2+y^2=1$ . But our parametrization works just as well over the rational numbers as it does over the real numbers, and we know that (aside from $(-1,0)$ ), every point has the form $\left(\frac{1-t^2}{1+t^2},,,\frac{2t}{1+t^2}\right)$ where $t=\frac{m}{n}$ is a rational number. This gives