1.1 A brief introduction to computational algebraic geometry  (Page 7/20)

It's natural to ask the reverse question as well: given a parametric rational curve $\left(x\right(t),y(t\left)\right)$ , can we find a polynomial $f(x,y)\in \mathbf{R}[x,y]$ so that $f\left(x\right(t),y(t\left)\right)=0$ ? (One of the homework problems asks you to do this for a non-rational parametrization of the cardioid.)

Proposition Given rational functions $x\left(t\right)$ and $y\left(t\right)$ , there exists a polynomial $f(x,y)\in \mathbf{R}[x,y]$ such that $f\left(x\right(t),y(t\left)\right)\equiv 0$ .

We can write the given rational functions as $x\left(t\right)=\frac{a\left(t\right)}{q\left(t\right)}$ and $y\left(t\right)=\frac{b\left(t\right)}{q\left(t\right)}$ for some polynomials $a\left(t\right),b\left(t\right),q\left(t\right)\in \mathbf{R}\left[t\right]$ . For some large degree $N$ , we'll try to find a polynomial $f(x,y)\in \mathbf{R}[x,y]$ of degree $N$ so that $f\left(x\right(t),y(t\left)\right)\equiv 0$ , or equivalently so that

Let $n=degq\left(t\right)$ and $m=max\{dega(t),degb(t\left)\right\}$ . Then $q{\left(t\right)}^{N}f(x\left(t\right),y\left(t\right))$ is a polynomial in $t$ of degree at most $Nn+Nm$ , whose coefficients are homogeneous linear functions of the coefficients of $f$ . Thus setting all of its coefficients equal to zero give at most $Nn+Nm+1$ homogeneous linear equations for the coefficients of $f$ , and if we can pick $N$ so that there are at least as many variables as there are equations (i.e. $f$ has at least $Nn+Nm+1$ coefficients), then we can solve the system and find an $f$ with $q{\left(t\right)}^{N}f(x\left(t\right),y\left(t\right))\equiv 0$ , and we'll be done.

The degree $k$ part of $f$ is

which has $k+1$ coefficients. Overall then, $f$ has $1+2+\cdots +(N+1)=\frac{N(N+1)}{2}$ coefficients, and since this is quadratic in $N$ and $Nn+Nm+1$ is linear, for sufficiently large $N$ we have $\frac{N(N+1)}{2}\ge Nn+Nm+1$ as desired.

While this proves that it is always possible to find a polynomial vanishing on a rationally parametrized curve, there are a few things that we might not like so much about it. For one thing, $N$ may be bigger than it has to be; for example, in the parametrization $\left(\frac{1-t^2}{1+t^2},,,\frac{2t}{1+t^2}\right)$ for the circle, the proof would use the value $N=8$ to find a degree $\le 8$ polynomial $f(x,y)$ vanishing on the circle, i.e. the solutions to the system would be the $f(x,y)=(x^2+y^2-1)g(x,y)$ for arbitrary polynomials $g(x,y)$ of degree $\le 6$ . This isn't really a big problem though, since we could always try to solve the systems for lower degrees first.

The bigger complaint we might have is that even for small degree examples, this involves solving a big system of linear equations in a big number of variables. Of course, computers are pretty good at solving systems of linear equations, but this certainly isn't something we'd want to do by hand, and even computer algebra systems might have some trouble dealing with huge numbers of variables.

In fact, there is a nicer way to do the computation, and it involves thinking about the problem more geometrically. We can think of the graph of the rational function $(x\left(t\right),y\left(t\right))=\left(\frac{a\left(t\right)}{p\left(t\right)},,,\frac{b\left(t\right)}{q\left(t\right)}\right)$ in ${\mathbf{R}}^3={\mathbf{R}}^2\times \mathbf{R}$ as being the common zeros of the two polynomials

in $\mathbf{R}[x,y,t]$ . Our goal then is to find a polynomial $f(x,y)$ in only the variables $x$ and $y$ so that whenever $g(x_0,y_0,t)=h(x_0,y_0,t)=0$ for some value of $t$ , we have $f(x_0,y_0)=0$ . If we could write some polynomial $f(x,y)\in \mathbf{R}[x,y]$ in the form

for polynomials $a(x,y,t),b(x,y,t)\in \mathbf{R}[x,y,t]$ , then $f(x,y)$ would certainly have this property, and thus be a polynomial vanishing on the parametrized curve. For example,