Link between root and parametric curve

Question

I have a problem with this assertion in Algebraic Geometry, by Justin Smith, p. 12

Quote

Suppose we are given parametric equations for a curve

\begin{equation} \begin{array}{cc} x =& \frac{f_1(t)}{g_1(t)} \\ y =& \frac{f_2(t)}{g_2(t)} \end{array} \end{equation}

where $f_i$ and $g_i$ are polynomials, and want an implicit equation for that curve, ie. one of the form

\begin{equation} F(x, y) = 0 \end{equation}

This is equivalent to finding $x$, $y$ such that the polynomials

\begin{equation} \begin{array}{cc} f_1(t) - x g_1(t) &= 0 \\ f_2(t) - y g_2(t) &= 0 \end{array} \end{equation}

have a common root (in $t$).

My question

These are the only indications from the book. I would like to have some clue about this equivalence. What are the theorem / general ideas used to prove it ?

Answer 1

What you have written says the following.

By definition, $(x,y)$ is a point on the curve with parametrization $\left(\frac{f_1(t)}{g_1(t)},\frac{f_2(t)}{g_2(t)}\right)$ if and only if there is some concrete value of $t$, call this value $a$ such that $x=\frac{f_1(a)}{g_1(a)}$ and $y=\frac{f_2(a)}{g_2(a)}$, which by rearranging is equivalent to looking for a value $a$ such that we simultaneously have $f_1(a)-xg_1(a)=0$ and $f_2(a)-yg_2(a)=0$. Thus $(x,y)$ is on the curve if and only if the equations $f_1(t)-xg_1(t)=0$ and $f_2(t)-yg_2(t)=0$ have a simultaneous solution (for $t$).

The next part isn't in what you've written above, but it's probably how the book proceeds? Maybe there's another way, but this is how I'd proceed anyway.

Next we can turn these two equations in $t$ into a single polynomial in the coefficients of $t$ in these equations by taking the resultant of these two polynomials. Since the coefficients are polynomials in $x$ and $y$, this produces a polynomial $R(x,y)$ which is zero precisely when these two polynomials have a simultaneous solution for $t$, i.e. precisely when $(x,y)$ is a point on the curve.

Answer 2

Starting from

$\begin{equation} \begin{array}{cc} x =& \frac{f_1(t)}{g_1(t)} \\ y =& \frac{f_2(t)}{g_2(t)} \end{array} \end{equation}$

we can re-arrange these equations to get

$\begin{equation} \begin{array}{cc} f_1(t) - xg_1(t) =& 0 \\ f_2(t) - yg_2(t) =& 0 \end{array} \end{equation}$

But if $(x,y)$ is a point on the curve then these equations must be simultaneously satisfied for the same value of $t$. Regarding them as polynomials in $t$ we can say that they must have a common root, and then we can find a relationship between $x$ and $y$ that makes this true.

For example, if

$\begin{array}\\ x = \frac{1+t^2}{2t}\\ y=\frac{1-t^2}{2t} \end{array}$

then

$t^2-2xt+1=0 \Rightarrow t=x \pm \sqrt{x^2-1}$

$t^2+2yt-1=0 \Rightarrow t=-y \pm \sqrt{y^2+1}$

Since these equations are satisfied by the same value of $t$ we can eliminate $t$ to get

$x+y = \mp \sqrt{x^2-1} \pm \sqrt{y^2+1}$

$\Rightarrow 2xy = -2\sqrt{(x^2-1)(y^2+1)}$

$\Rightarrow 4x^2y^2 = 4(x^2-1)(y^2+1)$

$\Rightarrow x^2-y^2=1$