UTM Ideals Varieties and Algorithm - Chapter 1 Section 3 Exercise 4

Problem:

Solution:

The first equation does not come with squares, so maybe it is easier to start with the first one.

$\begin{eqnarray*} x &=& \frac{t}{1+t} \\ x + tx &=& t \\ x &=& t - tx \\ &=& t(1 - x) \\ t &=& \frac{x}{1 - x} \end{eqnarray*}$

Now we can substitute $t$ into $y$ to eliminate $t$.

$\begin{eqnarray*} y &=& 1 - \frac{1}{t^2} \\ &=& 1 - \frac{1}{(\frac{x}{1 - x})^2} \\ &=& 1 - (\frac{1 - x}{x})^2 \\ &=& 1 - (\frac{1}{x} - 1)^2 \\ &=& 1 - (\frac{1}{x^2} - \frac{2}{x} + 1) \\ &=& \frac{2}{x} - \frac{1}{x^2} \\ x^2y &=& 2x - 1 \end{eqnarray*}$

For part (b), first of all note that $(1, 1)$ is in the variety. Note that by the equation $t = \frac{x}{1 - x}$, we can substitute any $x$ we want into the equation, except of course we cannot set $x = 1$. So for any point in the variety, we can find the corresponding $t$, therefore we know the parametric representation does cover all points but $(1, 1)$ on the variety.