UTM Ideals Varieties and Algorithm - Chapter 1 Section 2 Exercise 7

Problem:

Solution:

We know that any point in the four-leaved rose must satisfy $r = \sin(2\theta)$. Using the given formula, we have

$\begin{eqnarray*} & & (x^2 + y^2)^3 -4x^2y^2 \\ &=& r^6 - 4r^2\sin^2(\theta)r^2\cos^2(\theta) \\ &=& r^6 - r^4(4\sin^2(\theta)\cos^2(\theta)) \\ &=& r^6 - r^4(2\sin(\theta)\cos(\theta))^2 \\ &=& r^6 - r^4(\sin(2\theta))^2 \\ &=& r^6 - r^4(r)^2 \\ &=& r^6 - r^6 \\ &=& 0 \end{eqnarray*}$

Therefore, we proved part a, a point in the four-leaved rose is necessary in the affine variety.

Part (b) is mostly a language problem, once I figured out how do we represent the different polar coordinates, the problem is mostly solved.

Denote $r_u$, $\theta_u$ as the usual polar coordinate for a point, and $r_s$, $\theta_s$ be the special polar coordinates defined as follow:

If the point is in the 1st and 3rd quadrant, use the normal polar coordinate, otherwise take $r_s$ to be negative (and therefore $\theta_s$ need to take a phase shift of $\pi$ to compensate for the sign.

For example, the point $(1, 1)$ has the usual polar coordinates $(r_u, \theta_u) = (\sqrt{2}, \frac{\pi}{4})$ and special polar coordinates $(r_s, \theta_s) = (\sqrt{2}, \frac{\pi}{4})$, which is the same because (1, 1) is in 1st quadrant.

But the point $(-1, 1)$ will have usual polar coordinates $(r_u, \theta_u) = (\sqrt{2}, \frac{3\pi}{4})$ and special polar coordinates $(r_s, \theta_s) = (-\sqrt{2}, \frac{-\pi}{4})$, which is the different because (1, 1) is in 2nd quadrant.

The interest thing is these three formula still works for the special polar coordinates
 $x^2 + y^2 = r_s^2$
 $x = r_s \cos(\theta_s)$
 $y = r_s \sin(\theta_s)$

That can be argued on a case by case basis. If normal polar coordinate and special coordinate coincides, then it is obviously true, otherwise, the square of negative cancels, and the negative sign together with the phase shift also cancels.

$\begin{eqnarray*} & & (x^2 + y^2)^3 -4x^2y^2 \\ &=& r_s^6 - 4r_s^2\sin^2(\theta_s)r_s^2\cos^2(\theta_s) \\ &=& r_s^6 - r_s^4(4\sin^2(\theta_s)\cos^2(\theta_s)) \\ &=& r_s^6 - r_s^4(2\sin(\theta_s)\cos(\theta_s))^2 \\ &=& r_s^6 - r_s^4\sin^2(2\theta_s) \\ &=& r_s^4(r_s^2 - \sin^2(2\theta_s)) \end{eqnarray*}$

That implies either $r_s = 0$, the origin, or $r_s^2 = \sin^2(2\theta_s)$. Last but not least, we argue that $r_s = \sin^2(2\theta_s)$ when $r_s \ne 0$. That is true because by the way we defined the special polar coordinates, the two must match in sign and can be argued in more detail with a case by case analysis for each quadrant.