1. \((y^2z)x^5-(y^3)x^4+(z)x^2 + (y+2)x + (y^5-y^3z-5z+3)\)
2. \(y^5+(x^5z)y^2-(x^4-z)y^3+(x)y+(x^2z+2x-5z+3)\)
3. \((x^5y^2+x^2-y^3-5)z + (x^4y^3+y^5+xy+2x+3)\)

Recall that \(x=r\cos(\theta)\), \(y=r\sin(\theta)\), and \(r^2=x^2+y^2\) in polar coordinates and \(\sin(2\theta)=2\sin(\theta)\cos(\theta)\).

Suppose \((x,y) \in C\). Then \[\begin{align*} r &= \sin(2\theta)\\ r &= 2\sin(\theta)\cos(\theta)\\ r^3 &= 2(r\sin(\theta))(r\cos(\theta))\\ r^3 &= 2xy\\ r^6 &= 4x^2y^2\\ (x^2+y^2)^3-4x^2y^2 &= 0 \end{align*}\] Thus \((x,y) \in {\bf V}((x^2+y^2)^3-4x^2y^2)\).

Conversely, suppose \((x,y) \in {\bf V}((x^2+y^2)^3-4x^2y^2)\). Then \(r^6=4x^2y^2\) and so \(r^3=\pm 2xy\). Thus \(r^3 = \pm r^2\sin(2\theta)\). Observe that \(r=0\) if and only if \((x,y)=(0,0)\); thus it suffices to consider \(r \ne 0\). Thus, \(\pm r = \sin(2\theta)\). Note that \((-r,\theta)\) and \((r,\theta+\pi)\) represent the same point in polar coordinates and \(\sin(2(\theta+\pi))=\sin(2\theta)\). Thus \(-r = \sin(2\theta)\) has the same solution set as \(r = \sin(2\theta)\). Thus \((x,y) \in C\).

We proceed by contradiction. Suppose \(X= {\bf V}(f_1, \ldots, f_s)\) for some polynomials \(f_1,\ldots,f_s \in k[x,y]\). Then \(f_i(t,t)=0\) for all \(i \in \{1, \ldots, s\}\) and \(t \ne 1\). However, \(f_i(1,1) \ne 0\) for some \(i \in \{1, \ldots, s\}\). Let \(g\) be the polynomial in \(\mathbb{R}[t]\) such that \(g(t)=f_i(t,t)\). Since \(g(t)=0\) for all \(t \ne 1\), we see that \(g\) has infinitely many roots. Thus \(g\) is the zero polynomial. Thus \(f_i(1,1)=g(1,1)=0\); a contradiction.

Note that the parametrization is undefined when \(t=0\) or \(t=-1\). Solving the first equation, we find \(t = \frac{x}{1-x}\). There is no value of \(x\) so that \(t=-1\) and \(t=0\) if and only if \(x=0\); thus \(x \ne 0\). Substituting this into the second equation we find \(y = 1-\frac{(1-x)^2}{x^2} = \frac{2x-1}{x^2}\). Since \(x \ne 0\), we see that this is equivalent to \(yx^2-2x+1=0\). Observe that when \(x=0\) this equation has no solutions.

Note that the equation \(xy-1=0\) does not have a solution if \(x=0\) or \(y=0\). Thus we may assume \(x \ne 0\). Thus \(y=\frac{1}{x}\) and the first equation becomes \(x^2+\frac{1}{x^2}-1=0\). or \(x^4+1-x^2=0\). The polynomial is in the ideal since \[ x^4+1-x^2 = x^2(x^2+y^2-1) - (xy+1)(xy-1). \]

In view of Exercise 1.4.2, to show \(\langle f_1,\ldots f_s \rangle = \langle g_1,\ldots, g_t \rangle\) it suffices to show \(f_i \in \langle g_1,\ldots, g_t \rangle\) for all \(i \in \{1,\ldots, s\}\) and \(g_j \in \langle f_1,\ldots f_s \rangle\) for all \(j \in \{1,\ldots, t\}\). We do so explicitly in each case.

1. This case is a linear system.
   • \(x+y = (1)(x)+(1)(y)\)
   • \(x-y = (1)(x)+(-1)(y)\)
   • \(x=(\frac{1}{2})(x+y)+(\frac{1}{2})(x-y)\)
   • \(y=(\frac{1}{2})(x+y)-(\frac{1}{2})(x-y)\)
2. This case requires some experimentation.
   • \(x+xy = (1)x+(x)y\)
   • \(y+xy = (y)x+(1)y\)
   • \(x^2 = (x)x+(0)y\)
   • \(y^2 = (0)x+(y)y\)
   • \(x = (1)(x+xy)+(-x)(y+xy)+(y)(x^2)+(0)(y^2)\)
   • \(y = (-y)(x+xy)+(1)(y+xy)+(0)(x^2)+(x)(y^2)\)
3. This case is also a linear system in disguise!
   • \(2x^2+3y^2-11 = 2(x^2-4) + 3(y^2-1)\)
   • \(x^2-y^2-3 = 1(x^2-4) - 1(y^2-1)\)
   • \(x^2-4 = \frac{1}{5}(2x^2+3y^2-11) + \frac{3}{5}(x^2-y^2-3)\)
   • \(y^2-1 = \frac{1}{5}(2x^2+3y^2-11) + \frac{-2}{5}(x^2-y^2-3)\)

Using Sage (or by hand using the Euclidean algorithm) we find:

1. \(x^2+x+1\)
2. \(x-1\)

One can explicitly find the greatest common divisor of the generators and check that \(x^2-4\) is divisible by the gcd. However, the quickest way is simply to observe that \[ x^2-4 = \frac{1}{2}(x^3+x^2-4x-4) - \frac{1}{2}(x^3-x^2-4x+4) .\]