The following Sage code shows that \(I_1 = \langle y^2 + z^2 - 1 \rangle\).

R.<x,y,z> = PolynomialRing(QQ)
I = ideal(x^2+y^2+z^2+2, 3*x^2 + 4*y^2 + 4*z^2 + 5)
I.elimination_ideal([x])

Thus \(I_1 = \langle y^2+z^2-1 \rangle\). Over \(\mathbb{C}\) we have that \(\mathbf{V}(I_1)=\pi_1(V)\) since \((\sqrt{-3},y,z) \in V\) for all \((y,z) \in \mathbf{V}(I_1)\). However, over \(\mathbb{R}\), since \(x^2+y^2+z^2 \ge 0\) for all \((x,y,z)\), we see that \(V=\emptyset\) while \(\mathbf{V}(I_1)\) is the unit circle (and thus is infinite).

Using the following Sage code to find a lex Gröbner basis for the ideal \(I=\langle x-uv, y-uv^2, z-u^2 \rangle\).

sage: R.<u,v,x,y,z> = PolynomialRing(QQ, order="lex")
sage: I = ideal( x-u*v, y-u*v^2, z-u^2)
sage: I.groebner_basis()

The Gröbner basis is as follows: \[\begin{align*} u^{2} - z, u v - x, u x - v z, u y - x^{2},\\ v^{2} z - x^{2}, v x - y, v y z - x^{3},\\ x^{4} - y^{2} z \end{align*}\]

From this we find the elimination ideal \(I_2 = \langle x^4-y^2z \rangle\); thus \(V\) is defined by the equation \(x^4=y^2z\).

We determine when a point on \(V=\mathbf{V}(I_2)\) extends to \(X=\mathbf{V}(I_1)\) using the Extension Theorem. First, the highest coefficients of \(v\) in \(I_1\) generate the ideal \(J=\langle z,x,yz \rangle\). Thus every point of \(V\) extends to \(X\) except possibly those in \(\mathbf{V}(J)\). Observe that \(\mathbf{V}(J)\) consists of all points of the form \((0,y,0)\). The equation \(vx-y\) has no solution for points \((0,y,0)\) unless \(y=0\). However, \((0,0,0,0) \in \mathbf{V}(I_1)\), so this point extends.

Now, we determine when a point on \(X\) extends to \(Y=\mathbf{V}(I)\) using the Extension Theorem. The highest coefficients of \(u\) in \(I\) generate the ideal \(K= \langle 1,v,x,y \rangle = \emptyset\). Thus every point of \(X\) extends to \(Y\).

We conclude that \(V \setminus S\) is exactly the set of points \((0,y,0)\) where \(y \ne 0\).

Suppose that \(V \subseteq \mathbb{R}^n\) is a variety. Then \(V=\mathbf{V}( f_1, \ldots, f_s )\) for some polynomials \(f_1, \ldots, f_s \in \mathbb{R}[x_1,\ldots,x_n]\). Let \(g=f_1^2 + \cdots + f_s^2\). Note that \(f_i(x_1,\ldots,x_n)^2\ge 0\) for all \(i\) and all \((x_1,\ldots,x_n) \in \mathbb{R}^n\). Since a sum of nonnegative real numbers is zero if and only if every number in the sum is zero, \(g(x_1,\ldots,x_n)=0\) if and only if \(f_i(x_1,\ldots,x_n)=0\) for all \(i\). We conclude that \(V=\mathbf{V}(g)\).

The twisted cubic is clearly a special case of this construction.

We find that that \(\{ x^2, y-1 \}\) is a lex Gröbner basis for \(J\). Now \(\mathbf{V}(J)=\{(0,1)\}\) so \(f=x \in \mathbf{I}(\mathbf{V}(J))\). But \(x\) is its own remainder on division by \(J\), so \(x \notin J\).

Note that \(\mathbf{V}(x^2+1)=\emptyset\) since no real number \(x\) has a negative square. If \(f \in \sqrt{\langle x^2+1 \rangle}\) then \(f^n =g(x^2+1)\) for some \(g \in \mathbb{R}[x]\) and \(n \ge 1\). But then \(x^2+1\) divides \(f^n\). Since \(x^2+1\) is irreducible, \(x^2+1\) divides \(f\). Thus \(f \in \langle x^2+1 \rangle\) and we conclude that the ideal is radical.

The element \(x+y\) is in the radical of \(\langle x^3,\ y^3,\ xy(x+y)\rangle\) and the smallest power is \(3\). This can be demonstrated with the following Sage code:

R.<x,y,z> = PolynomialRing(QQ)
R.<x,y> = PolynomialRing(QQ)
I = ideal(x^3,y^3,x*y*(x+y))
(x+y)^2 in I
(x+y)^3 in I

The element \(x^2+3xy\) is in the radical of \(I=\langle x+z,\ x^2y,\ x-z^2\rangle\). To determine this, we use the radical membership algorithm. Let \(J = I + \langle \lambda(x^2+3xy)-1\). We find a Gröbner basis using the following Sage code:

R.<x,y,z,𝜆> = PolynomialRing(QQ)
J = ideal(x+z, x^2*y, x-z^2, 𝜆*(x^2+3*x*y))
J.groebner_basis()

We conclude that \(J \ne \langle 1 \rangle\), so the element is not in the ideal.