Determine the ideal of an affine variety Let $X=\{(r^2,r^3,r^4) : r\in\Bbb R\}\subset \Bbb R^3$. Show that 
1)  $X$ is an affine variety. 
2)  Determine the ideal of $X$. Every $f\in\Bbb R[x,y,z]$, can we write $f$ in the form $f=p(xz-y^2)+q(z-x^2) +r$ , where $p,q\in\Bbb R[x,y,z]$?
i) Affine variety is defined as the solution set to some polynomials in the affine space $k^n$.
ii) The ideal of $X$ is defined as 
$$I(X)=\{f\in \Bbb R[x,y,z]:f(x,y,z)=0, \forall (x,y,z)\in X\}$$
 A: 1) It is immediate that $X$ is included in the variety $V:=V(xz-y^2,z-x^2)$.     
2) To prove the converse inclusion $V\subset X$, we will take some point $P=(a,b,c)\in V$, i.e. a point satisfying $ac=b^2 $ and $c=a^2$,  and show that $P\in X$.
a) If $a=0$, then $ac-b^2=0$ forces $b=0$ and $c-a^2=0$ forces $c=0$.
So actually $P=(0,0,0)$ and $P\in X $, corresponding to $r=0$.
b) If $a\neq0$, a few little calculations (using $ac=b^2, c=a^2$) show that $P=(a,b,c)=(r^2,r^3,r^4)$ with $r=\frac ba$
[For example $r^2=\frac {b^2}{a^2}=\frac {ac}{a^2}=\frac {ac}{c}=a$ ]  
3) Hence $X=V(I)$ with $I=(xz-y^2,z-x^2)$, proving that $X$ is an affine variety.   
4) The calculations above are  valid over any field and the Nullstellensatz implies that $I (X)=\sqrt I$.
Now $I$ is prime since $k(x,y,z]/I=k[x,y]/(y^2-x^3)$ and $y^2-x^3$ is irrreducible. 
So finally $$I(X)=I=(xz-y^2,z-x^2)$$
A: For this simple example, you do not need to use computer algebra software to compute the ideal. The ideal $I$ is the kernel of the homomorphmism $$ k[x,y,z] \to k[r],$$ defined by $x \mapsto r^2, y \mapsto r^3, z \mapsto r^4$. Your task task is to find relations among the images of the homomorphism.
By inspection you see that $x^2$ and $z$ are sent to the same element, and also that $xz$ and $y^2$ are sent to the same element. Thus $I=(x^2-z,xz-y^2) \subseteq I(X)$. To show that this is an equality, you need to show that any point $(x,y,z) \in \mathbb A^3$ that satisfies these polynomial relations actually lie on the curve that any polynomial .
Hint: start with $(x,y,z) \in \mathbb A^3$. By the second equation, this is equal to $(x,y,x^2)$. Now, if $y$ is zero, then both $x$ and $z$ are. So assume $y > 0$. Then $y=\sqrt{x^3}$ (we take the positive square root always). The variables $x$ and $z$ are always positive because $xz > 0$ and $x^2=z$. Then $\sqrt{x}^2$ makes sense and is equal to $x$, and also $z=\sqrt{x}^4$. This means that $(x,y,z)=(\sqrt{x}^2,\sqrt{x}^3,\sqrt{x}^4)$, which have the required form. Now assume $y <0$. Then the same arguments give $(x,y,z)=(\sqrt{x}^2,-\sqrt{x}^3,\sqrt{x}^4)$. So $Z(I) \subset Z(I(X))$ also.
A: *

*Write down the ideal $J:=(x-r^2,y-r^3,z-r^4)$.

*Find a Groebner basis, using Buchberger's algorithm for example. Use the monomial order $r>x>y>z$ but such that the $r$ parts of the monomials are compared first.

*The elements of the basis that do not depend on $r$ give you a Groebner basis (a set of generators) of the ideal of $X$.

When we compute we get $J=(y^2-xz,x^2-z,rz-xy,ry-x^2,rx-y,r^2-x)$.
So, $I(X)=(y^2-xz,x^2-z)$.
Code in Singular:
 > ring R=0,(r,x,y,z),(dp(1),dp(3));
 > ideal J=(x-r^2,y-r^3,z-r^4);
 > J=groebner(J);
 > J;
 J[1]=y2-xz
 J[2]=x2-z
 J[3]=rz-xy
 J[4]=ry-x2
 J[5]=rx-y
 J[6]=r2-x
 >

Alternatively Singular has a command to do elimination directly.
> eliminate(J,r);
_[1]=y2-xz
_[2]=x2-z

What we did is one of the applications of Groebner bases. In this case, Implicitization.
