Ideal associated to $\mathbb{V}(X^3-Y^2)$ I have to find the ideal associated to the affine algebraic set $\mathbb{V}(X^3-Y^2)$ where the field is $\mathbb{R}$. I'm trying to show that it is $(X^3-Y^2)$ himself, but I can't conclude! Can anyone help me?
 A: You can "reduce to $\mathbb{C}$".
Note that we have a parametrization of $V(x^3-y^2)$, $t\mapsto(t^2, t^3)$, both for the $\mathbb{R}$ points and the $\mathbb{C}$ points.
Let $P\in \mathbb{R}[x,y]$, $P= 0$ on $V(x^3-y^2)$. We have $P(t^2, t^3) = 0$ for all $t \in \mathbb{R}$. Since $\mathbb{R}$ is infinite, it implies that the polynomial in $t$ $P(t^2, t^3)$ is the $0$ polynomial.
Now we conclude that $P(t^2, t^3) = 0$ for all $t \in \mathbb{C}$.  This implies that $P$ is $0$ on the complex $0$ locus of $x^3 - y^2$.  Now use nullstellensatz and get $(x^3-y^2) \mid P$ ( in $\mathbb{C}[x,y]$, so also in $\mathbb{R}[x,y]$).
A: Here's an outline of the solution. Try to fill in the details yourself - if you get stuck, please leave a comment.
Step 1: Observe $V(x^3-y^2)$ is the image of the map $\Bbb R\to \Bbb R^2$ by $t\mapsto (t^2,t^3)$.
Step 2: Prove if a polynomial $p(x,y)$ vanishes on $V(x^3-y^2)$, the polynomial $p(t^2,t^3)$ must be zero.
Step 3: Prove that any polynomial $p(x,y)$ can be written as $(y^2-x^3)\cdot a(x,y)+y\cdot b(x)+c(x)$ for polynomials $a,b,c$.
Step 4: Reason that if a polynomial $p(x,y)$ vanishes on $V(x^3-y^2)$, it must be divisible by $(y^2-x^3)$.
