In $\mathbb{C}[X,Y,Z,W]$ I have the following ideal $$I=(XZ-Y^2,XW-YZ,X^2-WY,YX-WZ,YW-Z^2,ZX-W^2).$$ I'm trying to find it's radical using Hilbert Nullstellensatz, i.e., trying to find the set of elements that vanishes at all the points of the variety of $I$. The bad news is that the variety of $I$ is not the easiest set to find. Since $\mathbb{C}$ is algebraically closed, it will be made of only elements of $\mathbb{C}$.
I found that the following sets are in the variety of $I$:
- $(a,a,a,a)\in\mathbb{C}^4$, $\forall a\in\mathbb{C}$
- $(-a,-a,-a,a)\in\mathbb{C}^4$, $\forall a\in\mathbb{C}$
- And the same as the above but with $a$ in different positions.
I managed to find these by seeing that $X^2-Z^2=(X-Z)(X+Z)$ and $Y^2-W^2=(Y-W)(Y+W)$ are in $I$ and trying to find a set of elements that vanishes these, but I have no idea how to find every single element. Besides, I don't know if this is the best approach, I could try to find the intersection of all ideals too.
Any tips are appreciated.
Here's one method. Note that, although your ideal turns out to already be radical, this method does not prove that.
The rational normal curve of degree 3 is $$C := \{[s^3:s^2t:st^2:t^3]\in\Bbb P^3 \mid [s:t]\in \Bbb P^1\}.$$
Step 1: Prove that $C = V(Z^2-YW, YZ-XW, Y^2-XZ)$ (inside $\Bbb P^3$).
Step 2: Prove that $V(I)\subseteq C$ (here, I'm using $V(I)$ to mean the vanishing locus in projective space).
Step 3: Use the definition of $C$ to find $V(I)$ explicitly. You'll end up with a finite set of points.
Step 4: Use the projective Hilbert Nullstellensatz to express $\sqrt{I}$ as an intersection of prime ideals.