Suppose you have the variety $X=V(z^2-xy,x^3-yz) \subset \mathbb{C}^3$, and that you want to find:
- The dimension of $X$;
- The cotangent Zariski space of the point $(1,1,1)$;
- A system of parameters for $A(X)_{(x-1,y-1,z-1)}$.
For the dimension, we have that $\dim{X}=\dim{A(X)}$, where $$A(X)=\mathbb{C}[x,y,z]/I(V(z^2-xy,x^3-yz))$$ is the ring of coordinates. Since $\mathbb{C}$ is algebraically closed, by the strong Nullstellensatz we have that $$I(V(z^2-xy,x^3-yz))=\sqrt{(z^2-xy,x^3-yz)}.$$ Ideally, I would like for $(z^2-xy,x^3-yz)$ to be a radical ideal, so that $$A(X)=\mathbb{C}[x,y,z]/(z^2-xy,x^3-yz),$$ and then I can check that $\dim A(X) =1$. Naturally, the ideal is contained in its radical, but I don't know how I could prove the reverse containment. So that is my first question:
Is there a general method to check if an ideal is radical?
For the second point, the cotangent space in this case would be $$(x-1,y-1,z-1)/(x-1,y-1,z-1)^2.$$ To find its dimension, I can find the kernel of the Jacobian matrix of the variety, which is
\begin{bmatrix} -y & -x & 2z\\ 3x^2 & -z & -y \end{bmatrix}
evaluated at $(1,1,1)$, which we can check that the matrix has rank 2, and so the dimension of the cotangent space is 1. That means there is only one element for the basis of $$(x-1,y-1,z-1)/(x-1,y-1,z-1)^2.$$ That would be my second question:
How can we find basis elements for a cotagent space?
My final question is related to the system of parameters. I know that it must be an ideal $(f)$ of only one element by the dimension of the variety, and $(x-1,y-1,z-1)$ must be a minimal prime over $(f)$.
How can I find a system of parameters?
My idea is that we could use Nakayama's lemma for local rings, and since we have one element that is a basis of the cotangent space $m/m^2$, this element generates $m=(x-1,y-1,z-1)$, and this would be a system of parameters.