Is it true that $V\not=\mathcal{V}(f)$?

Question

Let $\mathbb{K}$ be an algebraically closed field. Consider the cone $X\subseteq\mathbb{A}_{\mathbb{K}}^3$ given by $X=\mathcal{V}(x^2+y^2-z^2)$. Now, consider $V\subseteq X$ defined by $V=\mathcal{V}(x,y-z)$.

Is it true that $V\not=\mathcal{V}(f)$ for any $f\in\mathbb{K}[X]$? i.e., $V$ is not given by vanishing of a single element in the coordinate ring.

Edit: This is not a homework problem. I know for a UFD, the codimension one subvariety is given by the vanishing of one element of the coordinate ring. So, I'm looking for an example where the codimension one subvariety wouldn't be given by vanishing of one equation. Therefore, if I want to find a counterexample, I'll have to look at a non-UFD ring. The above ring is an example of a non-UFD that I know of. That's why I'm wondering if the above would give me the counterexample that I'm looking for.

I know that the above cone has a singularity at origin and it's not clear to me how to exploit this fact.

Answer 1

See 12.1.3 in Vakil's Foundations of Algebraic Geometry which is about $(x, z)$ in $k[x,y,z] / (xy-z^2)$. This problem can be solved in the same way.

At the vertex of the cone, the dimension of the Zariski tangent space of $X$ has dimension 3 since the vertex is a singular point. In $V(x, y-z) \subseteq X$, the dimension of the Zariski tangent space is 1 since $k[x,y,z]/(x, y-z, x^2 + y^2 - z^2) = k[y, z] / (y-z, y^2 - z^2) = k[y]$. Quotienting by a principal ideal can only drop the dimension down by 0 or 1, not by 2.

There is probably a generalization to whenever a singular prime ideal contains a regular prime ideal (see here, but the linked post doesn't contain an answer).